1289. Minimum Falling Path Sum II - Explanation

Description

Choose exactly one entry from each row of the square grid. Entries chosen in consecutive rows must be in different columns; the column may change by any nonzero distance. Return the smallest possible sum of the chosen entries.

Hatched cells show one minimum-sum path with different columns in adjacent rows.

Example 1:

Minimum Falling Path Sum II, example 1: illustration of the example

Input: grid = [[1,9],[2,8]]

Output: 9

Explanation: There are two valid choices: 1 followed by 8, or 9 followed by 2. The smaller sum is 9. Choosing 1 then 2 is forbidden because both occupy column 0.

Example 2:

Minimum Falling Path Sum II, example 2: illustration of the example

Input: grid = [[0,9,-2],[9,-3,9],[-4,9,9]]

Output: -9

Explanation: Choose -2 in column 2, then -3 in column 1, then -4 in column 0. Their sum is -9. Every adjacent pair uses different columns.

Example 3:

Minimum Falling Path Sum II, example 3: illustration of the example

Input: grid = [[5]]

Output: 5

Explanation: With one row, choose its only entry. There is no next row imposing a column restriction, so the result is 5.

Constraints:

1 <= grid.length <= 200
Every row has grid.length entries.
-99 <= grid[i][j] <= 99

Topics

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Prerequisites

Before attempting this problem, you should be comfortable with:

Dynamic Programming Fundamentals - Understanding how to break problems into overlapping subproblems and use memoization or tabulation
2D Grid/Matrix Traversal - Ability to navigate and process elements in a 2D array
Recursion with Memoization - Converting recursive solutions to top-down DP by caching computed results
Space Optimization in DP - Reducing 2D DP to 1D by only keeping track of the previous row's state

1. Recursion

Intuition

Unlike the standard falling path problem where we can move to adjacent columns, here we must pick a different column in each row. For each cell in a row, we recursively try all columns in the next row except the current one. This constraint increases complexity since we need to consider more transitions. The brute force approach explores all valid paths.

Algorithm

Define helper(r, c) that returns the minimum path sum from cell (r, c) to the last row.
Base case: if r == n-1, return grid[r][c] (we've reached the last row).
For each column next_col in the next row where next_col != c:
- Recursively compute the path sum and track the minimum.
Return grid[r][c] plus the minimum path sum found.
Try starting from each column in row 0 and return the minimum result.

class Solution:
    def minFallingPathSum(self, grid: List[List[int]]) -> int:
        N = len(grid)

        def helper(r, c):
            if r == N - 1:
                return grid[r][c]
            res = float("inf")
            for next_col in range(N):
                if c != next_col:
                    res = min(res, grid[r][c] + helper(r + 1, next_col))
            return res

        res = float("inf")
        for c in range(N):
            res = min(res, helper(0, c))
        return res

public class Solution {
    private int helper(int[][] grid, int r, int c) {
        int N = grid.length;
        if (r == N - 1) {
            return grid[r][c];
        }
        int res = Integer.MAX_VALUE;
        for (int nextCol = 0; nextCol < N; nextCol++) {
            if (c != nextCol) {
                res = Math.min(res, grid[r][c] + helper(grid, r + 1, nextCol));
            }
        }
        return res;
    }

    public int minFallingPathSum(int[][] grid) {
        int N = grid.length;
        int res = Integer.MAX_VALUE;
        for (int c = 0; c < N; c++) {
            res = Math.min(res, helper(grid, 0, c));
        }
        return res;
    }
}

class Solution {
    int helper(vector<vector<int>>& grid, int r, int c) {
        int N = grid.size();
        if (r == N - 1) {
            return grid[r][c];
        }
        int res = INT_MAX;
        for (int nextCol = 0; nextCol < N; nextCol++) {
            if (c != nextCol) {
                res = min(res, grid[r][c] + helper(grid, r + 1, nextCol));
            }
        }
        return res;
    }

public:
    int minFallingPathSum(vector<vector<int>>& grid) {
        int N = grid.size();
        int res = INT_MAX;
        for (int c = 0; c < N; c++) {
            res = min(res, helper(grid, 0, c));
        }
        return res;
    }
};

class Solution {
    /**
     * @param {number[][]} grid
     * @return {number}
     */
    minFallingPathSum(grid) {
        const N = grid.length;

        const helper = (r, c) => {
            if (r === N - 1) return grid[r][c];
            let res = Infinity;
            for (let nextCol = 0; nextCol < N; nextCol++) {
                if (c !== nextCol) {
                    res = Math.min(res, grid[r][c] + helper(r + 1, nextCol));
                }
            }
            return res;
        };

        let res = Infinity;
        for (let c = 0; c < N; c++) {
            res = Math.min(res, helper(0, c));
        }
        return res;
    }
}

public class Solution {
    public int MinFallingPathSum(int[][] grid) {
        int N = grid.Length;
        int res = int.MaxValue;
        for (int c = 0; c < N; c++) {
            res = Math.Min(res, Helper(grid, 0, c, N));
        }
        return res;
    }

    private int Helper(int[][] grid, int r, int c, int N) {
        if (r == N - 1) {
            return grid[r][c];
        }
        int res = int.MaxValue;
        for (int nextCol = 0; nextCol < N; nextCol++) {
            if (c != nextCol) {
                res = Math.Min(res, grid[r][c] + Helper(grid, r + 1, nextCol, N));
            }
        }
        return res;
    }
}

func minFallingPathSum(grid [][]int) int {
    n := len(grid)

    var helper func(r, c int) int
    helper = func(r, c int) int {
        if r == n-1 {
            return grid[r][c]
        }
        res := math.MaxInt32
        for nextCol := 0; nextCol < n; nextCol++ {
            if c != nextCol {
                res = min(res, grid[r][c]+helper(r+1, nextCol))
            }
        }
        return res
    }

    res := math.MaxInt32
    for c := 0; c < n; c++ {
        res = min(res, helper(0, c))
    }
    return res
}

class Solution {
    fun minFallingPathSum(grid: Array<IntArray>): Int {
        val n = grid.size

        fun helper(r: Int, c: Int): Int {
            if (r == n - 1) return grid[r][c]
            var res = Int.MAX_VALUE
            for (nextCol in 0 until n) {
                if (c != nextCol) {
                    res = minOf(res, grid[r][c] + helper(r + 1, nextCol))
                }
            }
            return res
        }

        var res = Int.MAX_VALUE
        for (c in 0 until n) {
            res = minOf(res, helper(0, c))
        }
        return res
    }
}

class Solution {
    func minFallingPathSum(_ grid: [[Int]]) -> Int {
        let n = grid.count

        func helper(_ r: Int, _ c: Int) -> Int {
            if r == n - 1 {
                return grid[r][c]
            }
            var res = Int.max
            for nextCol in 0..<n {
                if c != nextCol {
                    res = min(res, grid[r][c] + helper(r + 1, nextCol))
                }
            }
            return res
        }

        var res = Int.max
        for c in 0..<n {
            res = min(res, helper(0, c))
        }
        return res
    }
}

impl Solution {
    pub fn min_falling_path_sum(grid: Vec<Vec<i32>>) -> i32 {
        let n = grid.len();

        fn helper(grid: &Vec<Vec<i32>>, r: usize, c: usize, n: usize) -> i32 {
            if r == n - 1 {
                return grid[r][c];
            }
            let mut res = i32::MAX;
            for next_col in 0..n {
                if c != next_col {
                    res = res.min(grid[r][c] + helper(grid, r + 1, next_col, n));
                }
            }
            res
        }

        let mut res = i32::MAX;
        for c in 0..n {
            res = res.min(helper(&grid, 0, c, n));
        }
        res
    }
}

