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++

// C/C++ program to find a pair with the given difference #include <stdio.h> // 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) { printf("Pair Found: (%d, %d)", arr[i], arr[j]); return true; } else if (arr[j]-arr[i] < n) j++; else i++; } printf("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; }

## Java

// Java program to find a pair with the given difference import java.io.*; class PairDifference { // The function assumes that the array is sorted static boolean findPair(int arr[],int n) { int size = arr.length; // Initialize positions of two elements int i = 0, j = 1; // Search for a pair while (i < size && j < size) { if (i != j && arr[j]-arr[i] == n) { System.out.print("Pair Found: "+ "( "+arr[i]+", "+ arr[j]+" )"); return true; } else if (arr[j] - arr[i] < n) j++; else i++; } System.out.print("No such pair"); return false; } // Driver program to test above function public static void main (String[] args) { int arr[] = {1, 8, 30, 40, 100}; int n = 60; findPair(arr,n); } } /*This code is contributed by Devesh Agrawal*/

## Python

# Python program to find a pair with the given difference # The function assumes that the array is sorted def findPair(arr,n): size = len(arr) # Initialize positions of two elements i,j = 0,1 # Search for a pair while i < size and j < size: if i != j and arr[j]-arr[i] == n: print "Pair found (",arr[i],",",arr[j],")" return True elif arr[j] - arr[i] < n: j+=1 else: i+=1 print "No pair found" return False # Driver function to test above function arr = [1, 8, 30, 40, 100] n = 60 findPair(arr, n) # This code is contributed by Devesh Agrawal

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 write comments if you find any of the above codes/algorithms incorrect, or find other ways to solve the same problem.