931. Minimum Falling Path Sum

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/
// Author: github.com/lzl124631x
// 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/
// Author: github.com/lzl124631x
// 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());
    }
};