1937. Maximum Number of Points with Cost - Explanation

Description

You are given an m x n integer matrix points (0-indexed). Starting with 0 points, you want to maximize the number of points you can get from the matrix.

To gain points, you must pick one cell in each row. Picking the cell at coordinates (r, c) will add points[r][c] to your score.

However, you will lose points if you pick a cell too far from the cell that you picked in the previous row. For every two adjacent rows r and r + 1 (where 0 <= r < m - 1), picking cells at coordinates (r, c1) and (r + 1, c2) will subtract abs(c1 - c2) from your score.

Return the maximum number of points you can achieve.

abs(x) is defined as:

x for x >= 0.
-x for x < 0.

Example 1:

Input: points = [
    [1,2,3],
    [1,5,1],
    [3,1,1]
]

Output: 9

Explanation: The optimal cells to pick are (0, 2), (1, 1), and (2, 0).
You add 3 + 5 + 3 = 11 to your score.
However, you must subtract abs(2 - 1) + abs(1 - 0) = 2 from your score.
Your final score is 11 - 2 = 9.

Example 2:

Input: points = [
    [1,5],
    [2,3],
    [4,2]
]

Output: 11

Explanation: The optimal cells to pick are (0, 1), (1, 1), and (2, 0).
You add 5 + 3 + 4 = 12 to your score.
However, you must subtract abs(1 - 1) + abs(1 - 0) = 1 from your score.
Your final score is 12 - 1 = 11.

Constraints:

m == points.length
n == points[r].length
1 <= m, n <= 100,000
1 <= m * n <= 100,000
0 <= points[r][c] <= 100,000

Topics

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Prerequisites

Before attempting this problem, you should be comfortable with:

Dynamic Programming - Building solutions from overlapping subproblems with optimal substructure
Memoization - Caching recursive results to avoid redundant computation
Prefix/Suffix Maximum Arrays - Precomputing running maximums from left-to-right and right-to-left
2D Array Traversal - Processing matrices row by row with state transitions

1. Recursion

Intuition

We need to pick one cell from each row such that the total points are maximized. The tricky part is the penalty: if we pick column c in row r and column c' in row r+1, we lose |c - c'| points.

A natural approach is to try all possibilities. For each starting column in the first row, we recursively explore all column choices in subsequent rows, subtracting the movement penalty and adding the cell value. We return the maximum sum found.

Algorithm

Define a recursive function dfs(r, c) that returns the maximum points obtainable from row r+1 to the last row, given we are currently at column c in row r.
Base case: if r is the last row, return 0 (no more rows to process).
For each column col in the next row, compute points[r+1][col] - |col - c| + dfs(r+1, col).
Return the maximum over all column choices.
The answer is the maximum of points[0][c] + dfs(0, c) for all starting columns c.

class Solution:
    def maxPoints(self, points: List[List[int]]) -> int:
        m, n = len(points), len(points[0])

        def dfs(r, c):
            if r == m - 1:
                return 0

            res = 0
            for col in range(n):
                res = max(res, points[r + 1][col] - abs(col - c) + dfs(r + 1, col))
            return res

        ans = 0
        for c in range(n):
            ans = max(ans, points[0][c] + dfs(0, c))
        return ans

public class Solution {
    int m, n;
    int[][] points;

    long dfs(int r, int c) {
        if (r == m - 1) return 0;
        long res = 0;
        for (int col = 0; col < n; col++) {
            res = Math.max(res, points[r + 1][col] - Math.abs(col - c) + dfs(r + 1, col));
        }
        return res;
    }

    public long maxPoints(int[][] points) {
        this.m = points.length;
        this.n = points[0].length;
        this.points = points;
        long ans = 0;
        for (int c = 0; c < n; c++) {
            ans = Math.max(ans, points[0][c] + dfs(0, c));
        }
        return ans;
    }
}

class Solution {
public:
    int m, n;
    vector<vector<int>> points;

    long long dfs(int r, int c) {
        if (r == m - 1) return 0;
        long long res = 0;
        for (int col = 0; col < n; col++) {
            res = max(res, points[r + 1][col] - abs(col - c) + dfs(r + 1, col));
        }
        return res;
    }

