> For the complete documentation index, see [llms.txt](https://mayanktyagi3111.gitbook.io/interview-prep/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://mayanktyagi3111.gitbook.io/interview-prep/dynamic-programming/kingdom-war.md).

# Kingdom War

Two kingdoms are on a war right now, kingdom X and kingdom Y. As a war specialist of kingdom X, you scouted kingdom Y area.

A kingdom area is defined as a **N x M grid** with each cell denoting a village.\
Each cell has a value which denotes the strength of each corresponding village.\
The strength can also be **negative**, representing those warriors of your kingdom who were held hostages.

There’s also another thing to be noticed.

* The strength of any village on row larger than one (2<=r<=N) is stronger or equal to the strength of village which is exactly above it.
* The strength of any village on column larger than one (2<=c<=M) is stronger or equal to the strength of vilage which is exactly to its left.\
  (stronger means having higher value as defined above).

So your task is, find the **largest sum of strength** that you can erase by bombing **one sub-matrix** in the grid.

**Input format:**

```
First line consists of 2 integers N and M denoting the number of rows and columns in the grid respectively.
The next N lines, consists of M integers each denoting the strength of each cell.

1 <= N <= 1500
1 <= M <= 1500
-200 <= Cell Strength <= 200
```

Output:

```
The largest sum of strength that you can get by choosing one sub-matrix.
```

**Example:**

```
Input:
3 3
-5 -4 -1
-3 2 4
2 5 8

Output:
19

Explanation:
Bomb the sub-matrix from (2,2) to (3,3): 2 + 4 + 5 + 8 = 19
```

```java
public class Solution {
    public int solve(int[][] A) {
        if (A.length == 0)
            return 0;
        int m = A.length;
        int n = A[0].length;
        int dp[][] = new int[A.length][A[0].length];
        int max = Integer.MIN_VALUE;
        for (int i = m - 1; i >= 0; i--) {
            for (int j = n - 1; j >= 0; j--) {
                if (i == m - 1 && j == n - 1)
                    dp[i][j] = A[i][j];
                else if (i == m - 1)
                    dp[i][j] = A[i][j] + dp[i][j + 1];
                else if (j == n - 1)
                    dp[i][j] = A[i][j] + dp[i + 1][j];
                else
                    dp[i][j] = A[i][j] + dp[i + 1][j] + dp[i][j + 1] - dp[i + 1][j + 1];
                max = Math.max(max, dp[i][j]);
            }
        }
        return max;
    }
}
```


---

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