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 <= A.length == A[0].length <= 100
-100 <= A[i][j] <= 100
Solution: DP array[i][j] += Min(array[i - 1][j], array[i - 1][Max(0, j - 1)], array[i - 1][Min(columns - 1, j + 1)]); i > 1
then get the minimum element of the last row.
class Solution {
public int minFallingPathSum(int[][] A) {
int row = A.length;
int column = A[0].length;
for (int i = 1; i < row; i++) {
for (int j = 0; j < column; j++) {
A[i][j] += Math.min(A[i - 1][j],
Math.min(A[i - 1][Math.max(0, j - 1)],
A[i - 1][Math.min(column - 1, j + 1)]));
}
}
int minVal = A[row - 1][0];
for (int i = 0; i < column; i++) {
minVal = Math.min(minVal, A[row - 1][i]);
}
return minVal;
}
}