Time & Space Complexity

Time complexity: $O(n ^ n)$
Space complexity: $O(n)$ for recursion stack.

2. Dynamic Programming (Top-Down)

Intuition

The recursive solution recomputes the same (r, c) states multiple times. By storing computed results in a memoization table, we ensure each state is solved only once. This reduces time complexity from exponential to O(n^3), since for each of the n^2 cells, we may consider up to n transitions.

Algorithm

Create a memo table (dictionary or 2D array) to cache results.
Define helper(r, c) that returns the minimum path sum from (r, c) to the last row.
If (r, c) is already computed, return the cached value.
Base case: if r == n-1, return grid[r][c].
For each next_col != c, recursively compute and track the minimum.
Cache and return grid[r][c] plus the minimum found.
Return the minimum among all starting columns.

class Solution:
    def minFallingPathSum(self, grid: List[List[int]]) -> int:
        N = len(grid)
        cache = {}

        def helper(r, c):
            if r == N - 1:
                return grid[r][c]
            if (r, c) in cache:
                return cache[(r, c)]

            res = float("inf")
            for next_col in range(N):
                if c != next_col:
                    res = min(res, grid[r][c] + helper(r + 1, next_col))
            cache[(r, c)] = res
            return res

        res = float("inf")
        for c in range(N):
            res = min(res, helper(0, c))
        return res

public class Solution {
    private int[][] memo;

    private int helper(int[][] grid, int r, int c) {
        int N = grid.length;
        if (r == N - 1) {
            return grid[r][c];
        }
        if (memo[r][c] != Integer.MIN_VALUE) {
            return memo[r][c];
        }

        int res = Integer.MAX_VALUE;
        for (int nextCol = 0; nextCol < N; nextCol++) {
            if (c != nextCol) {
                res = Math.min(res, grid[r][c] + helper(grid, r + 1, nextCol));
            }
        }
        memo[r][c] = res;
        return res;
    }

    public int minFallingPathSum(int[][] grid) {
        int N = grid.length;
        memo = new int[N][N];
        for (int[] row : memo) {
            Arrays.fill(row, Integer.MIN_VALUE);
        }
        int res = Integer.MAX_VALUE;
        for (int c = 0; c < N; c++) {
            res = Math.min(res, helper(grid, 0, c));
        }
        return res;
    }
}

class Solution {
    vector<vector<int>> memo;

    int helper(vector<vector<int>>& grid, int r, int c) {
        int N = grid.size();
        if (r == N - 1) {
            return grid[r][c];
        }
        if (memo[r][c] != INT_MIN) {
            return memo[r][c];
        }
        int res = INT_MAX;
        for (int nextCol = 0; nextCol < N; nextCol++) {
            if (c != nextCol) {
                res = min(res, grid[r][c] + helper(grid, r + 1, nextCol));
            }
        }
        memo[r][c] = res;
        return res;
    }

public:
    int minFallingPathSum(vector<vector<int>>& grid) {
        int N = grid.size();
        memo.assign(N, vector<int>(N, INT_MIN));
        int res = INT_MAX;
        for (int c = 0; c < N; c++) {
            res = min(res, helper(grid, 0, c));
        }
        return res;
    }
};

class Solution {
    /**
     * @param {number[][]} grid
     * @return {number}
     */
    minFallingPathSum(grid) {
        const N = grid.length;
        const memo = Array.from({ length: N }, () => Array(N).fill(-Infinity));

        const helper = (r, c) => {
            if (r === N - 1) return grid[r][c];
            if (memo[r][c] !== -Infinity) return memo[r][c];
            let res = Infinity;
            for (let nextCol = 0; nextCol < N; nextCol++) {
                if (c !== nextCol) {
                    res = Math.min(res, grid[r][c] + helper(r + 1, nextCol));
                }
            }
            memo[r][c] = res;
            return res;
        };

        let res = Infinity;
        for (let c = 0; c < N; c++) {
            res = Math.min(res, helper(0, c));
        }
        return res;
    }
}

public class Solution {
    private int[,] memo;

    public int MinFallingPathSum(int[][] grid) {
        int N = grid.Length;
        memo = new int[N, N];
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                memo[i, j] = int.MinValue;
            }
        }
        int res = int.MaxValue;
        for (int c = 0; c < N; c++) {
            res = Math.Min(res, Helper(grid, 0, c, N));
        }
        return res;
    }

    private int Helper(int[][] grid, int r, int c, int N) {
        if (r == N - 1) return grid[r][c];
        if (memo[r, c] != int.MinValue) return memo[r, c];

        int res = int.MaxValue;
        for (int nextCol = 0; nextCol < N; nextCol++) {
            if (c != nextCol) {
                res = Math.Min(res, grid[r][c] + Helper(grid, r + 1, nextCol, N));
            }
        }
        memo[r, c] = res;
        return res;
    }
}

func minFallingPathSum(grid [][]int) int {
    n := len(grid)
    memo := make([][]int, n)
    for i := range memo {
        memo[i] = make([]int, n)
        for j := range memo[i] {
            memo[i][j] = math.MinInt32
        }
    }

    var helper func(r, c int) int
    helper = func(r, c int) int {
        if r == n-1 {
            return grid[r][c]
        }
        if memo[r][c] != math.MinInt32 {
            return memo[r][c]
        }
        res := math.MaxInt32
        for nextCol := 0; nextCol < n; nextCol++ {
            if c != nextCol {
                res = min(res, grid[r][c]+helper(r+1, nextCol))
            }
        }
        memo[r][c] = res
        return res
    }

    res := math.MaxInt32
    for c := 0; c < n; c++ {
        res = min(res, helper(0, c))
    }
    return res
}