    long long maxPoints(vector<vector<int>>& points) {
        this->points = points;
        m = points.size();
        n = points[0].size();
        long long ans = 0;
        for (int c = 0; c < n; c++) {
            ans = max(ans, points[0][c] + dfs(0, c));
        }
        return ans;
    }
};

class Solution {
    /**
     * @param {number[][]} points
     * @return {number}
     */
    maxPoints(points) {
        let m = points.length, n = points[0].length;

        function dfs(r, c) {
            if (r === m - 1) return 0;
            let res = 0;
            for (let col = 0; col < n; col++) {
                res = Math.max(res, points[r + 1][col] - Math.abs(col - c) + dfs(r + 1, col));
            }
            return res;
        }

        let ans = 0;
        for (let c = 0; c < n; c++) {
            ans = Math.max(ans, points[0][c] + dfs(0, c));
        }
        return ans;
    }
}

public class Solution {
    int m, n;
    int[][] points;

    long Dfs(int r, int c) {
        if (r == m - 1) return 0;
        long res = 0;
        for (int col = 0; col < n; col++) {
            res = Math.Max(res, points[r + 1][col] - Math.Abs(col - c) + Dfs(r + 1, col));
        }
        return res;
    }

    public long MaxPoints(int[][] points) {
        this.points = points;
        m = points.Length;
        n = points[0].Length;
        long ans = 0;
        for (int c = 0; c < n; c++) {
            ans = Math.Max(ans, points[0][c] + Dfs(0, c));
        }
        return ans;
    }
}

func maxPoints(points [][]int) int64 {
    m, n := len(points), len(points[0])

    var dfs func(r, c int) int64
    dfs = func(r, c int) int64 {
        if r == m-1 {
            return 0
        }
        var res int64 = 0
        for col := 0; col < n; col++ {
            diff := col - c
            if diff < 0 {
                diff = -diff
            }
            val := int64(points[r+1][col]) - int64(diff) + dfs(r+1, col)
            if val > res {
                res = val
            }
        }
        return res
    }

    var ans int64 = 0
    for c := 0; c < n; c++ {
        val := int64(points[0][c]) + dfs(0, c)
        if val > ans {
            ans = val
        }
    }
    return ans
}

class Solution {
    private var m = 0
    private var n = 0
    private lateinit var points: Array<IntArray>

    fun maxPoints(points: Array<IntArray>): Long {
        this.points = points
        m = points.size
        n = points[0].size
        var ans: Long = 0
        for (c in 0 until n) {
            ans = maxOf(ans, points[0][c].toLong() + dfs(0, c))
        }
        return ans
    }

    private fun dfs(r: Int, c: Int): Long {
        if (r == m - 1) return 0
        var res: Long = 0
        for (col in 0 until n) {
            res = maxOf(res, points[r + 1][col] - kotlin.math.abs(col - c) + dfs(r + 1, col))
        }
        return res
    }
}

class Solution {
    var m = 0
    var n = 0
    var points = [[Int]]()

    func maxPoints(_ points: [[Int]]) -> Int {
        self.points = points
        m = points.count
        n = points[0].count
        var ans: Int = 0
        for c in 0..<n {
            ans = max(ans, points[0][c] + dfs(0, c))
        }
        return ans
    }

    private func dfs(_ r: Int, _ c: Int) -> Int {
        if r == m - 1 { return 0 }
        var res = 0
        for col in 0..<n {
            res = max(res, points[r + 1][col] - abs(col - c) + dfs(r + 1, col))
        }
        return res
    }
}

Time & Space Complexity

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

Where $m$ is the number of rows, and $n$ is the number of columns.

2. Dynamic Programming (Top-Down)

Intuition

The recursive solution recalculates the same subproblems multiple times. For a given (row, column) pair, the maximum points from that position onward is always the same, regardless of how we got there.

By storing (memoizing) the result for each (r, c) pair, we avoid redundant computation. This transforms the exponential solution into a polynomial one.

