Minimum Falling Path Sum

Given an n x n array of integers matrix, return the minimum sum of any falling path through matrix.

A falling path starts at any element in the first row and chooses the element in the next row that is either directly below or diagonally left/right. Specifically, the next element from position (row, col) will be (row + 1, col - 1), (row + 1, col), or (row + 1, col + 1).

Example 1:

Input: matrix = [[2,1,3],[6,5,4],[7,8,9]]
Output: 13
Explanation: There are two falling paths with a minimum sum as shown.

Example 2:

Input: matrix = [[-19,57],[-40,-5]]
Output: -59
Explanation: The falling path with a minimum sum is shown.

Constraints:

• n == matrix.length == matrix[i].length
• 1 <= n <= 100
• -100 <= matrix[i][j] <= 100

Approach - Bottom Up 1D DP

• we have three options left mid and right and left start from infinity because we looping left to right
• Time & Space Complexity
  Time complexity: O(n^2)
  Space complexity: O(n)

class Solution {
    public int minFallingPathSum(int[][] matrix) {
        int m = matrix.length, n = matrix[0].length;
        int dp[] = new int[n];
        for (int i = 0; i < n; i++)
            dp[i] = matrix[0][i];
 
        for (int i = 1; i < m; i++) {
            int left = Integer.MAX_VALUE;
            for (int j = 0; j < n; j++) {
                int mid = dp[j];
                int right = (j < n - 1) ? dp[j + 1] : Integer.MAX_VALUE;
                dp[j] = matrix[i][j] + Math.min(left, Math.min(mid, right));
                left = mid;
            }
        }
 
        int ans = Integer.MAX_VALUE;
        for (int val : dp) {
            ans = Math.min(ans, val);
        }
 
        return ans;
    }
}