class Solution {
    fun minFallingPathSum(grid: Array<IntArray>): Int {
        val n = grid.size
        val memo = Array(n) { IntArray(n) { Int.MIN_VALUE } }

        fun helper(r: Int, c: Int): Int {
            if (r == n - 1) return grid[r][c]
            if (memo[r][c] != Int.MIN_VALUE) return memo[r][c]

            var res = Int.MAX_VALUE
            for (nextCol in 0 until n) {
                if (c != nextCol) {
                    res = minOf(res, grid[r][c] + helper(r + 1, nextCol))
                }
            }
            memo[r][c] = res
            return res
        }

        var res = Int.MAX_VALUE
        for (c in 0 until n) {
            res = minOf(res, helper(0, c))
        }
        return res
    }
}

class Solution {
    func minFallingPathSum(_ grid: [[Int]]) -> Int {
        let n = grid.count
        var memo = Array(repeating: Array(repeating: Int.min, count: n), count: n)

        func helper(_ r: Int, _ c: Int) -> Int {
            if r == n - 1 {
                return grid[r][c]
            }
            if memo[r][c] != Int.min {
                return memo[r][c]
            }
            var res = Int.max
            for nextCol in 0..<n {
                if c != nextCol {
                    res = min(res, grid[r][c] + helper(r + 1, nextCol))
                }
            }
            memo[r][c] = res
            return res
        }

        var res = Int.max
        for c in 0..<n {
            res = min(res, helper(0, c))
        }
        return res
    }
}

impl Solution {
    pub fn min_falling_path_sum(grid: Vec<Vec<i32>>) -> i32 {
        let n = grid.len();
        let mut memo = vec![vec![i32::MIN; n]; n];

        fn helper(
            grid: &Vec<Vec<i32>>,
            memo: &mut Vec<Vec<i32>>,
            r: usize,
            c: usize,
            n: usize,
        ) -> i32 {
            if r == n - 1 {
                return grid[r][c];
            }
            if memo[r][c] != i32::MIN {
                return memo[r][c];
            }
            let mut res = i32::MAX;
            for next_col in 0..n {
                if c != next_col {
                    res = res.min(grid[r][c] + helper(grid, memo, r + 1, next_col, n));
                }
            }
            memo[r][c] = res;
            res
        }

        let mut res = i32::MAX;
        for c in 0..n {
            res = res.min(helper(&grid, &mut memo, 0, c, n));
        }
        res
    }
}

Time & Space Complexity

Time complexity: $O(n ^ 3)$
Space complexity: $O(n ^ 2)$

3. Dynamic Programming (Bottom-Up)

Intuition

We can build the solution iteratively from the last row upward. For each cell, we compute the minimum path sum by considering all columns in the next row except the current one. This bottom-up approach fills a 2D DP table where dp[r][c] represents the minimum path sum from cell (r, c) to any cell in the last row.

Algorithm

Create a 2D DP table initialized to infinity.
Fill the last row: dp[n-1][c] = grid[n-1][c] for all columns.
For each row r from n-2 down to 0:
- For each column c:
  - For each next_col != c:
    - Update dp[r][c] = min(dp[r][c], grid[r][c] + dp[r+1][next_col]).
Return the minimum value in dp[0].

class Solution:
    def minFallingPathSum(self, grid: List[List[int]]) -> int:
        N = len(grid)
        dp = [[float("inf")] * N for _ in range(N)]

        for c in range(N):
            dp[N - 1][c] = grid[N - 1][c]

        for r in range(N - 2, -1, -1):
            for c in range(N):
                for next_col in range(N):
                    if c != next_col:
                        dp[r][c] = min(dp[r][c], grid[r][c] + dp[r + 1][next_col])

        return min(dp[0])

class Solution {
public:
    int minFallingPathSum(vector<vector<int>>& grid) {
        int N = grid.size();
        vector<vector<int>> dp(N, vector<int>(N, INT_MAX));

        for (int c = 0; c < N; c++) {
            dp[N - 1][c] = grid[N - 1][c];
        }

        for (int r = N - 2; r >= 0; r--) {
            for (int c = 0; c < N; c++) {
                dp[r][c] = INT_MAX;
                for (int nextCol = 0; nextCol < N; nextCol++) {
                    if (c != nextCol) {
                        dp[r][c] = min(dp[r][c], grid[r][c] + dp[r + 1][nextCol]);
                    }
                }
            }
        }

        int res = INT_MAX;
        for (int c = 0; c < N; c++) {
            res = min(res, dp[0][c]);
        }
        return res;
    }
};

class Solution {
    /**
     * @param {number[][]} grid
     * @return {number}
     */
    minFallingPathSum(grid) {
        const N = grid.length;
        const dp = Array.from({ length: N }, () => Array(N).fill(Infinity));

        for (let c = 0; c < N; c++) {
            dp[N - 1][c] = grid[N - 1][c];
        }

        for (let r = N - 2; r >= 0; r--) {
            for (let c = 0; c < N; c++) {
                for (let nextCol = 0; nextCol < N; nextCol++) {
                    if (c !== nextCol) {
                        dp[r][c] = Math.min(
                            dp[r][c],
                            grid[r][c] + dp[r + 1][nextCol],
                        );
                    }
                }
            }
        }

        return Math.min(...dp[0]);
    }
}

public class Solution {
    public int MinFallingPathSum(int[][] grid) {
        int N = grid.Length;
        int[,] dp = new int[N, N];

        for (int c = 0; c < N; c++) {
            dp[N - 1, c] = grid[N - 1][c];
        }

        for (int r = N - 2; r >= 0; r--) {
            for (int c = 0; c < N; c++) {
                dp[r, c] = int.MaxValue;
                for (int nextCol = 0; nextCol < N; nextCol++) {
                    if (c != nextCol) {
                        dp[r, c] = Math.Min(dp[r, c], grid[r][c] + dp[r + 1, nextCol]);
                    }
                }
            }
        }

        int res = int.MaxValue;
        for (int c = 0; c < N; c++) {
            res = Math.Min(res, dp[0, c]);
        }
        return res;
    }
}

func minFallingPathSum(grid [][]int) int {
    n := len(grid)
    dp := make([][]int, n)
    for i := range dp {
        dp[i] = make([]int, n)
        for j := range dp[i] {
            dp[i][j] = math.MaxInt32
        }
    }

    for c := 0; c < n; c++ {
        dp[n-1][c] = grid[n-1][c]
    }

    for r := n - 2; r >= 0; r-- {
        for c := 0; c < n; c++ {
            for nextCol := 0; nextCol < n; nextCol++ {
                if c != nextCol {
                    dp[r][c] = min(dp[r][c], grid[r][c]+dp[r+1][nextCol])
                }
            }
        }
    }

