leetcode 1289 - Minimum Falling Path Sum II

https://leetcode.com/problems/minimum-falling-path-sum-ii/

Given a square grid of integers arr, a falling path with non-zero shifts is a choice of exactly one element from each row of arr, such that no two elements chosen in adjacent rows are in the same column.

Return the minimum sum of a falling path with non-zero shifts.

Example 1:

Input: arr = [[1,2,3],[4,5,6],[7,8,9]]
Output: 13
Explanation:
The possible falling paths are:
[1,5,9], [1,5,7], [1,6,7], [1,6,8],
[2,4,8], [2,4,9], [2,6,7], [2,6,8],
[3,4,8], [3,4,9], [3,5,7], [3,5,9]
The falling path with the smallest sum is [1,5,7], so the answer is 13.

Constraints:

1 <= arr.length == arr[i].length <= 200
-99 <= arr[i][j] <= 99

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class Solution {
public:
int minFallingPathSum(vector<vector<int>>& arr) {
int row = arr.size(), col = arr[0].size();

int prev_min1 = INT_MAX, prev_min2 = INT_MAX;
int prev_min1_idx, prev_min2_idx;

int dp[row][col];

for (int i = 0; i < row; i++) {
int min1 = INT_MAX, min2 = INT_MAX;
int min1_idx, min2_idx;

for (int j = 0; j < col; j++) {
if (i == 0) {
dp[i][j] = arr[i][j];
} else {
if (j != prev_min1_idx)
dp[i][j] = dp[i-1][prev_min1_idx] + arr[i][j];
else
dp[i][j] = dp[i-1][prev_min2_idx] + arr[i][j];
}

if (dp[i][j] <= min1) {
min2 = min1;
min2_idx = min1_idx;
min1 = dp[i][j];
min1_idx = j;
} else if (dp[i][j] < min2) {
min2 = dp[i][j];
min2_idx = j;
}
}

prev_min1 = min1;
prev_min1_idx = min1_idx;
prev_min2 = min2;
prev_min2_idx = min2_idx;
}

return prev_min1;
}
};