# Find a specific pair in matrix in C

## Find a specific pair in Matrix in C

Here, in this page we will discuss the program to find a specific pair in matrix in C Programming language. We are given with 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 :Using Linear Search
• Method 2 : Using Binary Search

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 and 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 C

Run
```#include <stdio.h>
#include <limits.h>
#define N 5

int findMaxValue(int mat[][N])
{
int maxValue = INT_MIN;

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;
}

int main()
{
int mat[N][N] = {
{ 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 }
};
printf("Maximum Value is %d", findMaxValue(mat));

return 0;
}```

### Output :

`Maximum Value is 18`

## Method 2:

• Check to see if the middle element is Fixed Point.
• If it is, return it; otherwise, if the middle + 1 element’s index is less than or equal to the value at the high index, Fixed Point(s) may be on the right side of the middle point (obviously only if there is a Fixed Point).
• If the middle – 1 element’s index is larger than or equal to the value at the low index, Fixed Point(s) may be on the left side of the middle point.

### Method 2 : Code in C

Run
```#include <stdio.h>
#include <limits.h>
#define N 5

int max(int a, int b){
if(a>b)
return a;
return b;
}

int findMaxValue(int mat[][N])
{
int maxValue = INT_MIN;

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

return maxValue;
}

int main()
{
int mat[N][N] = {
{ 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 }
};
printf("Maximum Value is %d", findMaxValue(mat));

return 0;
}
```

### Output :

`Maximum Value is 18`