# 64. Minimum Path Sum
## Problem
{% embed url="" %}
## Intuition
## Time Complexity
## Space Complexity
## Solution
```java
class Solution {
public int minPathSum(int[][] grid) {
// R -> No. of rows, C -> No. of columns
// Time Complexity: O(R*C), need to visit all the cells only once
// Space Complexity: O(1), no extra space is used.
// Input only used to store subproblem results
int R = grid.length; // Number of rows
int C = grid[0].length; // Number of columns
// Iterate over each cell in the grid
for (int row = 0; row < R; row++) {
for (int col = 0; col < C; col++) {
if (row == 0 && col == 0) {
// Skip the top-left cell since it has no incoming paths
continue;
} else if (row == 0) {
// For cells in the first row, only consider the path from the left
grid[0][col] += grid[0][col - 1];
} else if (col == 0) {
// For cells in the first column, only consider the path from above
grid[row][0] += grid[row - 1][0];
} else {
// For other cells, choose the minimum path from above and from the left,
// and add it to the current cell's value
int up = grid[row - 1][col];
int left = grid[row][col - 1];
grid[row][col] += Math.min(up, left);
}
}
}
return grid[R - 1][C - 1]; // Return the minimum path sum at the bottom-right cell
}
}
```
