Minimum Falling Path Sum
Minimum Falling Path Sum is a medium dynamic programming problem solved with the 2d dp - max/min of last row pattern.
The best approach, optimal (dp row by row, answer = min of the last row), runs in O(n²) time and O(n) space.
Below are 2 approaches in Java, from recursion from every start up.
Problem
A falling path starts at any cell of the first row and moves down one row at a time, to the cell directly below or diagonally left or right. Return the minimum sum of any falling path through the n × n matrix.
Examples
Example 1
- Input
matrix = [[2,1,3],[6,5,4],[7,8,9]]- Output
13- Why
- 1 → 5 → 7 or 1 → 4 → 8.
Example 2
- Input
matrix = [[-19,57],[-40,-5]]- Output
-59
Constraints
1 <= n <= 100; square matrix; values from -100 to 100.- From (r, c) you can move to (r + 1, c - 1), (r + 1, c) or (r + 1, c + 1).
Animated walkthrough
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Solutions
Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.
| Approach | Time | Space |
|---|---|---|
| Recursion from every start | O(n · 3ⁿ) | O(n) |
| Optimal (DP row by row, answer = min of the last row) | O(n²) | O(n) |
1Recursion from every start
O(n · 3ⁿ)O(n)For each starting column in the top row, try all three moves recursively.
- fall(r, c) = matrix[r][c] + min over the three columns below.
- Answer = min over c of fall(0, c).
class Solution {
public int minFallingPathSum(int[][] matrix) {
int best = Integer.MAX_VALUE;
for (int c = 0; c < matrix.length; c++) best = Math.min(best, fall(matrix, 0, c));
return best;
}
private int fall(int[][] m, int r, int c) {
if (c < 0 || c >= m.length) return Integer.MAX_VALUE;
if (r == m.length - 1) return m[r][c];
int below = Math.min(fall(m, r + 1, c), Math.min(fall(m, r + 1, c - 1), fall(m, r + 1, c + 1)));
return m[r][c] + below;
}
}2Optimal (DP row by row, answer = min of the last row)
O(n²)O(n)dp[c] = cheapest falling path ending at column c of the current row. Each new row takes the best of the three cells above, then the answer is the minimum over the last row.
- dp = first row.
- next[c] = matrix[r][c] + min(dp[c - 1], dp[c], dp[c + 1]) within bounds.
- Return min(dp).
class Solution {
public int minFallingPathSum(int[][] matrix) {
int n = matrix.length;
int[] dp = matrix[0].clone();
for (int r = 1; r < n; r++) {
int[] next = new int[n];
for (int c = 0; c < n; c++) {
int best = dp[c];
if (c > 0) best = Math.min(best, dp[c - 1]);
if (c < n - 1) best = Math.min(best, dp[c + 1]);
next[c] = matrix[r][c] + best;
}
dp = next;
}
int ans = Integer.MAX_VALUE;
for (int x : dp) ans = Math.min(ans, x);
return ans;
}
}Edge cases to test
- Edge columns have only two moves
- n = 1
Hints
Hint 1
Unlike Minimum Path Sum there is no single end cell: the answer is the minimum over the whole last row.
FAQ
What is the best time complexity for Minimum Falling Path Sum?
Optimal (DP row by row, answer = min of the last row) runs in O(n²) time and O(n) extra space.
Which pattern does Minimum Falling Path Sum use?
It is a dynamic programming problem that uses the 2d dp - max/min of last row pattern. Other problems with the same pattern: Triangle, Geek's Training.
Is there a brute force solution for Minimum Falling Path Sum?
Yes. Recursion from every start takes O(n · 3ⁿ) time and O(n) space. For each starting column in the top row, try all three moves recursively.
Which edge cases should I test for Minimum Falling Path Sum?
Edge columns have only two moves; n = 1.