Algorithm

Create a memoization table memo to cache results for each (r, c) pair.
Define dfs(r, c) as before, but first check if the result is already cached.
If cached, return it immediately.
Otherwise, compute the result by trying all columns in the next row, cache it, and return.
The final answer is computed the same way as the recursive approach.

class Solution:
    def maxPoints(self, points: List[List[int]]) -> int:
        m, n = len(points), len(points[0])
        memo = {}

        def dfs(r, c):
            if (r, c) in memo:
                return memo[(r, c)]
            if r == m - 1:
                return 0

            res = 0
            for col in range(n):
                res = max(res, points[r + 1][col] - abs(col - c) + dfs(r + 1, col))

            memo[(r, c)] = res
            return res

        ans = 0
        for c in range(n):
            ans = max(ans, points[0][c] + dfs(0, c))
        return ans

class Solution {
public:
    int m, n;
    vector<vector<int>> points;
    vector<vector<long long>> memo;

    long long dfs(int r, int c) {
        if (memo[r][c] != -1) return memo[r][c];
        if (r == m - 1) return 0;

        long long res = 0;
        for (int col = 0; col < n; col++) {
            res = max(res, (long long)points[r + 1][col] - abs(col - c) + dfs(r + 1, col));
        }
        return memo[r][c] = res;
    }

    long long maxPoints(vector<vector<int>>& points) {
        this->points = points;
        m = points.size();
        n = points[0].size();
        memo.assign(m, vector<long long>(n, -1));
        long long ans = 0;
        for (int c = 0; c < n; c++) {
            ans = max(ans, (long long)points[0][c] + dfs(0, c));
        }
        return ans;
    }
};

class Solution {
    /**
     * @param {number[][]} points
     * @return {number}
     */
    maxPoints(points) {
        let m = points.length, n = points[0].length;
        let memo = {};

        function dfs(r, c) {
            let key = r + "," + c;
            if (key in memo) return memo[key];
            if (r === m - 1) return 0;

            let res = 0;
            for (let col = 0; col < n; col++) {
                res = Math.max(res, points[r + 1][col] - Math.abs(col - c) + dfs(r + 1, col));
            }
            return memo[key] = res;
        }

        let ans = 0;
        for (let c = 0; c < n; c++) {
            ans = Math.max(ans, points[0][c] + dfs(0, c));
        }
        return ans;
    }
}

func maxPoints(points [][]int) int64 {
    m, n := len(points), len(points[0])
    memo := make([][]int64, m)
    for i := range memo {
        memo[i] = make([]int64, n)
        for j := range memo[i] {
            memo[i][j] = -1
        }
    }

    var dfs func(r, c int) int64
    dfs = func(r, c int) int64 {
        if memo[r][c] != -1 {
            return memo[r][c]
        }
        if r == m-1 {
            return 0
        }
        var res int64 = 0
        for col := 0; col < n; col++ {
            diff := col - c
            if diff < 0 {
                diff = -diff
            }
            val := int64(points[r+1][col]) - int64(diff) + dfs(r+1, col)
            if val > res {
                res = val
            }
        }
        memo[r][c] = res
        return res
    }

    var ans int64 = 0
    for c := 0; c < n; c++ {
        val := int64(points[0][c]) + dfs(0, c)
        if val > ans {
            ans = val
        }
    }
    return ans
}

class Solution {
    private var m = 0
    private var n = 0
    private lateinit var points: Array<IntArray>
    private lateinit var memo: Array<LongArray>

    fun maxPoints(points: Array<IntArray>): Long {
        this.points = points
        m = points.size
        n = points[0].size
        memo = Array(m) { LongArray(n) { -1 } }
        var ans: Long = 0
        for (c in 0 until n) {
            ans = maxOf(ans, points[0][c].toLong() + dfs(0, c))
        }
        return ans
    }

    private fun dfs(r: Int, c: Int): Long {
        if (memo[r][c] != -1L) return memo[r][c]
        if (r == m - 1) return 0
        var res: Long = 0
        for (col in 0 until n) {
            res = maxOf(res, points[r + 1][col] - kotlin.math.abs(col - c) + dfs(r + 1, col))
        }
        memo[r][c] = res
        return res
    }
}

class Solution {
    var m = 0
    var n = 0
    var points = [[Int]]()
    var memo = [[Int]]()

    func maxPoints(_ points: [[Int]]) -> Int {
        self.points = points
        m = points.count
        n = points[0].count
        memo = Array(repeating: Array(repeating: -1, count: n), count: m)
        var ans = 0
        for c in 0..<n {
            ans = max(ans, points[0][c] + dfs(0, c))
        }
        return ans
    }

    private func dfs(_ r: Int, _ c: Int) -> Int {
        if memo[r][c] != -1 { return memo[r][c] }
        if r == m - 1 { return 0 }
        var res = 0
        for col in 0..<n {
            res = max(res, points[r + 1][col] - abs(col - c) + dfs(r + 1, col))
        }
        memo[r][c] = res
        return res
    }
}

