Minimum Path Sum
Minimum Path Sum is a medium dynamic programming problem solved with the 2d dp pattern.
The best approach, optimal (one row), runs in O(m · n) time and O(n) space.
Below are 3 approaches in Java, from recursion up.
Problem
Find a path from the top-left to the bottom-right of a grid of non-negative numbers, moving only right or down, that minimises the sum of the numbers along it.
Examples
Example 1
- Input
grid = [[1,3,1],[1,5,1],[4,2,1]]- Output
7- Why
- 1 → 3 → 1 → 1 → 1.
Example 2
- Input
grid = [[1,2,3],[4,5,6]]- Output
12
Constraints
1 <= m, n <= 200; values 0 to 200.- Move only right or down.
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 | O(2^(m + n)) | O(m + n) |
| Tabulation (2D table) | O(m · n) | O(m · n) |
| Optimal (one row) | O(m · n) | O(n) |
1Recursion
O(2^(m + n))O(m + n)The cheapest path into (i, j) comes from above or from the left.
- cost(0, 0) = grid[0][0]; out of bounds = infinity.
- cost(i, j) = grid[i][j] + min(cost(i - 1, j), cost(i, j - 1)).
class Solution {
public int minPathSum(int[][] grid) {
return cost(grid, grid.length - 1, grid[0].length - 1);
}
private int cost(int[][] g, int i, int j) {
if (i < 0 || j < 0) return Integer.MAX_VALUE;
if (i == 0 && j == 0) return g[0][0];
return g[i][j] + Math.min(cost(g, i - 1, j), cost(g, i, j - 1));
}
}2Tabulation (2D table)
O(m · n)O(m · n)Fill the table row by row. The first row and first column each have only one way in.
- dp[0][0] = grid[0][0]; fill the first row and column with running sums.
- dp[i][j] = grid[i][j] + min(up, left).
class Solution {
public int minPathSum(int[][] grid) {
int m = grid.length, n = grid[0].length;
int[][] dp = new int[m][n];
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++) {
if (i == 0 && j == 0) dp[i][j] = grid[0][0];
else if (i == 0) dp[i][j] = dp[i][j - 1] + grid[i][j];
else if (j == 0) dp[i][j] = dp[i - 1][j] + grid[i][j];
else dp[i][j] = grid[i][j] + Math.min(dp[i - 1][j], dp[i][j - 1]);
}
return dp[m - 1][n - 1];
}
}3Optimal (one row)
O(m · n)O(n)Keep a single row. Before updating, dp[j] still holds the value from the row above, and dp[j - 1] already holds the new value to the left.
- dp[j] = grid[i][j] + min(dp[j] (up), dp[j - 1] (left)).
class Solution {
public int minPathSum(int[][] grid) {
int m = grid.length, n = grid[0].length;
int[] dp = new int[n];
Arrays.fill(dp, Integer.MAX_VALUE);
dp[0] = 0;
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
dp[j] = grid[i][j] + (j == 0 ? dp[0] : Math.min(dp[j], dp[j - 1]));
return dp[n - 1];
}
}Edge cases to test
- Single row or single column
Hints
Hint 1
dp[i][j] = grid[i][j] + min(dp[i - 1][j], dp[i][j - 1]).
FAQ
What is the best time complexity for Minimum Path Sum?
Optimal (one row) runs in O(m · n) time and O(n) extra space.
Which pattern does Minimum Path Sum use?
It is a dynamic programming problem that uses the 2d dp pattern. Other problems with the same pattern: Unique Paths, Unique Paths II.
Is there a brute force solution for Minimum Path Sum?
Yes. Recursion takes O(2^(m + n)) time and O(m + n) space. The cheapest path into (i, j) comes from above or from the left.
Which edge cases should I test for Minimum Path Sum?
Single row or single column.