# Find a specific pair in Matrix in Java

## Find a specific pair in Matrix in Java

Here, in this page we will discuss the program to find a specific pair in matrix in Java Programming language. Given an n x n matrix mat[n][n] of integers, find the maximum value of mat(c, d) – mat(a, b) over all choices of indexes such that both c > a and d > b.

## Method Discussed :

• Method 1 : Naive Approach
• Method 2 : Efficient Approach

Let’s discuss them one by one in brief,

## Method 1:

• For all values mat(a, b) in the matrix
• Find mat(c, d) that has maximum value such that c > a and d > b.
• Keeps on updating maximum value found so far.
• Finally return the maximum value.

## Time and Space Complexity :

• Time complexity: O(N*N)
• Space complexity: O(1)

### Method 1 : Code in Java

Run
```import java.io.*;
import java.util.*;

class Main
{
static int findMaxValue(int N,int mat[][])
{
int maxValue = Integer.MIN_VALUE;

for (int a = 0; a < N - 1; a++)
for (int b = 0; b < N - 1; b++)
for (int d = a + 1; d < N; d++)
for (int e = b + 1; e < N; e++)
if (maxValue < (mat[d][e] - mat[a][b]))
maxValue = mat[d][e] - mat[a][b];

return maxValue;
}

public static void main (String[] args)
{
int N = 5;

int mat[][] = {
{ 1, 2, -1, -4, -20 },
{ -8, -3, 4, 2, 1 },
{ 3, 8, 6, 1, 3 },
{ -4, -1, 1, 7, -6 },
{ 0, -4, 10, -5, 1 }
};

System.out.print("Maximum Value is " + findMaxValue(N,mat));
}
}```

### Output :

`Maximum Value is 18`

## Method 2:

In this method we pre-process the matrix such that index(i, j) stores max of elements in matrix from (i, j) to (N-1, N-1) and in the process keeps on updating maximum value found so far. Then finally return the maximum value.

### Method 2 : Code in Java

Run
```import java.io.*;
import java.util.*;

class Main
{
static int findMaxValue(int N,int mat[][])
{

int maxValue = Integer.MIN_VALUE;

int maxArr[][] = new int[N][N];

maxArr[N-1][N-1] = mat[N-1][N-1];

int maxv = mat[N-1][N-1];
for (int j = N - 2; j >= 0; j--)
{
if (mat[N-1][j] > maxv)
maxv = mat[N - 1][j];
maxArr[N-1][j] = maxv;
}

maxv = mat[N - 1][N - 1];
for (int i = N - 2; i >= 0; i--)
{
if (mat[i][N - 1] > maxv)
maxv = mat[i][N - 1];
maxArr[i][N - 1] = maxv;
}

for (int i = N-2; i >= 0; i--)
{
for (int j = N-2; j >= 0; j--)
{

if (maxArr[i+1][j+1] - mat[i][j] > maxValue)
maxValue = maxArr[i + 1][j + 1] - mat[i][j];

maxArr[i][j] = Math.max(mat[i][j],Math.max(maxArr[i][j + 1], maxArr[i + 1][j]) );
}
}

return maxValue;
}

// Driver code
public static void main (String[] args)
{
int N = 5;

int mat[][] = {
{ 1, 2, -1, -4, -20 },
{ -8, -3, 4, 2, 1 },
{ 3, 8, 6, 1, 3 },
{ -4, -1, 1, 7, -6 },
{ 0, -4, 10, -5, 1 }
};

System.out.print("Maximum Value is " + findMaxValue(N,mat));
}
}
```

### Output :

`Maximum Value is 18`