Q17:Given an array Arr[] of size L and a number N, you need to write a program to find if there exists a pair of elements in the array whose difference is N.

Example :

Input:
L = 6, N = 78
arr[] = {5, 20, 3, 2, 5, 80}
Output: 1
Explanation: (2, 80) have difference of 78.

Solution:
bool findPair(int arr[], int size, int n){ 
    sort(arr, arr + size);
    int i = 0, j = 1;
    while (i<size && j< size){
        if(arr[j] - arr[i] == n && j != i){
            return true;
        }
        if(arr[j] - arr[i] > n){
            i++;
        }
        else{
            j++;
        }
    }
    return false;   
}

Explantion:
>Sorting: The array arr is sorted in ascending order.
>Pointers: Two pointers, i and j, are initialized to 0 and 1, respectively.
>Comparison Loop:
 >While i and j are within the array bounds:
 >If arr[j] - arr[i] equals n and j is not equal to i, return true.
 >If arr[j] - arr[i] is greater than n, increment i.
 >If arr[j] - arr[i] is less than or equal to n, increment j.

>Termination: If no such pair is found, return false