    res := math.MaxInt32
    for c := 0; c < n; c++ {
        res = min(res, dp[0][c])
    }
    return res
}

class Solution {
    fun minFallingPathSum(grid: Array<IntArray>): Int {
        val n = grid.size
        val dp = Array(n) { IntArray(n) { Int.MAX_VALUE } }

        for (c in 0 until n) {
            dp[n - 1][c] = grid[n - 1][c]
        }

        for (r in n - 2 downTo 0) {
            for (c in 0 until n) {
                for (nextCol in 0 until n) {
                    if (c != nextCol) {
                        dp[r][c] = minOf(dp[r][c], grid[r][c] + dp[r + 1][nextCol])
                    }
                }
            }
        }

        return dp[0].minOrNull()!!
    }
}

class Solution {
    func minFallingPathSum(_ grid: [[Int]]) -> Int {
        let n = grid.count
        var dp = Array(repeating: Array(repeating: Int.max, count: n), count: n)

        for c in 0..<n {
            dp[n - 1][c] = grid[n - 1][c]
        }

        for r in stride(from: n - 2, through: 0, by: -1) {
            for c in 0..<n {
                for nextCol in 0..<n {
                    if c != nextCol {
                        dp[r][c] = min(dp[r][c], grid[r][c] + dp[r + 1][nextCol])
                    }
                }
            }
        }

        return dp[0].min()!
    }
}

impl Solution {
    pub fn min_falling_path_sum(grid: Vec<Vec<i32>>) -> i32 {
        let n = grid.len();
        let mut dp = vec![vec![i32::MAX; n]; n];

        for c in 0..n {
            dp[n - 1][c] = grid[n - 1][c];
        }

        for r in (0..n - 1).rev() {
            for c in 0..n {
                for next_col in 0..n {
                    if c != next_col {
                        dp[r][c] = dp[r][c].min(grid[r][c] + dp[r + 1][next_col]);
                    }
                }
            }
        }

        *dp[0].iter().min().unwrap()
    }
}

Time & Space Complexity

Time complexity: $O(n ^ 3)$
Space complexity: $O(n ^ 2)$

4. Dynamic Programming (Space Optimized)

Intuition

Since each row only depends on the next row's values, we can reduce space from O(n^2) to O(n) by using a single 1D array. We process rows from top to bottom, updating the array for each row while referencing the values from the previous iteration.

Algorithm

Initialize DP array with the first row of the grid.
For each subsequent row r:
- Create a new DP array for the current row.
- For each column curr_c:
  - For each prev_c != curr_c, track the minimum from the previous row.
  - Set next_dp[curr_c] = grid[r][curr_c] + min_prev.
- Replace the DP array with the new array.
Return the minimum value in the final DP array.

class Solution:
    def minFallingPathSum(self, grid: List[List[int]]) -> int:
        N = len(grid)
        dp = grid[0]

        for r in range(1, N):
            next_dp = [float("inf")] * N
            for curr_c in range(N):
                for prev_c in range(N):
                    if prev_c != curr_c:
                        next_dp[curr_c] = min(
                            next_dp[curr_c],
                            grid[r][curr_c] + dp[prev_c]
                        )
            dp = next_dp

        return min(dp)

public class Solution {
    public int minFallingPathSum(int[][] grid) {
        int N = grid.length;
        int[] dp = grid[0];

        for (int r = 1; r < N; r++) {
            int[] nextDp = new int[N];
            Arrays.fill(nextDp, Integer.MAX_VALUE);

            for (int currC = 0; currC < N; currC++) {
                for (int prevC = 0; prevC < N; prevC++) {
                    if (prevC != currC) {
                        nextDp[currC] = Math.min(
                            nextDp[currC],
                            grid[r][currC] + dp[prevC]
                        );
                    }
                }
            }
            dp = nextDp;
        }

        int res = Integer.MAX_VALUE;
        for (int i : dp) res = Math.min(res, i);
        return res;
    }
}

class Solution {
public:
    int minFallingPathSum(vector<vector<int>>& grid) {
        int N = grid.size();
        vector<int> dp = grid[0];

        for (int r = 1; r < N; r++) {
            vector<int> nextDp(N, INT_MAX);
            for (int currC = 0; currC < N; currC++) {
                for (int prevC = 0; prevC < N; prevC++) {
                    if (prevC != currC) {
                        nextDp[currC] = min(
                            nextDp[currC],
                            grid[r][currC] + dp[prevC]
                        );
                    }
                }
            }
            dp = nextDp;
        }

        return *min_element(dp.begin(), dp.end());
    }
};

class Solution {
    /**
     * @param {number[][]} grid
     * @return {number}
     */
    minFallingPathSum(grid) {
        const N = grid.length;
        let dp = grid[0];

        for (let r = 1; r < N; r++) {
            const nextDp = Array(N).fill(Infinity);
            for (let currC = 0; currC < N; currC++) {
                for (let prevC = 0; prevC < N; prevC++) {
                    if (prevC !== currC) {
                        nextDp[currC] = Math.min(
                            nextDp[currC],
                            grid[r][currC] + dp[prevC],
                        );
                    }
                }
            }
            dp = nextDp;
        }

        return Math.min(...dp);
    }
}

public class Solution {
    public int MinFallingPathSum(int[][] grid) {
        int N = grid.Length;
        int[] dp = (int[])grid[0].Clone();

        for (int r = 1; r < N; r++) {
            int[] nextDp = new int[N];
            Array.Fill(nextDp, int.MaxValue);

            for (int currC = 0; currC < N; currC++) {
                for (int prevC = 0; prevC < N; prevC++) {
                    if (prevC != currC) {
                        nextDp[currC] = Math.Min(nextDp[currC], grid[r][currC] + dp[prevC]);
                    }
                }
            }
            dp = nextDp;
        }

        int res = int.MaxValue;
        foreach (int val in dp) {
            res = Math.Min(res, val);
        }
        return res;
    }
}

func minFallingPathSum(grid [][]int) int {
    n := len(grid)
    dp := make([]int, n)
    copy(dp, grid[0])

    for r := 1; r < n; r++ {
        nextDp := make([]int, n)
        for i := range nextDp {
            nextDp[i] = math.MaxInt32
        }
        for currC := 0; currC < n; currC++ {
            for prevC := 0; prevC < n; prevC++ {
                if prevC != currC {
                    nextDp[currC] = min(nextDp[currC], grid[r][currC]+dp[prevC])
                }
            }
        }
        dp = nextDp
    }

    res := math.MaxInt32
    for _, val := range dp {
        res = min(res, val)
    }
    return res
}

class Solution {
    fun minFallingPathSum(grid: Array<IntArray>): Int {
        val n = grid.size
        var dp = grid[0].clone()