Time & Space Complexity

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

Where $m$ is the number of rows, and $n$ is the number of columns.

3. Dynamic Programming (Bottom-Up)

Intuition

Computing max(dp[c'] - |c - c'|) for each column c normally takes O(n) time per cell, making the total O(m * n^2). We can optimize this using prefix maximums.

The penalty |c - c'| splits into two cases:

If c' <= c: the penalty is c - c', so we want max(dp[c'] + c') for all c' <= c.
If c' > c: the penalty is c' - c, so we want max(dp[c'] - c') for all c' > c.

By precomputing a left array (max of dp[c'] + c' from the left) and a right array (max of dp[c'] - c' from the right), we can answer each column's query in O(1) time.

Algorithm

Initialize dp with the first row's values.
For each subsequent row:
- Build left[c] = max over all c' <= c of dp[c']. Propagate left-to-right, subtracting 1 at each step.
- Build right[c] = max over all c' >= c of dp[c']. Propagate right-to-left, subtracting 1 at each step.
- For each column c, set nextDp[c] = points[r][c] + max(left[c], right[c]).
After processing all rows, return the maximum value in dp.

class Solution:
    def maxPoints(self, points: List[List[int]]) -> int:
        ROWS, COLS = len(points), len(points[0])
        dp = points[0]

        for r in range(1, ROWS):
            left = [0] * COLS
            left[0] = dp[0]
            for c in range(1, COLS):
                left[c] = max(dp[c], left[c - 1] - 1)

            right = [0] * COLS
            right[COLS - 1] = dp[COLS - 1]
            for c in range(COLS - 2, -1, -1):
                right[c] = max(dp[c], right[c + 1] - 1)

            nextDp = points[r][:]
            for c in range(COLS):
                nextDp[c] += max(left[c], right[c])

            dp = nextDp

        return max(dp)

public class Solution {
    public long maxPoints(int[][] points) {
        int ROWS = points.length, COLS = points[0].length;
        long[] dp = new long[COLS];
        for (int c = 0; c < COLS; c++) dp[c] = points[0][c];

        for (int r = 1; r < ROWS; r++) {
            long[] left = new long[COLS];
            left[0] = dp[0];
            for (int c = 1; c < COLS; c++)
                left[c] = Math.max(dp[c], left[c - 1] - 1);

            long[] right = new long[COLS];
            right[COLS - 1] = dp[COLS - 1];
            for (int c = COLS - 2; c >= 0; c--)
                right[c] = Math.max(dp[c], right[c + 1] - 1);

            long[] nextDp = new long[COLS];
            for (int c = 0; c < COLS; c++)
                nextDp[c] = points[r][c] + Math.max(left[c], right[c]);

            dp = nextDp;
        }

        long ans = 0;
        for (long val : dp) ans = Math.max(ans, val);
        return ans;
    }
}

class Solution {
public:
    long long maxPoints(vector<vector<int>>& points) {
        int ROWS = points.size(), COLS = points[0].size();
        vector<long long> dp(points[0].begin(), points[0].end());

        for (int r = 1; r < ROWS; r++) {
            vector<long long> left(COLS), right(COLS);
            left[0] = dp[0];
            for (int c = 1; c < COLS; c++)
                left[c] = max(dp[c], left[c - 1] - 1);

            right[COLS - 1] = dp[COLS - 1];
            for (int c = COLS - 2; c >= 0; c--)
                right[c] = max(dp[c], right[c + 1] - 1);

            vector<long long> nextDp(COLS);
            for (int c = 0; c < COLS; c++)
                nextDp[c] = points[r][c] + max(left[c], right[c]);

