931. Minimum Falling Path Sum
Explanation
To solve this problem, we can use dynamic programming. We will iterate through the matrix starting from the second row, and for each cell, we will update the value to be the minimum of the adjacent cells in the row above plus the current cell's value. Finally, we return the minimum value in the last row, which represents the minimum falling path sum.
- Time complexity: O(n^2) where n is the size of the matrix
- Space complexity: O(1)
class Solution {
public int minFallingPathSum(int[][] matrix) {
int n = matrix.length;
for (int i = 1; i < n; i++) {
for (int j = 0; j < n; j++) {
int minPrev = matrix[i - 1][j];
if (j > 0) {
minPrev = Math.min(minPrev, matrix[i - 1][j - 1]);
}
if (j < n - 1) {
minPrev = Math.min(minPrev, matrix[i - 1][j + 1]);
}
matrix[i][j] += minPrev;
}
}
int minSum = Integer.MAX_VALUE;
for (int num : matrix[n - 1]) {
minSum = Math.min(minSum, num);
}
return minSum;
}
}Code Editor (Testing phase)
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