Leetcode 931. Minimum Falling Path Sum
Question Description
Original Question: 931. Minimum Falling Path Sum
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:
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.
Solution
- Java
- Python
class Solution {
public int minFallingPathSum(int[][] A) {
int m = A.length, n = A[0].length;
int[][] dp = new int[m][n];
for(int i = 0; i < n; i++){
dp[0][i] = A[0][i];
}
for(int i = 1; i < m; i++){
for(int j = 0; j < n; j++){
int topLeft = j-1 < 0 ? Integer.MAX_VALUE : dp[i-1][j-1];
int topMid = dp[i-1][j];
int topRight = j+1 >= n ? Integer.MAX_VALUE : dp[i-1][j+1];
dp[i][j] = A[i][j] + Math.min(topLeft, Math.min(topMid, topRight));
}
}
int smallestPath = Integer.MAX_VALUE;
for(int i = 0; i < n; i++){
smallestPath = Math.min(smallestPath, dp[m-1][i]);
}
return smallestPath;
}
}
class Solution:
def minFallingPathSum(self, A: List[List[int]]) -> int:
for i in range(1,len(A)):
for j in range(len(A[0])):
if j == 0:
A[i][j] = min((A[i][j] + A[i - 1][j]), (A[i][j] + A[i - 1][j + 1]) )
elif (j == len(A[0]) - 1):
A[i][j] = min((A[i][j] + A[i - 1][j]), (A[i][j] + A[i - 1][j - 1]) )
else:
A[i][j] = min(A[i][j] + A[i - 1][j],A[i][j] + A[i - 1][j + 1], A[i][j] + A[i - 1][j - 1])
return min(A[len(A) - 1])