# Find a pair with the given difference

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

Recommended Practice

Method 1: 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(n2).

Algorithm:

1.    Start iterating through each element of the array using an outer loop.
2.    For each element, start iterating again through each of the elements of the array except the one picked in outer loop using an inner loop.
3.    If the difference between the current element and any of the elements of it is equal to the given difference, print both elements.
4.    Continue the process until all possible pairs of elements are compared.
5.    If no pair found, print “No such pair”.

Below is the implementation of the approach:

## C++

 // C++ code for the approach   #include using namespace std;   // Function to find if there exists a pair // of elements in the array whose difference is n void findPair(int arr[], int n, int diff) {     // Nested loop to compare all possible       // pairs of elements     for(int i=0; i

## Javascript

 function findPair(arr, diff) {   // Nested loop to compare   // all  possible pairs of elements   for (let i = 0; i < arr.length; i++) {     for (let j = 0; j < arr.length; j++) {       if (i === j) {         continue;       }       // If the difference between the two elements is 'diff'       // print them and return       if (arr[j] - arr[i] === diff) {         console.log(`Pair Found: (\${arr[i]}, \${arr[j]})`);         return;       }     }   }   console.log('No such pair'); }   // Input array and 'diff' const arr = [1, 8, 30, 40, 100]; const diff = -60; // Function call findPair(arr, diff);

Output

Pair Found: (100, 40)

Time Complexity: O(n*n) as two nested for loops are executing both from 1 to n where n is size of input array.
Space Complexity: O(1) as no extra space has been taken.

Method 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).
Method 3: The second step of the Method -2 can be improved to O(n). The first step remains the same. The idea for the second step is to take two index variables i and j, and 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 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 || arr[i] - arr[j] == 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

## C

 // C program to find a pair with the given difference #include   // The function assumes that the array is sorted int findPair(int arr[], int size, int n) {     // Initialize positions of two elements     int i = 0;      int j = 1;       // Search for a pair     while (i

## 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 || arr[i] - arr[j] == 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

## C#

 // C# program to find a pair with the given difference using System;   class GFG {           // The function assumes that the array is sorted     static bool 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)             {                 Console.Write("Pair Found: "                 + "( " + arr[i] + ", " + arr[j] +" )");                                   return true;             }             else if (arr[j] - arr[i] < n)                 j++;             else                 i++;         }           Console.Write("No such pair");                   return false;     }       // Driver program to test above function     public static void Main ()     {         int []arr = {1, 8, 30, 40, 100};         int n = 60;                   findPair(arr, n);     } }   // This code is contributed by Sam007.



## PHP



Output

Pair Found: (100, 40)

Time Complexity: O(n*log(n)) [Sorting is still required as first step], Where n is number of element in given array
Auxiliary Space: O(1)

The above code can be simplified and can be made more understandable by reducing bunch of If-Else checks . Thanks to Nakshatra Chhillar for suggesting this simplification. We will understand simplifications through following code:

## C++

 // C++ program to find a pair with the given difference #include using namespace std; bool findPair(int arr[], int size, int n) {     // Step-1 Sort the array     sort(arr, arr + size);       // Initialize positions of two elements     int l = 0;     int r = 1;       // take absolute value of difference     // this does not affect the pair as A-B=diff is same as     // B-A= -diff     n = abs(n);       // Search for a pair       // These loop running conditions are sufficient     while (l <= r and r < size) {         int diff = arr[r] - arr[l];         if (diff == n             and l != r) // we need distinct elements in pair                         // so l!=r         {             cout << "Pair Found: (" << arr[l] << ", "                  << arr[r] << ")";             return true;         }         else if (diff > n) // try to reduce the diff             l++;         else // Note if l==r then r will be advanced thus no              // pair will be missed             r++;     }     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);     cout << endl;     n = 20;     findPair(arr, size, n);     return 0; }   // This code is contributed by Nakshatra Chhillar