        for (r in 1 until n) {
            val nextDp = IntArray(n) { Int.MAX_VALUE }
            for (currC in 0 until n) {
                for (prevC in 0 until n) {
                    if (prevC != currC) {
                        nextDp[currC] = minOf(nextDp[currC], grid[r][currC] + dp[prevC])
                    }
                }
            }
            dp = nextDp
        }

        return dp.minOrNull()!!
    }
}

class Solution {
    func minFallingPathSum(_ grid: [[Int]]) -> Int {
        let n = grid.count
        var dp = grid[0]

        for r in 1..<n {
            var nextDp = Array(repeating: Int.max, count: n)
            for currC in 0..<n {
                for prevC in 0..<n {
                    if prevC != currC {
                        nextDp[currC] = min(nextDp[currC], grid[r][currC] + dp[prevC])
                    }
                }
            }
            dp = nextDp
        }

        return dp.min()!
    }
}

impl Solution {
    pub fn min_falling_path_sum(grid: Vec<Vec<i32>>) -> i32 {
        let n = grid.len();
        let mut dp = grid[0].clone();

        for r in 1..n {
            let mut next_dp = vec![i32::MAX; n];
            for curr_c in 0..n {
                for prev_c in 0..n {
                    if prev_c != curr_c {
                        next_dp[curr_c] =
                            next_dp[curr_c].min(grid[r][curr_c] + dp[prev_c]);
                    }
                }
            }
            dp = next_dp;
        }

        *dp.iter().min().unwrap()
    }
}

Time & Space Complexity

Time complexity: $O(n ^ 3)$
Space complexity: $O(n)$

5. Dynamic Programming (Time Optimized)

Intuition

The O(n^3) complexity comes from checking all n columns for each of n^2 cells. We can optimize by observing that when transitioning to a cell, we usually want the minimum from the previous row, unless that minimum is in the same column. So we only need to track the two smallest values from each row: if the current column matches the smallest, use the second smallest; otherwise, use the smallest.

Algorithm

Initialize DP by computing the two smallest (value, index) pairs from the first row.
For each subsequent row:
- For each column, compute the path sum using the smallest previous value if columns differ, else the second smallest.
- Collect all results and find the new two smallest pairs.
Return the minimum value from the final DP.

class Solution:
    def minFallingPathSum(self, grid: List[List[int]]) -> int:
        def get_min_two(row):
            two_smallest = []
            for val, idx in row:
                if len(two_smallest) < 2:
                    two_smallest.append((val, idx))
                elif two_smallest[1][0] > val:
                    two_smallest.pop()
                    two_smallest.append((val, idx))
                two_smallest.sort()
            return two_smallest

        N = len(grid)
        first_row = [(val, idx) for idx, val in enumerate(grid[0])]
        dp = get_min_two(first_row)

        for r in range(1, N):
            next_dp = []
            for curr_c in range(N):
                curr_val = grid[r][curr_c]
                min_val = float("inf")
                for prev_val, prev_c in dp:
                    if curr_c != prev_c:
                        min_val = min(min_val, curr_val + prev_val)
                next_dp.append((min_val, curr_c))
            dp = get_min_two(next_dp)

        return min(val for val, idx in dp)

public class Solution {
    public List<int[]> getMinTwo(List<int[]> row) {
        List<int[]> twoSmallest = new ArrayList<>();
        for (int[] entry : row) {
            if (twoSmallest.size() < 2) {
                twoSmallest.add(entry);
            } else if (twoSmallest.get(1)[0] > entry[0]) {
                twoSmallest.remove(1);
                twoSmallest.add(entry);
            }
            twoSmallest.sort((a, b) -> a[0] - b[0]);
        }
        return twoSmallest;
    }

    public int minFallingPathSum(int[][] grid) {
        int N = grid.length;

        List<int[]> firstRow = new ArrayList<>();
        for (int i = 0; i < grid[0].length; i++) {
            firstRow.add(new int[]{grid[0][i], i});
        }

        List<int[]> dp = getMinTwo(firstRow);

        for (int r = 1; r < N; r++) {
            List<int[]> nextDp = new ArrayList<>();
            for (int c = 0; c < grid[0].length; c++) {
                int currVal = grid[r][c];
                int minVal = Integer.MAX_VALUE;
                for (int[] prev : dp) {
                    if (prev[1] != c) {
                        minVal = Math.min(minVal, currVal + prev[0]);
                    }
                }
                nextDp.add(new int[]{minVal, c});
            }
            dp = getMinTwo(nextDp);
        }

        return dp.stream().mapToInt(a -> a[0]).min().getAsInt();
    }
}

class Solution {
public:
    int minFallingPathSum(vector<vector<int>>& grid) {
        int N = grid.size();

        auto getMinTwo = [](vector<pair<int, int>>& row) {
            vector<pair<int, int>> twoSmallest;
            for (auto& entry : row) {
                if (twoSmallest.size() < 2) {
                    twoSmallest.push_back(entry);
                } else if (twoSmallest[1].first > entry.first) {
                    twoSmallest.pop_back();
                    twoSmallest.push_back(entry);
                }
                sort(twoSmallest.begin(), twoSmallest.end());
            }
            return twoSmallest;
        };

        vector<pair<int, int>> firstRow;
        for (int i = 0; i < grid[0].size(); i++) {
            firstRow.push_back({grid[0][i], i});
        }

        vector<pair<int, int>> dp = getMinTwo(firstRow);

        for (int r = 1; r < N; r++) {
            vector<pair<int, int>> nextDp;
            for (int c = 0; c < grid[0].size(); c++) {
                int currVal = grid[r][c];
                int minVal = INT_MAX;
                for (auto& prev : dp) {
                    if (prev.second != c) {
                        minVal = min(minVal, currVal + prev.first);
                    }
                }
                nextDp.push_back({minVal, c});
            }
            dp = getMinTwo(nextDp);
        }

        int result = INT_MAX;
        for (auto& entry : dp) {
            result = min(result, entry.first);
        }
        return result;
    }
};

class Solution {
    /**
     * @param {number[][]} grid
     * @return {number}
     */
    minFallingPathSum(grid) {
        const N = grid.length;

        const getMinTwo = (row) => {
            const twoSmallest = [];
            for (const [val, idx] of row) {
                if (twoSmallest.length < 2) {
                    twoSmallest.push([val, idx]);
                } else if (twoSmallest[1][0] > val) {
                    twoSmallest.pop();
                    twoSmallest.push([val, idx]);
                }
                twoSmallest.sort((a, b) => a[0] - b[0]);
            }
            return twoSmallest;
        };

        const firstRow = grid[0].map((val, idx) => [val, idx]);
        let dp = getMinTwo(firstRow);

        for (let r = 1; r < N; r++) {
            const nextDp = [];
            for (let c = 0; c < grid[0].length; c++) {
                const currVal = grid[r][c];
                let minVal = Infinity;
                for (const [prevVal, prevC] of dp) {
                    if (c !== prevC) {
                        minVal = Math.min(minVal, currVal + prevVal);
                    }
                }
                nextDp.push([minVal, c]);
            }
            dp = getMinTwo(nextDp);
        }

        return Math.min(...dp.map(([val]) => val));
    }
}

public class Solution {
    public int MinFallingPathSum(int[][] grid) {
        int N = grid.Length;

