Question
Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.
Note: You can only move either down or right at any point in time.
Stats
| Frequency | 3 |
| Difficulty | 3 |
| Adjusted Difficulty | 2 |
| Time to use | -------- |
Ratings/Color = 1(white) 2(lime) 3(yellow) 4/5(red)
Analysis
This is a very easy DP problem.
But this is not a boring question because I found a few interesting solutions.
Solution
Code 1 is 2-D DP.
2nd code is from this blog. Instead of 2-D array, it simply uses a “rotational array” (滚动数组), or 1-D array.
3rd code is using no extra space. It works in place.
My code
2D array
public class Solution {
public int minPathSum(int[][] grid) {
if (grid == null || grid.length == 0) {
return 0;
}
int m = grid.length;
int 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[i][j];
} else if (i == 0) {
dp[i][j] = grid[i][j] + dp[i][j - 1];
} else if (j == 0) {
dp[i][j] = grid[i][j] + dp[i - 1][j];
} else {
dp[i][j] = grid[i][j] + Math.min(dp[i - 1][j], dp[i][j - 1]);
}
}
}
return dp[m - 1][n - 1];
}
}
1-D array
public int minPathSum(int[][] grid) {
if (grid.length == 0) return 0;
if (grid[0].length == 0) return 0;
int[] dp = new int[grid[0].length];
for (int i = 0; i < grid.length; i ++)
for (int j = 0; j < grid[0].length; j ++)
if (i == 0 && j == 0) dp[j] = grid[0][0];
else if (i == 0) dp[j] = dp[j-1] + grid[i][j];
else if (j == 0) dp[j] += grid[i][j];
else dp[j] = Math.min(dp[j-1], dp[j]) + grid[i][j];
return dp[dp.length-1];
}
in-place
public int minPathSum(int[][] grid) {
if (grid.length == 0) return 0;
if (grid[0].length == 0) return 0;
for (int i = 0; i < grid.length; i ++)
for (int j = 0; j < grid[0].length; j ++)
if (i == 0 && j == 0) continue;
else if (i == 0) grid[i][j] = grid[i][j] + grid[i][j-1];
else if (j == 0) grid[i][j] = grid[i][j] + grid[i-1][j];
else grid[i][j] = Math.min(grid[i-1][j], grid[i][j-1]) + grid[i][j];
return grid[grid.length-1][grid[0].length-1];
}