            dp = move(nextDp);
        }

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

class Solution {
    /**
     * @param {number[][]} points
     * @return {number}
     */
    maxPoints(points) {
        let ROWS = points.length, COLS = points[0].length;
        let dp = [...points[0]];

        for (let r = 1; r < ROWS; r++) {
            let left = Array(COLS).fill(0);
            left[0] = dp[0];
            for (let c = 1; c < COLS; c++)
                left[c] = Math.max(dp[c], left[c - 1] - 1);

            let right = Array(COLS).fill(0);
            right[COLS - 1] = dp[COLS - 1];
            for (let c = COLS - 2; c >= 0; c--)
                right[c] = Math.max(dp[c], right[c + 1] - 1);

            let nextDp = [...points[r]];
            for (let c = 0; c < COLS; c++)
                nextDp[c] += Math.max(left[c], right[c]);

            dp = nextDp;
        }

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

public class Solution {
    public long MaxPoints(int[][] points) {
        int ROWS = points.Length, COLS = points[0].Length;
        long[] dp = new long[COLS];
        for (int c = 0; c < COLS; c++) dp[c] = points[0][c];

        for (int r = 1; r < ROWS; r++) {
            long[] left = new long[COLS];
            left[0] = dp[0];
            for (int c = 1; c < COLS; c++)
                left[c] = Math.Max(dp[c], left[c - 1] - 1);

            long[] right = new long[COLS];
            right[COLS - 1] = dp[COLS - 1];
            for (int c = COLS - 2; c >= 0; c--)
                right[c] = Math.Max(dp[c], right[c + 1] - 1);

            long[] nextDp = new long[COLS];
            for (int c = 0; c < COLS; c++)
                nextDp[c] = points[r][c] + Math.Max(left[c], right[c]);

            dp = nextDp;
        }

        long ans = 0;
        foreach (long val in dp) ans = Math.Max(ans, val);
        return ans;
    }
}

class Solution {
    fun maxPoints(points: Array<IntArray>): Long {
        val ROWS = points.size
        val COLS = points[0].size
        var dp = LongArray(COLS) { points[0][it].toLong() }

        for (r in 1 until ROWS) {
            val left = LongArray(COLS)
            left[0] = dp[0]
            for (c in 1 until COLS) {
                left[c] = maxOf(dp[c], left[c - 1] - 1)
            }

            val right = LongArray(COLS)
            right[COLS - 1] = dp[COLS - 1]
            for (c in COLS - 2 downTo 0) {
                right[c] = maxOf(dp[c], right[c + 1] - 1)
            }

            val nextDp = LongArray(COLS)
            for (c in 0 until COLS) {
                nextDp[c] = points[r][c] + maxOf(left[c], right[c])
            }
            dp = nextDp
        }

        return dp.maxOrNull() ?: 0
    }
}

class Solution {
    func maxPoints(_ points: [[Int]]) -> Int {
        let ROWS = points.count
        let COLS = points[0].count
        var dp = points[0].map { Int($0) }

        for r in 1..<ROWS {
            var left = [Int](repeating: 0, count: COLS)
            left[0] = dp[0]
            for c in 1..<COLS {
                left[c] = max(dp[c], left[c - 1] - 1)
            }

            var right = [Int](repeating: 0, count: COLS)
            right[COLS - 1] = dp[COLS - 1]
            for c in stride(from: COLS - 2, through: 0, by: -1) {
                right[c] = max(dp[c], right[c + 1] - 1)
            }

            var nextDp = [Int](repeating: 0, count: COLS)
            for c in 0..<COLS {
                nextDp[c] = points[r][c] + max(left[c], right[c])
            }
            dp = nextDp
        }

        return dp.max() ?? 0
    }
}

Time & Space Complexity

Time complexity: $O(m * n)$
Space complexity: $O(n)$

Where $m$ is the number of rows, and $n$ is the number of columns.

4. Dynamic Programming (Space Optimized)

Intuition

We can reduce memory usage by combining the left and right sweeps into a single pass. Instead of storing separate left and right arrays, we build the left maximum in a cur array during the left-to-right pass. Then, during the right-to-left pass, we update cur[c] by taking the max of the current left value and the running right maximum, adding the cell value as we go.