        List<int[]> GetMinTwo(List<int[]> row) {
            var twoSmallest = new List<int[]>();
            foreach (var entry in row) {
                if (twoSmallest.Count < 2) {
                    twoSmallest.Add(entry);
                } else if (twoSmallest[1][0] > entry[0]) {
                    twoSmallest.RemoveAt(1);
                    twoSmallest.Add(entry);
                }
                twoSmallest.Sort((a, b) => a[0].CompareTo(b[0]));
            }
            return twoSmallest;
        }

        var firstRow = new List<int[]>();
        for (int i = 0; i < grid[0].Length; i++) {
            firstRow.Add(new int[] { grid[0][i], i });
        }

        var dp = GetMinTwo(firstRow);

        for (int r = 1; r < N; r++) {
            var nextDp = new List<int[]>();
            for (int c = 0; c < grid[0].Length; c++) {
                int currVal = grid[r][c];
                int minVal = int.MaxValue;
                foreach (var prev in dp) {
                    if (prev[1] != c) {
                        minVal = Math.Min(minVal, currVal + prev[0]);
                    }
                }
                nextDp.Add(new int[] { minVal, c });
            }
            dp = GetMinTwo(nextDp);
        }

        return dp.Min(a => a[0]);
    }
}

func minFallingPathSum(grid [][]int) int {
    n := len(grid)

    getMinTwo := func(row [][2]int) [][2]int {
        twoSmallest := make([][2]int, 0, 2)
        for _, entry := range row {
            if len(twoSmallest) < 2 {
                twoSmallest = append(twoSmallest, entry)
            } else if twoSmallest[1][0] > entry[0] {
                twoSmallest[1] = entry
            }
            sort.Slice(twoSmallest, func(i, j int) bool {
                return twoSmallest[i][0] < twoSmallest[j][0]
            })
        }
        return twoSmallest
    }

    firstRow := make([][2]int, n)
    for i, val := range grid[0] {
        firstRow[i] = [2]int{val, i}
    }
    dp := getMinTwo(firstRow)

    for r := 1; r < n; r++ {
        nextDp := make([][2]int, n)
        for c := 0; c < n; c++ {
            currVal := grid[r][c]
            minVal := math.MaxInt32
            for _, prev := range dp {
                if c != prev[1] {
                    minVal = min(minVal, currVal+prev[0])
                }
            }
            nextDp[c] = [2]int{minVal, c}
        }
        dp = getMinTwo(nextDp)
    }

    res := math.MaxInt32
    for _, entry := range dp {
        res = min(res, entry[0])
    }
    return res
}

class Solution {
    fun minFallingPathSum(grid: Array<IntArray>): Int {
        val n = grid.size

        fun getMinTwo(row: List<Pair<Int, Int>>): List<Pair<Int, Int>> {
            val twoSmallest = mutableListOf<Pair<Int, Int>>()
            for (entry in row) {
                if (twoSmallest.size < 2) {
                    twoSmallest.add(entry)
                } else if (twoSmallest[1].first > entry.first) {
                    twoSmallest.removeAt(1)
                    twoSmallest.add(entry)
                }
                twoSmallest.sortBy { it.first }
            }
            return twoSmallest
        }

        val firstRow = grid[0].mapIndexed { idx, value -> Pair(value, idx) }
        var dp = getMinTwo(firstRow)

        for (r in 1 until n) {
            val nextDp = mutableListOf<Pair<Int, Int>>()
            for (c in 0 until n) {
                val currVal = grid[r][c]
                var minVal = Int.MAX_VALUE
                for ((prevVal, prevC) in dp) {
                    if (c != prevC) {
                        minVal = minOf(minVal, currVal + prevVal)
                    }
                }
                nextDp.add(Pair(minVal, c))
            }
            dp = getMinTwo(nextDp)
        }

        return dp.minOfOrNull { it.first }!!
    }
}

class Solution {
    func minFallingPathSum(_ grid: [[Int]]) -> Int {
        let n = grid.count

        func getMinTwo(_ row: [(Int, Int)]) -> [(Int, Int)] {
            var twoSmallest: [(Int, Int)] = []
            for entry in row {
                if twoSmallest.count < 2 {
                    twoSmallest.append(entry)
                } else if twoSmallest[1].0 > entry.0 {
                    twoSmallest.removeLast()
                    twoSmallest.append(entry)
                }
                twoSmallest.sort { $0.0 < $1.0 }
            }
            return twoSmallest
        }

        let firstRow = grid[0].enumerated().map { ($0.element, $0.offset) }
        var dp = getMinTwo(firstRow)

        for r in 1..<n {
            var nextDp: [(Int, Int)] = []
            for c in 0..<n {
                let currVal = grid[r][c]
                var minVal = Int.max
                for (prevVal, prevC) in dp {
                    if c != prevC {
                        minVal = min(minVal, currVal + prevVal)
                    }
                }
                nextDp.append((minVal, c))
            }
            dp = getMinTwo(nextDp)
        }

        return dp.map { $0.0 }.min()!
    }
}

impl Solution {
    pub fn min_falling_path_sum(grid: Vec<Vec<i32>>) -> i32 {
        let n = grid.len();

        fn get_min_two(row: &[(i32, usize)]) -> Vec<(i32, usize)> {
            let mut two: Vec<(i32, usize)> = Vec::with_capacity(2);
            for &entry in row {
                if two.len() < 2 {
                    two.push(entry);
                } else if two[1].0 > entry.0 {
                    two[1] = entry;
                }
                two.sort_by_key(|x| x.0);
            }
            two
        }

        let first_row: Vec<(i32, usize)> =
            grid[0].iter().enumerate().map(|(i, &v)| (v, i)).collect();
        let mut dp = get_min_two(&first_row);

        for r in 1..n {
            let mut next_dp: Vec<(i32, usize)> = Vec::with_capacity(n);
            for c in 0..n {
                let curr_val = grid[r][c];
                let mut min_val = i32::MAX;
                for &(prev_val, prev_c) in &dp {
                    if c != prev_c {
                        min_val = min_val.min(curr_val + prev_val);
                    }
                }
                next_dp.push((min_val, c));
            }
            dp = get_min_two(&next_dp);
        }

        dp.iter().map(|x| x.0).min().unwrap()
    }
}

Time & Space Complexity

Time complexity: $O(n ^ 2)$
Space complexity: $O(n)$

6. Dynamic Programming (Optimal)

Intuition

We can further optimize by not storing an entire array of pairs. Instead, we only track four values: the index and value of the smallest element, and the index and value of the second smallest element from the previous row. For each cell in the current row, we add the appropriate previous minimum based on column matching. Then we update our tracked values for the next iteration. This achieves O(n^2) time with O(1) extra space.

