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.

**Example:**

Input:[ [1,3,1], [1,5,1], [4,2,1] ]Output:7Explanation:Because the path 1→3→1→1→1 minimizes the sum.

The core challenge of this problem is to find the path from the top-left corner to the bottom-right corner of a grid that results in the minimum sum of the numbers along the path. This problem is significant in various applications such as robotics (finding the shortest path in a grid), game development, and pathfinding algorithms.

Potential pitfalls include misunderstanding the movement constraints (only right and down) and not considering all possible paths.

To solve this problem, we can use dynamic programming. The idea is to build a 2D array `dp`

where `dp[i][j]`

represents the minimum path sum to reach cell `(i, j)`

. We can initialize the first cell with the value of the grid's first cell. For each cell, we can only come from the left or from above, so we take the minimum of these two possible paths and add the current cell's value.

A naive solution would involve exploring all possible paths from the top-left to the bottom-right, which is not optimal due to its exponential time complexity.

The optimized solution uses dynamic programming to store the results of subproblems and build up the solution iteratively. This approach has a time complexity of O(m*n) and a space complexity of O(m*n).

- Create a 2D array
`dp`

of the same size as the grid. - Initialize
`dp[0][0]`

with`grid[0][0]`

. - Fill the first row and first column of
`dp`

since they can only be reached from one direction. - For each cell
`(i, j)`

, calculate the minimum path sum by taking the minimum of the cell above and the cell to the left, then add the current cell's value. - The value at
`dp[m-1][n-1]`

will be the minimum path sum.

```
public class MinimumPathSum {
public int minPathSum(int[][] grid) {
int m = grid.length;
int n = grid[0].length;
// Create a dp array to store the minimum path sums
int[][] dp = new int[m][n];
// Initialize the first cell
dp[0][0] = grid[0][0];
// Fill the first row
for (int i = 1; i < n; i++) {
dp[0][i] = dp[0][i - 1] + grid[0][i];
}
// Fill the first column
for (int i = 1; i < m; i++) {
dp[i][0] = dp[i - 1][0] + grid[i][0];
}
// Fill the rest of the dp array
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
}
}
// The bottom-right cell contains the minimum path sum
return dp[m - 1][n - 1];
}
}
```

The time complexity of this approach is O(m*n) because we iterate through each cell of the grid once. The space complexity is also O(m*n) due to the additional 2D array used to store the minimum path sums.

Potential edge cases include:

- Empty grid: The function should handle this gracefully.
- Single row or single column grid: The function should correctly compute the path sum.

For example, for a single row grid `[[1, 2, 3]]`

, the output should be `6`

.

To test the solution comprehensively, consider the following test cases:

- General case with multiple rows and columns.
- Single row grid.
- Single column grid.
- Grid with all zeros.
- Grid with large numbers.

Using a testing framework like JUnit can help automate and manage these tests effectively.

When approaching such problems, it's essential to:

- Understand the constraints and requirements thoroughly.
- Break down the problem into smaller subproblems.
- Consider edge cases and test your solution against them.
- Practice similar problems to improve your problem-solving skills.

In this blog post, we discussed the problem of finding the minimum path sum in a grid. We explored a dynamic programming approach to solve the problem efficiently and provided a detailed explanation of the algorithm and its implementation in Java. Understanding and solving such problems is crucial for developing strong problem-solving skills and preparing for technical interviews.

For further reading and practice, consider the following resources: