# Find a pair with the given difference

• Difficulty Level : Easy
• Last Updated : 08 Feb, 2023

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

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).

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 sortedbool 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 functionint 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 sortedint 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 differenceimport 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 sorteddef 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 functionarr = [1, 8, 30, 40, 100]n = 60findPair(arr, n) # This code is contributed by Devesh Agrawal

## C#

 // C# program to find a pair with the given differenceusing 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.



## Javascript



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 functionint 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 differenceimport 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 functionpublic 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 differencedef 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 functionarr = [ 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 implementationusing 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 functionconst 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 sortedbool 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 functionint 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 approachimport 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 Codepublic 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 sorteddef 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 functionarr = [ 1, 8, 30, 40, 100 ]size = len(arr)n = -60findPair(arr, size, n) # This code is contributed by shinjanpatra

## C#

 // C# program for the above approachusing System;using System.Collections.Generic; public class GFG{ // The function assumes that the array is sortedstatic 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 Codepublic 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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