Algorithm

Initialize dpIdx1, dpIdx2 (indices of first and second smallest) and dpVal1, dpVal2 (their values) to represent the previous row's state.
For each row:
- Track new smallest and second smallest values and indices.
- For each column j:
  - If j != dpIdx1, add dpVal1; otherwise add dpVal2.
  - Update the running smallest and second smallest for this row.
After processing all rows, return dpVal1 (the overall minimum).

class Solution:
    def minFallingPathSum(self, grid: List[List[int]]) -> int:
        n = len(grid)
        if n == 1:
            return grid[0][0]

        dp_idx1 = dp_idx2 = -1
        dp_val1 = dp_val2 = 0

        for i in range(n):
            nextDp_idx1 = nextDp_idx2 = -1
            nextDp_val1 = nextDp_val2 = float("inf")

            for j in range(n):
                cur = dp_val1 if j != dp_idx1 else dp_val2
                cur += grid[i][j]

                if nextDp_idx1 == -1 or cur < nextDp_val1:
                    nextDp_idx2, nextDp_val2 = nextDp_idx1, nextDp_val1
                    nextDp_idx1, nextDp_val1 = j, cur
                elif nextDp_idx2 == -1 or cur < nextDp_val2:
                    nextDp_idx2, nextDp_val2 = j, cur

            dp_idx1, dp_idx2, dp_val1, dp_val2 = nextDp_idx1, nextDp_idx2, nextDp_val1, nextDp_val2

        return dp_val1

public class Solution {
    public int minFallingPathSum(int[][] grid) {
        int n = grid.length;
        if (n == 1) {
            return grid[0][0];
        }

        int dpIdx1 = -1, dpIdx2 = -1;
        int dpVal1 = 0, dpVal2 = 0;

        for (int i = 0; i < n; i++) {
            int nextDpIdx1 = -1, nextDpIdx2 = -1;
            int nextDpVal1 = Integer.MAX_VALUE, nextDpVal2 = Integer.MAX_VALUE;

            for (int j = 0; j < n; j++) {
                int cur = (j != dpIdx1) ? dpVal1 : dpVal2;
                cur += grid[i][j];

                if (nextDpIdx1 == -1 || cur < nextDpVal1) {
                    nextDpIdx2 = nextDpIdx1;
                    nextDpVal2 = nextDpVal1;
                    nextDpIdx1 = j;
                    nextDpVal1 = cur;
                } else if (nextDpIdx2 == -1 || cur < nextDpVal2) {
                    nextDpIdx2 = j;
                    nextDpVal2 = cur;
                }
            }

            dpIdx1 = nextDpIdx1;
            dpIdx2 = nextDpIdx2;
            dpVal1 = nextDpVal1;
            dpVal2 = nextDpVal2;
        }

        return dpVal1;
    }
}

class Solution {
public:
    int minFallingPathSum(vector<vector<int>>& grid) {
        int n = grid.size();
        if (n == 1) {
            return grid[0][0];
        }

        int dpIdx1 = -1, dpIdx2 = -1;
        int dpVal1 = 0, dpVal2 = 0;

        for (int i = 1; i < n; i++) {
            int nextDpIdx1 = -1, nextDpIdx2 = -1;
            int nextDpVal1 = INT_MAX, nextDpVal2 = INT_MAX;

            for (int j = 0; j < n; j++) {
                int cur = (j != dpIdx1) ? dpVal1 : dpVal2;
                cur += grid[i][j];

                if (nextDpIdx1 == -1 || cur < nextDpVal1) {
                    nextDpIdx2 = nextDpIdx1;
                    nextDpVal2 = nextDpVal1;
                    nextDpIdx1 = j;
                    nextDpVal1 = cur;
                } else if (nextDpIdx2 == -1 || cur < nextDpVal2) {
                    nextDpIdx2 = j;
                    nextDpVal2 = cur;
                }
            }

            dpIdx1 = nextDpIdx1;
            dpIdx2 = nextDpIdx2;
            dpVal1 = nextDpVal1;
            dpVal2 = nextDpVal2;
        }

        return dpVal1;
    }
};

class Solution {
    /**
     * @param {number[][]} grid
     * @return {number}
     */
    minFallingPathSum(grid) {
        const n = grid.length;
        if (n === 1) return grid[0][0];

        let dpIdx1 = -1,
            dpIdx2 = -1;
        let dpVal1 = 0,
            dpVal2 = 0;

        for (let i = 0; i < n; i++) {
            let nextDpIdx1 = -1,
                nextDpIdx2 = -1;
            let nextDpVal1 = Infinity,
                nextDpVal2 = Infinity;

            for (let j = 0; j < n; j++) {
                let cur = j !== dpIdx1 ? dpVal1 : dpVal2;
                cur += grid[i][j];

                if (nextDpIdx1 === -1 || cur < nextDpVal1) {
                    nextDpIdx2 = nextDpIdx1;
                    nextDpVal2 = nextDpVal1;
                    nextDpIdx1 = j;
                    nextDpVal1 = cur;
                } else if (nextDpIdx2 === -1 || cur < nextDpVal2) {
                    nextDpIdx2 = j;
                    nextDpVal2 = cur;
                }
            }

            dpIdx1 = nextDpIdx1;
            dpIdx2 = nextDpIdx2;
            dpVal1 = nextDpVal1;
            dpVal2 = nextDpVal2;
        }

        return dpVal1;
    }
}

public class Solution {
    public int MinFallingPathSum(int[][] grid) {
        int n = grid.Length;
        if (n == 1) return grid[0][0];

        int dpIdx1 = -1, dpIdx2 = -1;
        int dpVal1 = 0, dpVal2 = 0;

        for (int i = 0; i < n; i++) {
            int nextDpIdx1 = -1, nextDpIdx2 = -1;
            int nextDpVal1 = int.MaxValue, nextDpVal2 = int.MaxValue;

            for (int j = 0; j < n; j++) {
                int cur = (j != dpIdx1) ? dpVal1 : dpVal2;
                cur += grid[i][j];