## Java

 // Java program to find a pair with the given difference import java.io.*; import java.util.Arrays;    class GFG { static boolean findPair(int arr[], int size, int n) {     // Step-1 Sort the array     Arrays.sort(arr);       // Initialize positions of two elements     int l = 0;     int r = 1;       // take absolute value of difference     // this does not affect the pair as A-B=diff is same as     // B-A= -diff     n = Math.abs(n);       // Search for a pair       // These loop running conditions are sufficient     while (l <= r && r < size) {         int diff = arr[r] - arr[l];         if (diff == n             && l != r) // we need distinct elements in pair                         // so l!=r         {             System.out.print("Pair Found: (" + arr[l] + ", "                 + arr[r] + ")");             return true;         }         else if (diff > n) // try to reduce the diff             l++;         else // Note if l==r then r will be advanced thus no             // pair will be missed             r++;     }     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 size = arr.length;     int n = -60;     findPair(arr, size, n);     System.out.println();     n = 20;     findPair(arr, size, n); } }   // This code is contributed by Pushpesh Raj

## Python3

 # Python program to find a pair with the given difference def findPair( arr, size, n):     # Step-1 Sort the array     arr.sort();           # Initialize positions of two elements     l = 0;     r = 1;       # take absolute value of difference     # this does not affect the pair as A-B=diff is same as     # B-A= -diff     n = abs(n);       # Search for a pair       # These loop running conditions are sufficient     while (l <= r and r < size) :         diff = arr[r] - arr[l];         if (diff == n and l != r):         # we need distinct elements in pair         # so l!=r             print("Pair Found: (" , arr[l] , ", "                  , arr[r] , ")");             return True;           elif (diff > n):# try to reduce the diff             l += 1;         else :# Note if l==r then r will be advanced thus no              # pair will be missed             r+=1;       print("No such pair");     return False;   # Driver program to test above function arr = [ 1, 8, 30, 40, 100 ]; size = len(arr); n = -60; findPair(arr, size, n); n = 20; findPair(arr, size, n);   # This code is contributed by agrawalpoojaa976.

## C#

 // C# code implementation using System; using System.Collections; public class GFG {     static bool findPair(int[] arr, int size, int n)   {       // Step-1 Sort the array     Array.Sort(arr);       // Initialize positions of two elements     int l = 0;     int r = 1;       // take absolute value of difference     // this does not affect the pair as A-B=diff is same     // as B-A= -diff     n = Math.Abs(n);       // Search for a pair       // These loop running conditions are sufficient     while (l <= r && r < size) {       int diff = arr[r] - arr[l];       if (diff == n           && l != r) // we need distinct elements in         // pair so l!=r       {         Console.Write("Pair Found: (" + arr[l]                       + ", " + arr[r] + ")");         return true;       }       else if (diff > n) // try to reduce the diff         l++;       else // Note if l==r then r will be advanced         // thus no         // pair will be missed         r++;     }     Console.Write("No such pair");     return false;   }     static public void Main()   {       // Code     int[] arr = { 1, 8, 30, 40, 100 };     int size = arr.Length;     int n = -60;     findPair(arr, size, n);     Console.WriteLine();     n = 20;     findPair(arr, size, n);   } }   // This code is contributed by lokesh.

## Javascript

 // JavaScript program to find a pair with the given difference   const findPair = (arr, size, n) => {     // Step-1 Sort the array     arr.sort((a, b) => a - b);       // Initialize positions of two elements     let l = 0;     let r = 1;       // take absolute value of difference     // this does not affect the pair as A-B=diff is same as     // B-A= -diff     n = Math.abs(n);       // Search for a pair       // These loop running conditions are sufficient     while (l <= r && r < size) {         let diff = arr[r] - arr[l];         if (diff === n             && l !== r) // we need distinct elements in pair                         // so l!==r         {             console.log("Pair Found: (" + arr[l] + ", "                 + arr[r] + ")");             return true;         }         else if (diff > n) // try to reduce the diff             l++;         else // Note if l==r then r will be advanced thus no              // pair will be missed             r++;     }     console.log("No such pair");     return false; }   // Driver program to test above function const main = () => {     let arr = [1, 8, 30, 40, 100];     let size = arr.length;     let n = -60;     findPair(arr, size, n);     console.log();     n = 20;     findPair(arr, size, n); }   main();

Output

Pair Found: (40, 100)
No such pair

Time Complexity: O(n*log(n)) [Sorting is still required as first step], Where n is number of element in given array
Auxiliary Space: O(1)

Method 4 :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.

