120. Triangle - Explanation

Description

You are given a triangle array, return the minimum path sum from top to bottom.

For each step, you may move to an adjacent number of the row below. More formally, if you are on index i on the current row, you may move to either index i or index i + 1 on the next row.

Example 1:

Input: triangle = [
    [2],
   [3,4],
  [6,5,7],
 [4,1,8,3]
]

Output: 11

Explanation: The minimum path sum from top to bottom is 2 + 3 + 5 + 1 = 11

Example 2:

Input: triangle = [[-1]]

Output: -1

Constraints:

1 <= triangle.length <= 200
triangle[0].length == 1
triangle[i].length == triangle[i - 1].length + 1
-10,000 <= triangle[i][j] <= 10,000

Follow up: Could you do this using only O(n) extra space, where n is the total number of rows in the triangle?

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1. Recursion

class Solution:
    def minimumTotal(self, triangle: List[List[int]]) -> int:
        def dfs(row, col):
            if row >= len(triangle):
                return 0
            return triangle[row][col] + min(dfs(row + 1, col), dfs(row + 1, col + 1))

        return dfs(0, 0)

Time & Space Complexity

Time complexity: $O(2 ^ n)$
Space complexity: $O(n)$ for recursion stack.

2. Dynamic Programming (Top-Down)

class Solution:
    def minimumTotal(self, triangle: List[List[int]]) -> int:
        memo = [[0] * len(triangle[r]) for r in range(len(triangle))]
        INF = float("inf")
        for r in range(len(triangle)):
            for c in range(len(triangle[r])):
                memo[r][c] = INF

        def dfs(row, col):
            if row >= len(triangle):
                return 0
            if memo[row][col] != INF:
                return memo[row][col]

            memo[row][col] = triangle[row][col] + min(dfs(row + 1, col), dfs(row + 1, col + 1))
            return memo[row][col]

        return dfs(0, 0)

Time & Space Complexity

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

3. Dynamic Programming (Bottom-Up)

class Solution:
    def minimumTotal(self, triangle: List[List[int]]) -> int:
        n = len(triangle)
        dp = [[0] * len(triangle[row]) for row in range(n)]
        dp[-1] = triangle[-1][:]

        for row in range(n - 2, -1, -1):
            for col in range(len(triangle[row])):
                dp[row][col] = triangle[row][col] + min(dp[row + 1][col], dp[row + 1][col + 1])

        return dp[0][0]

Time & Space Complexity

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

4. Dynamic Programming (Space Optimized) - I

class Solution:
    def minimumTotal(self, triangle: List[List[int]]) -> int:
        n = len(triangle)
        dp = triangle[0][:]

        for row in range(1, n):
            nxtDp = [0] * len(triangle[row])
            nxtDp[0] = dp[0] + triangle[row][0]
            for col in range(1, len(triangle[row]) - 1):
                nxtDp[col] = triangle[row][col] + min(dp[col], dp[col - 1])
            nxtDp[-1] = dp[-1] + triangle[row][-1]
            dp = nxtDp

        return min(dp)

Time & Space Complexity

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

5. Dynamic Programming (Space Optimized) - II

class Solution:
    def minimumTotal(self, triangle: List[List[int]]) -> int:
        n = len(triangle)
        dp = triangle[-1][:]

        for row in range(n - 2, -1, -1):
            for col in range(len(triangle[row])):
                dp[col] = triangle[row][col] + min(dp[col], dp[col + 1])

        return dp[0]

Time & Space Complexity

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

6. Dynamic Programming (In-Place)

class Solution:
    def minimumTotal(self, triangle: List[List[int]]) -> int:
        for row in range(len(triangle) - 2, -1, -1):
            for col in range(len(triangle[row])):
                triangle[row][col] += min(triangle[row + 1][col], triangle[row + 1][col + 1])

        return triangle[0][0]

Time & Space Complexity

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