C++ Program to Find a pair with the given difference - GeeksforGeeks

21 Dec, 2021
Given an unsorted array and a number n, find if there exists a pair of elements in the array whose difference is n. Examples:  
Input: arr[] = {5, 20, 3, 2, 50, 80}, n = 78
Output: Pair Found: (2, 80)

Input: arr[] = {90, 70, 20, 80, 50}, n = 45
Output: No Such Pair
 
The simplest method is to run two loops, the outer loop picks the first element (smaller element) and the inner loop looks for the element picked by outer loop plus n. Time complexity of this method is O(n^2).We can use sorting and Binary Search to improve time complexity to O(nLogn). The first step is to sort the array in ascending order. Once the array is sorted, traverse the array from left to right, and for each element arr[i], binary search for arr[i] + n in arr[i+1..n-1]. If the element is found, return the pair. Both first and second steps take O(nLogn). So overall complexity is O(nLogn). The second step of the above algorithm can be improved to O(n). The first step remain same. The idea for second step is take two index variables i and j, initialize them as 0 and 1 respectively. Now run a linear loop. If arr[j] – arr[i] is smaller than n, we need to look for greater arr[j], so increment j. If arr[j] – arr[i] is greater than n, we need to look for greater arr[i], so increment i. Thanks to Aashish Barnwal for suggesting this approach. The following code is only for the second step of the algorithm, it assumes that the array is already sorted.  
C++
// C++ program to find a pair with the given difference #include <bits/stdc++.h>using namespace std;  // The function assumes that the array is sorted bool findPair(int arr[], int size, int n) {     // Initialize positions of two elements     int i = 0;     int j = 1;       // Search for a pair     while (i < size && j < size)     {         if (i != j && arr[j] - arr[i] == n)         {             cout << "Pair Found: (" << arr[i] <<                        ", " << arr[j] << ")";             return true;         }         else if (arr[j]-arr[i] < n)             j++;         else            i++;     }       cout << "No such pair";     return false; }   // Driver program to test above function int main() {     int arr[] = {1, 8, 30, 40, 100};     int size = sizeof(arr)/sizeof(arr[0]);     int n = 60;     findPair(arr, size, n);     return 0; }   // This is code is contributed by rathbhupendra
Output:  
Pair Found: (40, 100)
Hashing can also be used to solve this problem. Create an empty hash table HT. Traverse the array, use array elements as hash keys and enter them in HT. Traverse the array again look for value n + arr[i] in HT.  
Please refer complete article on Find a pair with the given difference for more details!
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