This avoids allocating a separate right array, reducing the constant factor in space usage.

Algorithm

Initialize prev with the first row's values.
For each subsequent row:
- Create a cur array and build left maximums by iterating left-to-right: cur[c] = max(prev[c], cur[c-1] - 1).
- Iterate right-to-left, tracking rightMax. For each column, update cur[c] = max(cur[c], rightMax) + points[r][c], then update rightMax = max(prev[c], rightMax - 1).
- Handle the last column separately since it was not processed in the right-to-left loop.
- Set prev = cur for the next iteration.
Return the maximum value in prev.

class Solution:
    def maxPoints(self, points: List[List[int]]) -> int:
        ROWS, COLS = len(points), len(points[0])
        prev = points[0]

        for r in range(1, ROWS):
            cur = [0] * COLS
            cur[0] = prev[0]
            for c in range(1, COLS):
                cur[c] = max(prev[c], cur[c - 1] - 1)

            rightMax = prev[COLS - 1]
            for c in range(COLS - 2, -1 , -1):
                rightMax = max(prev[c], rightMax - 1)
                cur[c] = max(cur[c], rightMax) + points[r][c]

            cur[COLS - 1] += points[r][COLS - 1]
            prev = cur

        return max(prev)

public class Solution {
    public long maxPoints(int[][] points) {
        int rows = points.length, cols = points[0].length;
        long[] prev = new long[cols];
        for (int c = 0; c < cols; c++) prev[c] = points[0][c];

        for (int r = 1; r < rows; r++) {
            long[] cur = new long[cols];
            cur[0] = prev[0];
            for (int c = 1; c < cols; c++)
                cur[c] = Math.max(prev[c], cur[c - 1] - 1);

            long rightMax = prev[cols - 1];
            for (int c = cols - 2; c >= 0; c--) {
                rightMax = Math.max(prev[c], rightMax - 1);
                cur[c] = Math.max(cur[c], rightMax) + points[r][c];
            }
            cur[cols - 1] += points[r][cols - 1];
            prev = cur;
        }

        long ans = 0;
        for (long val : prev) ans = Math.max(ans, val);
        return ans;
    }
}

class Solution {
public:
    long long maxPoints(vector<vector<int>>& points) {
        int rows = points.size(), cols = points[0].size();
        vector<long long> prev(cols);
        for (int c = 0; c < cols; c++) prev[c] = points[0][c];

        for (int r = 1; r < rows; r++) {
            vector<long long> cur(cols);
            cur[0] = prev[0];
            for (int c = 1; c < cols; c++)
                cur[c] = max(prev[c], cur[c - 1] - 1);

            long long rightMax = prev[cols - 1];
            for (int c = cols - 2; c >= 0; c--) {
                rightMax = max(prev[c], rightMax - 1);
                cur[c] = max(cur[c], rightMax) + points[r][c];
            }
            cur[cols - 1] += points[r][cols - 1];
            prev = cur;
        }
        return *max_element(prev.begin(), prev.end());
    }
};

class Solution {
    /**
     * @param {number[][]} points
     * @return {number}
     */
    maxPoints(points) {
        let rows = points.length, cols = points[0].length;
        let prev = points[0].map(v => v);

        for (let r = 1; r < rows; r++) {
            let cur = Array(cols).fill(0);
            cur[0] = prev[0];
            for (let c = 1; c < cols; c++)
                cur[c] = Math.max(prev[c], cur[c - 1] - 1);

            let rightMax = prev[cols - 1];
            for (let c = cols - 2; c >= 0; c--) {
                rightMax = Math.max(prev[c], rightMax - 1);
                cur[c] = Math.max(cur[c], rightMax) + points[r][c];
            }
            cur[cols - 1] += points[r][cols - 1];
            prev = cur;
        }
        return Math.max(...prev);
    }
}

public class Solution {
    public long MaxPoints(int[][] points) {
        int rows = points.Length, cols = points[0].Length;
        long[] prev = new long[cols];
        for (int c = 0; c < cols; c++) prev[c] = points[0][c];

