Minimum Path Sum
LeetCode #64 · Medium · Dynamic Programming · C++ solution with worked-out approach and complexity analysis.
Approach
Bottom-Up DP
Key insight
The minimum sum to reach any cell (i,j) is the cell's value plus the smaller of the minimum sums to reach (i-1,j) or (i,j-1).
Time
O(m * n)
Space
O(m * n)
uses copy of grid as dp table
Problem description(from LeetCode)
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.
Examples
Example 1
- Input:
- grid = [[1,3,1],[1,5,1],[4,2,1]]
- Output:
- 7
- Note:
- The path 1 → 3 → 1 → 1 → 1 minimizes the sum. Input: grid = [[1,2,3],[4,5,6]] Output: 12
Constraints
- •m == grid.length
- •n == grid[i].length
- •1 <= m, n <= 200
- •0 <= grid[i][j] <= 200
C++ Solution
solution.cpp
class Solution {
public:
int minPathSum(vector<vector<int>> &grid) {
int m = grid.size(), n = grid[0].size();
vector<vector<int>> dp = grid;
for (int i = 1; i < m; i++) {
dp[i][0] += dp[i - 1][0];
}
for (int j = 1; j < n; j++) {
dp[0][j] += dp[0][j - 1];
}
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
dp[i][j] += min(dp[i - 1][j], dp[i][j - 1]);
}
}
return dp[m - 1][n - 1];
}
};