## C++

 // C++ program to find a pair with the given difference #include using namespace std;   // The function assumes that the array is sorted bool findPair(int arr[], int size, int n) {     unordered_map mpp;     for (int i = 0; i < size; i++) {         mpp[arr[i]]++;           // Check if any element whose frequency         // is greater than 1 exist or not for n == 0         if (n == 0 && mpp[arr[i]] > 1)             return true;     }       // Check if difference is zero and     // we are unable to find any duplicate or     // element whose frequency is greater than 1     // then no such pair found.     if (n == 0)         return false;       for (int i = 0; i < size; i++) {         if (mpp.find(n + arr[i]) != mpp.end()) {             cout << "Pair Found: (" << arr[i] << ", "                  << n + arr[i] << ")";             return true;         }     }       cout << "No Pair found";     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 for the above approach import java.io.*; import java.util.*;   class GFG {     // The function assumes that the array is sorted   static boolean findPair(int[] arr, int size, int n)   {     HashMap mpp = new HashMap();        // Traverse the array     for(int i = 0; i < size; i++)     {                    // Update frequency         // of arr[i]         mpp.put(arr[i],                mpp.getOrDefault(arr[i], 0) + 1);                 // Check if any element whose frequency         // is greater than 1 exist or not for n == 0         if (n == 0 && mpp.get(arr[i]) > 1)             return true;     }         // Check if difference is zero and     // we are unable to find any duplicate or     // element whose frequency is greater than 1     // then no such pair found.     if (n == 0)         return false;       for (int i = 0; i < size; i++) {       if (mpp.containsKey(n + arr[i])) {         System.out.print("Pair Found: (" + arr[i] + ", " +                       + (n + arr[i]) + ")");         return true;       }     }     System.out.print("No Pair found");     return false;   }     // Driver Code public static void main(String[] args) {     int[] arr = { 1, 8, 30, 40, 100 };     int size = arr.length;     int n = -60;     findPair(arr, size, n); } }   // This code is contributed by code_hunt.

## Python3

 # Python program to find a pair with the given difference   # The function assumes that the array is sorted def findPair(arr, size, n):       mpp = {}       for i in range(size):         if arr[i] in mpp.keys():              mpp[arr[i]] += 1              if(n == 0 and mpp[arr[i]] > 1):                 return true;         else:              mpp[arr[i]] = 1           if(n == 0):       return false;       for i in range(size):          if n + arr[i] in mpp.keys():             print("Pair Found: (" + str(arr[i]) + ", " + str(n + arr[i]) + ")")             return True           print("No Pair found")     return False   # Driver program to test above function arr = [ 1, 8, 30, 40, 100 ] size = len(arr) n = -60 findPair(arr, size, n)   # This code is contributed by shinjanpatra

## C#

 // C# program for the above approach using System; using System.Collections.Generic;   public class GFG {   // The function assumes that the array is sorted static bool findPair(int[] arr, int size, int n) {     Dictionary mpp = new Dictionary();       // Traverse the array     for(int i = 0; i < size; i++)     {                   // Update frequency         // of arr[i]         mpp[arr[i]]=mpp.GetValueOrDefault(arr[i], 0) + 1;               // Check if any element whose frequency         // is greater than 1 exist or not for n == 0         if (n == 0 && mpp[arr[i]] > 1)             return true;     }       // Check if difference is zero and     // we are unable to find any duplicate or     // element whose frequency is greater than 1     // then no such pair found.     if (n == 0)         return false;       for (int i = 0; i < size; i++) {     if (mpp.ContainsKey(n + arr[i])) {         Console.WriteLine("Pair Found: (" + arr[i] + ", " +                     + (n + arr[i]) + ")");         return true;     }     }     Console.WriteLine("No Pair found");     return false; }   // Driver Code public static void Main(string []args) {     int[] arr = { 1, 8, 30, 40, 100 };     int size = arr.Length;     int n = -60;     findPair(arr, size, n); } }   // This code is contributed by Aman Kumar

## Javascript



Output

Pair Found: (100, 40)

Time Complexity: O(n), Where n is number of element in given array
Auxiliary Space: O(n)

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

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