# Minimum Falling Path Sum Problem

## Description

LeetCode Problem 931.

Given an n x n array of integers matrix, return the minimum sum of any falling path through matrix.

A falling path starts at any element in the first row and chooses the element in the next row that is either directly below or diagonally left/right. Specifically, the next element from position (row, col) will be (row + 1, col - 1), (row + 1, col), or (row + 1, col + 1).

Example 1:

``````1
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Input: matrix = [[2,1,3],[6,5,4],[7,8,9]]
Output: 13
Explanation: There are two falling paths with a minimum sum as shown.
``````

Example 2:

``````1
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Input: matrix = [[-19,57],[-40,-5]]
Output: -59
Explanation: The falling path with a minimum sum is shown.
``````

Constraints:

• n == matrix.length == matrix[i].length
• 1 <= n <= 100
• -100 <= matrix[i][j] <= 100

## Sample C++ Code

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class Solution {
public:
int minFallingPathSum(vector<vector<int>>& A) {
if (A.size() == 0)
return 0;
int r = A.size(), c = A[0].size();
vector<vector<int>> dp(r+1, vector<int>(c+2, 0));

for (int i = 1; i <= r; i ++) {
for (int j = 1; j <= c; j ++) {
if (j == 1) {
dp[i][j] = min(dp[i-1][j], dp[i-1][j+1]) + A[i-1][j-1];
} else if (j == c) {
dp[i][j] = min(dp[i-1][j-1], dp[i-1][j]) + A[i-1][j-1];
} else {
dp[i][j] = min(dp[i-1][j-1], dp[i-1][j]);
dp[i][j] = min(dp[i][j], dp[i-1][j+1]);
dp[i][j] += A[i-1][j-1];
}
}
}
int min_sum = INT_MAX;
for (int i = 1; i <= c; i ++) {
if (dp[r][i] < min_sum) {
min_sum = dp[r][i];
}
}
return min_sum;
}
};
``````