                if (nextDpIdx1 == -1 || cur < nextDpVal1) {
                    nextDpIdx2 = nextDpIdx1;
                    nextDpVal2 = nextDpVal1;
                    nextDpIdx1 = j;
                    nextDpVal1 = cur;
                } else if (nextDpIdx2 == -1 || cur < nextDpVal2) {
                    nextDpIdx2 = j;
                    nextDpVal2 = cur;
                }
            }

            dpIdx1 = nextDpIdx1;
            dpIdx2 = nextDpIdx2;
            dpVal1 = nextDpVal1;
            dpVal2 = nextDpVal2;
        }

        return dpVal1;
    }
}

func minFallingPathSum(grid [][]int) int {
    n := len(grid)
    if n == 1 {
        return grid[0][0]
    }

    dpIdx1, dpIdx2 := -1, -1
    dpVal1, dpVal2 := 0, 0

    for i := 0; i < n; i++ {
        nextDpIdx1, nextDpIdx2 := -1, -1
        nextDpVal1, nextDpVal2 := math.MaxInt32, math.MaxInt32

        for j := 0; j < n; j++ {
            cur := dpVal1
            if j == dpIdx1 {
                cur = dpVal2
            }
            cur += grid[i][j]

            if nextDpIdx1 == -1 || cur < nextDpVal1 {
                nextDpIdx2, nextDpVal2 = nextDpIdx1, nextDpVal1
                nextDpIdx1, nextDpVal1 = j, cur
            } else if nextDpIdx2 == -1 || cur < nextDpVal2 {
                nextDpIdx2, nextDpVal2 = j, cur
            }
        }

        dpIdx1, dpIdx2, dpVal1, dpVal2 = nextDpIdx1, nextDpIdx2, nextDpVal1, nextDpVal2
    }

    return dpVal1
}

class Solution {
    fun minFallingPathSum(grid: Array<IntArray>): Int {
        val n = grid.size
        if (n == 1) return grid[0][0]

        var dpIdx1 = -1
        var dpIdx2 = -1
        var dpVal1 = 0
        var dpVal2 = 0

        for (i in 0 until n) {
            var nextDpIdx1 = -1
            var nextDpIdx2 = -1
            var nextDpVal1 = Int.MAX_VALUE
            var nextDpVal2 = Int.MAX_VALUE

            for (j in 0 until n) {
                var cur = if (j != dpIdx1) dpVal1 else dpVal2
                cur += grid[i][j]

                if (nextDpIdx1 == -1 || cur < nextDpVal1) {
                    nextDpIdx2 = nextDpIdx1
                    nextDpVal2 = nextDpVal1
                    nextDpIdx1 = j
                    nextDpVal1 = cur
                } else if (nextDpIdx2 == -1 || cur < nextDpVal2) {
                    nextDpIdx2 = j
                    nextDpVal2 = cur
                }
            }

            dpIdx1 = nextDpIdx1
            dpIdx2 = nextDpIdx2
            dpVal1 = nextDpVal1
            dpVal2 = nextDpVal2
        }

        return dpVal1
    }
}

class Solution {
    func minFallingPathSum(_ grid: [[Int]]) -> Int {
        let n = grid.count
        if n == 1 { return grid[0][0] }

        var dpIdx1 = -1, dpIdx2 = -1
        var dpVal1 = 0, dpVal2 = 0

        for i in 0..<n {
            var nextDpIdx1 = -1, nextDpIdx2 = -1
            var nextDpVal1 = Int.max, nextDpVal2 = Int.max

            for j in 0..<n {
                var cur = (j != dpIdx1) ? dpVal1 : dpVal2
                cur += grid[i][j]

                if nextDpIdx1 == -1 || cur < nextDpVal1 {
                    nextDpIdx2 = nextDpIdx1
                    nextDpVal2 = nextDpVal1
                    nextDpIdx1 = j
                    nextDpVal1 = cur
                } else if nextDpIdx2 == -1 || cur < nextDpVal2 {
                    nextDpIdx2 = j
                    nextDpVal2 = cur
                }
            }

            dpIdx1 = nextDpIdx1
            dpIdx2 = nextDpIdx2
            dpVal1 = nextDpVal1
            dpVal2 = nextDpVal2
        }

        return dpVal1
    }
}

impl Solution {
    pub fn min_falling_path_sum(grid: Vec<Vec<i32>>) -> i32 {
        let n = grid.len();
        if n == 1 {
            return grid[0][0];
        }

        let (mut dp_idx1, mut dp_idx2): (i32, i32) = (-1, -1);
        let (mut dp_val1, mut dp_val2): (i32, i32) = (0, 0);

        for i in 0..n {
            let (mut next_idx1, mut next_idx2): (i32, i32) = (-1, -1);
            let (mut next_val1, mut next_val2) = (i32::MAX, i32::MAX);

            for j in 0..n {
                let cur = if j as i32 != dp_idx1 { dp_val1 } else { dp_val2 }
                    + grid[i][j];

                if next_idx1 == -1 || cur < next_val1 {
                    next_idx2 = next_idx1;
                    next_val2 = next_val1;
                    next_idx1 = j as i32;
                    next_val1 = cur;
                } else if next_idx2 == -1 || cur < next_val2 {
                    next_idx2 = j as i32;
                    next_val2 = cur;
                }
            }

            dp_idx1 = next_idx1;
            dp_idx2 = next_idx2;
            dp_val1 = next_val1;
            dp_val2 = next_val2;
        }

        dp_val1
    }
}

Time & Space Complexity

Time complexity: $O(n ^ 2)$
Space complexity: $O(1)$ extra space.

Common Pitfalls

Confusing with Standard Falling Path Sum

Unlike the standard falling path problem where you move to adjacent columns, this variant requires choosing a different column in each row. Using the adjacent-column logic from the standard problem will produce incorrect results.

O(n^3) Time Complexity Trap

A naive approach checks all n columns from the previous row for each of n^2 cells, resulting in O(n^3) time. The key optimization is tracking only the two smallest values from each row, since you only need the second smallest when the current column matches the smallest.

Edge Case with Single Column Grid

When n = 1, there is only one column, making it impossible to pick a different column in each row after the first. The answer is simply grid[0][0] since a 1x1 grid has only one cell. Failing to handle this edge case causes index errors or infinite loops.

Incorrect Second Minimum Tracking

When optimizing to O(n^2), you must correctly track both the minimum and second minimum values along with their column indices. Common mistakes include not updating both values properly, using the wrong index for comparison, or forgetting to handle ties when multiple columns have the same minimum value.

Integer Overflow with Large Negative Values

Grid values can be negative, so initializing with INT_MIN as a sentinel can cause overflow when added to grid values. Use a distinct sentinel value or nullable type to distinguish uncomputed states from valid negative sums.