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

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<bits/stdc++.h>
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<n; i++) {
        for(int j=0; j<n; j++) {
              if(i == j)
                  continue;

            // If the difference between the
              // two elements is n, print them
            if((arr[j] - arr[i]) == diff) {
                cout << "Pair Found: (" << arr[i] << ", " << arr[j] << ")";
                  return;
            }
        }
    }

      cout << "No such pair";
}

int main() {
      // Input array and diff
    int arr[] = { 1, 8, 30, 40, 100 };
    int n = sizeof(arr)/sizeof(arr[0]);
    int diff = -60;

      // Function call
    findPair(arr, n, diff);
    return 0;
}

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 <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 || 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 <stdio.h>

// 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<size && j<size)
    {
        if (i != j && (arr[j] - arr[i] == n || arr[i] - arr[j] == n))
        {
            printf("Pair Found: (%d, %d)", arr[i], arr[j]);
            return 1;
        }
        else if (arr[j]-arr[i] < n)
            j++;
        else
            i++;
    }

    printf("No such pair");
    return 0;
}

// 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 || 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

<?php
// PHP program to find a pair with
// the given difference

// The function assumes that the
// array is sorted
function findPair(&$arr, $size, $n)
{
    // Initialize positions of
    // two elements
    $i = 0;
    $j = 1;

    // Search for a pair
    while ($i < $size && $j < $size)
    {
        if ($i != $j && $arr[$j] -
                        $arr[$i] == $n)
        {
            echo "Pair Found: " . "(" .
                  $arr[$i] . ", " . $arr[$j] . ")";
            return true;
        }
        else if ($arr[$j] - $arr[$i] < $n)
            $j++;
        else
            $i++;
    }

    echo "No such pair";
    return false;
}

// Driver Code
$arr = array(1, 8, 30, 40, 100);
$size = sizeof($arr);
$n = 60;
findPair($arr, $size, $n);

// This code is contributed
// by ChitraNayal
?>

Javascript

<script>

       // JavaScript program for the above approach

       // The function assumes that the array is sorted
       function findPair(arr, size, n) {
           // Initialize positions of two elements
           let i = 0;
           let j = 1;

           // Search for a pair
           while (i < size && j < size) {
               if (i != j && arr[j] - arr[i] == n) {
                   document.write("Pair Found: (" + arr[i] + ", " +
                   arr[j] + ")");
                   return true;
               }
               else if (arr[j] - arr[i] < n)
                   j++;
               else
                   i++;
           }

           document.write("No such pair");
           return false;
       }

       // Driver program to test above function

       let arr = [1, 8, 30, 40, 100];
       let size = arr.length;
       let n = 60;
       findPair(arr, size, n);


   // This code is contributed by Potta Lokesh

   </script>

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 <bits/stdc++.h>
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 <bits/stdc++.h>
using namespace std;

// The function assumes that the array is sorted
bool findPair(int arr[], int size, int n)
{
    unordered_map<int, int> 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<Integer,
            Integer> mpp = new HashMap<Integer,
                                      Integer>();

     // 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<int, int> mpp = new Dictionary<int, int>();

    // 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

<script>
// Javascript program for the above approach

// The function assumes that the array is sorted
function findPair(arr, size, n) {
  let mpp = new Map();

  // Traverse the array
  for (let i = 0; i < size; i++) {

    // Update frequency
    // of arr[i]
    if (mpp.has(arr[i]))
      mpp.set(arr[i], mpp.get(arr[i]) + 1);
    else
      mpp.set(arr[i], 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 (let i = 0; i < size; i++) {
    if (mpp.has(n + arr[i])) {
      document.write("Pair Found: (" + arr[i] + ", " +
        + (n + arr[i]) + ")");
      return true;
    }
  }
  document.write("No Pair found");
  return false;
}

// Driver Code
let arr = [1, 8, 30, 40, 100];
let size = arr.length;
let n = -60;
findPair(arr, size, n);

// This code is contributed by Saurabh Jaiswal
</script>

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.