Formatted question description: https://leetcode.ca/all/931.html
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
Companies:
Goldman Sachs, Google
Related Topics:
Dynamic Programming
Solution 1. DP
// OJ: https://leetcode.com/problems/minimum-falling-path-sum/
// Time: O(N^2)
// Space: O(N)
class Solution {
public:
int minFallingPathSum(vector<vector<int>>& A) {
int N = A.size();
vector<vector<int>> dp(2, vector<int>(N, 0));
for (int i = 0; i < N; ++i) {
for (int j = 0; j < N; ++j) {
dp[(i + 1) % 2][j] = A[i][j] + min({ j - 1 >= 0 ? dp[i % 2][j - 1] : INT_MAX,
dp[i % 2][j],
j + 1 < N ? dp[i % 2][j + 1] : INT_MAX });
}
}
return *min_element(dp[N % 2].begin(), dp[N % 2].end());
}
};
Solution 2.
Same idea as Solution 1, but use A to cache DP values.
// OJ: https://leetcode.com/problems/minimum-falling-path-sum/
// Time: O(N^2)
// Space: O(1)
class Solution {
public:
int minFallingPathSum(vector<vector<int>>& A) {
int N = A.size();
for (int i = 1; i < N; ++i) {
for (int j = 0; j < N; ++j) {
A[i][j] += min({ j - 1 >= 0 ? A[i - 1][j - 1] : INT_MAX,
A[i - 1][j],
j + 1 < N ? A[i - 1][j + 1] : INT_MAX });
}
}
return *min_element(A.back().begin(), A.back().end());
}
};
Java
class Solution {
public int minFallingPathSum(int[][] A) {
int side = A.length;
int[][] dp = new int[side][side];
for (int i = 0; i < side; i++)
dp[0][i] = A[0][i];
for (int i = 1; i < side; i++) {
for (int j = 0; j < side; j++) {
if (j == 0)
dp[i][j] = Math.min(dp[i - 1][j], dp[i - 1][j + 1]) + A[i][j];
else if (j == side - 1)
dp[i][j] = Math.min(dp[i - 1][j - 1], dp[i - 1][j]) + A[i][j];
else
dp[i][j] = Math.min(Math.min(dp[i - 1][j - 1], dp[i - 1][j]), dp[i - 1][j + 1]) + A[i][j];
}
}
int minSum = Integer.MAX_VALUE;
for (int i = 0; i < side; i++)
minSum = Math.min(minSum, dp[side - 1][i]);
return minSum;
}
}