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.

Convert the previous solution to an **iterative** one. Your algorithm should run in **O(n * m)** time and use **O(n * m)** extra space.

The core challenge of this problem is to find the minimum path sum from the top-left corner to the bottom-right corner of a grid, where you can only move right or down. This problem is significant in various applications such as robotics, game development, and pathfinding algorithms.

Potential pitfalls include misunderstanding the movement constraints (only right or down) and not properly handling the grid boundaries.

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

where `dp[i][j]`

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

.

1. **Naive Solution**: A naive approach would involve exploring all possible paths using recursion, which is highly inefficient with exponential time complexity.

2. **Optimized Solution**: Using dynamic programming, we can iteratively build up the solution by considering the minimum path sum to reach each cell from its top or left neighbor.

1. Initialize a 2D array `dp`

with the same dimensions as the input grid.

2. Set `dp[0][0]`

to `grid[0][0]`

since that's the starting point.

3. Fill in the first row and first column of `dp`

since there's only one way to reach those cells (either from the left or from above).

4. For each remaining cell, set `dp[i][j]`

to the minimum of `dp[i-1][j]`

and `dp[i][j-1]`

plus `grid[i][j]`

.

5. 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 2D dp array
int[][] dp = new int[m][n];
// Initialize the starting point
dp[0][0] = grid[0][0];
// Fill the first row
for (int j = 1; j < n; j++) {
dp[0][j] = dp[0][j - 1] + grid[0][j];
}
// 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 corner contains the minimum path sum
return dp[m - 1][n - 1];
}
}
```

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

.

1. **Single Cell Grid**: The grid has only one cell. The output should be the value of that cell.

2. **Single Row or Column**: The grid has only one row or one column. The path is straightforward, either all right or all down.

3. **Large Values**: Ensure the algorithm handles large values without overflow.

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

- Simple 3x3 grid as given in the example.
- Single cell grid:
`[[5]]`

. - Single row grid:
`[[1, 2, 3]]`

. - Single column grid:
`[[1], [2], [3]]`

. - Large grid with large values to test performance and overflow handling.

1. **Break Down the Problem**: Understand the constraints and break down the problem into smaller subproblems.

2. **Dynamic Programming**: Use dynamic programming to store intermediate results and avoid redundant calculations.

3. **Practice**: Solve similar problems to get a better grasp of dynamic programming techniques.

Understanding and solving the Minimum Path Sum problem using dynamic programming is crucial for developing efficient pathfinding algorithms. Practice and familiarity with dynamic programming will help in solving similar problems effectively.