Formatted question description: https://leetcode.ca/all/931.html

931. Minimum Falling Path Sum (Medium)

Given a square array of integers A, we want the minimum sum of a falling path through A.

A falling path starts at any element in the first row, and chooses one element from each row.  The next row's choice must be in a column that is different from the previous row's column by at most one.

 

Example 1:

Input: [[1,2,3],[4,5,6],[7,8,9]]
Output: 12
Explanation: 
The possible falling paths are:
  • [1,4,7], [1,4,8], [1,5,7], [1,5,8], [1,5,9]
  • [2,4,7], [2,4,8], [2,5,7], [2,5,8], [2,5,9], [2,6,8], [2,6,9]
  • [3,5,7], [3,5,8], [3,5,9], [3,6,8], [3,6,9]

The falling path with the smallest sum is [1,4,7], so the answer is 12.

 

Note:

  1. 1 <= A.length == A[0].length <= 100
  2. -100 <= A[i][j] <= 100

Companies:
Goldman Sachs, Google

Related Topics:
Dynamic Programming

Solution 1. DP

// OJ: https://leetcode.com/problems/minimum-falling-path-sum/
// Time: O(N^2)
// Space: O(N)
class Solution {
public:
    int minFallingPathSum(vector<vector<int>>& A) {
        int N = A.size();
        vector<vector<int>> dp(2, vector<int>(N, 0));
        for (int i = 0; i < N; ++i) {
            for (int j = 0; j < N; ++j) {
                dp[(i + 1) % 2][j] = A[i][j] + min({ j - 1 >= 0 ? dp[i % 2][j - 1] : INT_MAX,
                                                     dp[i % 2][j],
                                                     j + 1 < N ? dp[i % 2][j + 1] : INT_MAX });
            }
        }
        return *min_element(dp[N % 2].begin(), dp[N % 2].end());
    }
};

Solution 2.

Same idea as Solution 1, but use A to cache DP values.

// OJ: https://leetcode.com/problems/minimum-falling-path-sum/
// Time: O(N^2)
// Space: O(1)
class Solution {
public:
    int minFallingPathSum(vector<vector<int>>& A) {
        int N = A.size();
        for (int i = 1; i < N; ++i) {
            for (int j = 0; j < N; ++j) {
                A[i][j] += min({ j - 1 >= 0 ? A[i - 1][j - 1] : INT_MAX,
                                 A[i - 1][j],
                                 j + 1 < N ? A[i - 1][j + 1] : INT_MAX });
            }
        }
        return *min_element(A.back().begin(), A.back().end());
    }
};

Java

  • class Solution {
        public int minFallingPathSum(int[][] A) {
            int side = A.length;
            int[][] dp = new int[side][side];
            for (int i = 0; i < side; i++)
                dp[0][i] = A[0][i];
            for (int i = 1; i < side; i++) {
                for (int j = 0; j < side; j++) {
                    if (j == 0)
                        dp[i][j] = Math.min(dp[i - 1][j], dp[i - 1][j + 1]) + A[i][j];
                    else if (j == side - 1)
                        dp[i][j] = Math.min(dp[i - 1][j - 1], dp[i - 1][j]) + A[i][j];
                    else
                        dp[i][j] = Math.min(Math.min(dp[i - 1][j - 1], dp[i - 1][j]), dp[i - 1][j + 1]) + A[i][j];
                }
            }
            int minSum = Integer.MAX_VALUE;
            for (int i = 0; i < side; i++)
                minSum = Math.min(minSum, dp[side - 1][i]);
            return minSum;
        }
    }
    
  • // OJ: https://leetcode.com/problems/minimum-falling-path-sum/
    // Time: O(N^2)
    // Space: O(N)
    class Solution {
    public:
        int minFallingPathSum(vector<vector<int>>& A) {
            int N = A.size();
            vector<vector<int>> dp(2, vector<int>(N, 0));
            for (int i = 0; i < N; ++i) {
                for (int j = 0; j < N; ++j) {
                    dp[(i + 1) % 2][j] = A[i][j] + min({ j - 1 >= 0 ? dp[i % 2][j - 1] : INT_MAX,
                                                         dp[i % 2][j],
                                                         j + 1 < N ? dp[i % 2][j + 1] : INT_MAX });
                }
            }
            return *min_element(dp[N % 2].begin(), dp[N % 2].end());
        }
    };
    
  • class Solution(object):
        def minFallingPathSum(self, A):
            """
            :type A: List[List[int]]
            :rtype: int
            """
            M, N = len(A), len(A[0])
            dp = [[0] * (N + 2) for _ in range(M)]
            for i in range(M):
                dp[i][0] = dp[i][-1] = float('inf')
                for j in range(1, N + 1):
                    dp[i][j] = A[i][j - 1]
            for i in range(1, M):
                for j in range(1, N + 1):
                    dp[i][j] = A[i][j - 1] + min(dp[i - 1][j - 1], dp[i - 1][j], dp[i - 1][j + 1])
            return min(dp[-1])
    

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