        for (int r = 1; r < rows; r++) {
            long[] cur = new long[cols];
            cur[0] = prev[0];
            for (int c = 1; c < cols; c++)
                cur[c] = Math.Max(prev[c], cur[c - 1] - 1);

            long rightMax = prev[cols - 1];
            for (int c = cols - 2; c >= 0; c--) {
                rightMax = Math.Max(prev[c], rightMax - 1);
                cur[c] = Math.Max(cur[c], rightMax) + points[r][c];
            }
            cur[cols - 1] += points[r][cols - 1];
            prev = cur;
        }
        long ans = long.MinValue;
        foreach (long val in prev) ans = Math.Max(ans, val);
        return ans;
    }
}

func maxPoints(points [][]int) int64 {
    rows, cols := len(points), len(points[0])
    prev := make([]int64, cols)
    for c := 0; c < cols; c++ {
        prev[c] = int64(points[0][c])
    }

    for r := 1; r < rows; r++ {
        cur := make([]int64, cols)
        cur[0] = prev[0]
        for c := 1; c < cols; c++ {
            cur[c] = max64(prev[c], cur[c-1]-1)
        }

        rightMax := prev[cols-1]
        for c := cols - 2; c >= 0; c-- {
            rightMax = max64(prev[c], rightMax-1)
            cur[c] = max64(cur[c], rightMax) + int64(points[r][c])
        }
        cur[cols-1] += int64(points[r][cols-1])
        prev = cur
    }

    var ans int64 = prev[0]
    for _, val := range prev {
        if val > ans {
            ans = val
        }
    }
    return ans
}

func max64(a, b int64) int64 {
    if a > b {
        return a
    }
    return b
}

class Solution {
    fun maxPoints(points: Array<IntArray>): Long {
        val rows = points.size
        val cols = points[0].size
        var prev = LongArray(cols) { points[0][it].toLong() }

        for (r in 1 until rows) {
            val cur = LongArray(cols)
            cur[0] = prev[0]
            for (c in 1 until cols) {
                cur[c] = maxOf(prev[c], cur[c - 1] - 1)
            }

            var rightMax = prev[cols - 1]
            for (c in cols - 2 downTo 0) {
                rightMax = maxOf(prev[c], rightMax - 1)
                cur[c] = maxOf(cur[c], rightMax) + points[r][c]
            }
            cur[cols - 1] += points[r][cols - 1]
            prev = cur
        }

        return prev.maxOrNull() ?: 0
    }
}

class Solution {
    func maxPoints(_ points: [[Int]]) -> Int {
        let rows = points.count
        let cols = points[0].count
        var prev = points[0].map { $0 }

        for r in 1..<rows {
            var cur = [Int](repeating: 0, count: cols)
            cur[0] = prev[0]
            for c in 1..<cols {
                cur[c] = max(prev[c], cur[c - 1] - 1)
            }

            var rightMax = prev[cols - 1]
            for c in stride(from: cols - 2, through: 0, by: -1) {
                rightMax = max(prev[c], rightMax - 1)
                cur[c] = max(cur[c], rightMax) + points[r][c]
            }
            cur[cols - 1] += points[r][cols - 1]
            prev = cur
        }

        return prev.max() ?? 0
    }
}

Time & Space Complexity

Time complexity: $O(m * n)$
Space complexity: $O(n)$

Where $m$ is the number of rows, and $n$ is the number of columns.

Common Pitfalls

Integer Overflow

With grid dimensions up to 10^5 and cell values up to 10^5, the maximum total points can exceed the 32-bit integer limit. Using int instead of long (or equivalent 64-bit type) will cause overflow and incorrect results.

Missing One Direction in Prefix Maximum

The optimized solution requires computing both left-to-right and right-to-left prefix maximums. Forgetting one direction or incorrectly combining them will miss optimal transitions from certain columns, leading to suboptimal answers.

Off-by-One in Penalty Propagation

When propagating the prefix maximum, the penalty decreases by 1 for each column of distance. A common error is applying the penalty incorrectly, such as subtracting the penalty before comparing or not decrementing for each step. The update should be left[c] = max(dp[c], left[c-1] - 1), not left[c] = max(dp[c] - 1, left[c-1]).