Print all pairs with given sum

Given an array of integers, and a number ‘sum’, print all pairs in the array whose sum is equal to ‘sum’.

Examples :
Input  :  arr[] = {1, 5, 7, -1, 5}, 
          sum = 6
Output : (1, 5) (7, -1) (1, 5)

Input  :  arr[] = {2, 5, 17, -1}, 
          sum = 7
Output :  (2, 5)

A simple solution is be traverse each element and check if there’s another number in the array which can be added to it to give sum.

C++

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// C++ implementation of simple method to
// find print pairs with given sum.
#include <bits/stdc++.h>
using namespace std;
  
// Returns number of pairs in arr[0..n-1]
// with sum equal to 'sum'
int printPairs(int arr[], int n, int sum)
{
    int count = 0; // Initialize result
  
    // Consider all possible pairs and check
    // their sums
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
            if (arr[i] + arr[j] == sum)
                cout << "(" << arr[i] << ", "
                     << arr[j] << ")" << endl;
}
  
// Driver function to test the above function
int main()
{
    int arr[] = { 1, 5, 7, -1, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int sum = 6;
    printPairs(arr, n, sum);
    return 0;
}

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Java

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// Java implementation of
// simple method to find
// print pairs with given sum.
  
class GFG {
  
    // Returns number of pairs
    // in arr[0..n-1] with sum
    // equal to 'sum'
    static void printPairs(int arr[],
                           int n, int sum)
    {
        // int count = 0;
  
        // Consider all possible pairs
        // and check their sums
        for (int i = 0; i < n; i++)
            for (int j = i + 1; j < n; j++)
                if (arr[i] + arr[j] == sum)
                    System.out.println("(" + arr[i] + ", " + arr[j] + ")");
    }
  
    // Driver Code
    public static void main(String[] arg)
    {
        int arr[] = { 1, 5, 7, -1, 5 };
        int n = arr.length;
        int sum = 6;
        printPairs(arr, n, sum);
    }
}
  
// This code is contributed
// by Smitha

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Python 3

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# Python 3 implementation 
# of simple method to find
# print pairs with given sum.
  
# Returns number of pairs 
# in arr[0..n-1] with sum
# equal to 'sum'
def printPairs(arr, n, sum):
  
    # count = 0 
  
    # Consider all possible 
    # pairs and check their sums
    for i in range(0, n ):
        for j in range(i + 1, n ):
            if (arr[i] + arr[j] == sum):
                print("(", arr[i], 
                      ", ", arr[j], 
                      ")", sep = "")
  
  
# Driver Code
arr = [1, 5, 7, -1, 5]
n = len(arr)
sum = 6
printPairs(arr, n, sum)
  
# This code is contributed 
# by Smitha

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C#

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// C# implementation of simple
// method to find print pairs
// with given sum.
using System;
  
class GFG {
    // Returns number of pairs
    // in arr[0..n-1] with sum
    // equal to 'sum'
    static void printPairs(int[] arr,
                           int n, int sum)
    {
        // int count = 0;
  
        // Consider all possible pairs
        // and check their sums
        for (int i = 0; i < n; i++)
            for (int j = i + 1; j < n; j++)
                if (arr[i] + arr[j] == sum)
                    Console.Write("(" + arr[i] + ", " + arr[j] + ")"
                                  + "\n");
    }
  
    // Driver Code
    public static void Main()
    {
        int[] arr = { 1, 5, 7, -1, 5 };
        int n = arr.Length;
        int sum = 6;
        printPairs(arr, n, sum);
    }
}
  
// This code is contributed
// by Smitha

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PHP

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<?php
// PHP implementation of simple 
// method to find print pairs 
// with given sum.
  
// Returns number of pairs in 
// arr[0..n-1] with sum equal
// to 'sum'
function printPairs($arr, $n, $sum)
{
    // Initialize result
    $count = 0; 
  
    // Consider all possible 
    // pairs and check their sums
    for ($i = 0; $i < $n; $i++)
        for ( $j = $i + 1; $j < $n; $j++)
            if ($arr[$i] + $arr[$j] == $sum)
                echo "(", $arr[$i], ", ",
                           $arr[$j], ")", "\n";
}
  
// Driver Code
$arr = array (1, 5, 7, -1, 5);
$n = sizeof($arr);
$sum = 6;
printPairs($arr, $n, $sum);
  
// This code is contributed by m_kit
?>

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Output :



(1, 5)
(1, 5)
(7, -1)

Method 2 (Use hashing).
We create an empty hash table. Now we traverse through the array and check for pairs in hash table. If a matching element is found, we print the pair number of times equal to number of occurrences of the matching element.

Note that the worst case of time complexity of this solution is O(c + n) where c is count of pairs with given sum.

C++

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// C++ implementation of simple method to
// find count of pairs with given sum.
#include <bits/stdc++.h>
using namespace std;
  
// Returns number of pairs in arr[0..n-1]
// with sum equal to 'sum'
void printPairs(int arr[], int n, int sum)
{
    // Store counts of all elements in map m
    unordered_map<int, int> m;
  
    // Traverse through all elements
    for (int i = 0; i < n; i++) {
  
        // Search if a pair can be formed with
        // arr[i].
        int rem = sum - arr[i];
        if (m.find(rem) != m.end()) {
            int count = m[rem];
            for (int j = 0; j < count; j++)
                cout << "(" << rem << ", "
                     << arr[i] << ")" << endl;
        }
        m[arr[i]]++;
    }
}
  
// Driver function to test the above function
int main()
{
    int arr[] = { 1, 5, 7, -1, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int sum = 6;
    printPairs(arr, n, sum);
    return 0;
}

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Output :

(1, 5)
(7, -1)
(1, 5)

Method 3.
Another method to Print all pairs with the given sum is given as follows:

Java

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import java.util.Arrays;
  
/**
 * Created by sampat.
 */
public class SumOfPairs {
  
    public void pairedElements(int arr[], int sum)
    {
        int low = 0;
        int high = arr.length - 1;
  
        while (low < high) {
            if (arr[low] + arr[high] == sum) {
                System.out.println("The pair is : ("
                                   + arr[low] + ", " + arr[high] + ")");
            }
            if (arr[low] + arr[high] > sum) {
                high--;
            }
            else {
                low++;
            }
        }
    }
  
    public static void main(String[] args)
    {
        int arr[] = { 2, 3, 4, -2, 6, 8, 9, 11 };
        Arrays.sort(arr);
  
        SumOfPairs sp = new SumOfPairs();
        sp.pairedElements(arr, 6);
    }
}

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C#

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// C# program to find triplets in a given
// array whose sum is equal to given sum.
using System;
  
public class SumOfPairs 
{
  
    public void pairedElements(int []arr, int sum)
    {
        int low = 0;
        int high = arr.Length - 1;
  
        while (low < high) 
        {
            if (arr[low] + arr[high] == sum)
            {
                Console.WriteLine("The pair is : ("
                                + arr[low] + ", " + arr[high] + ")");
            }
            if (arr[low] + arr[high] > sum)
            {
                high--;
            }
            else 
            {
                low++;
            }
        }
    }
  
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = { 2, 3, 4, -2, 6, 8, 9, 11 };
        Array.Sort(arr);
  
        SumOfPairs sp = new SumOfPairs();
        sp.pairedElements(arr, 6);
    }
}
  
// This code is contributed by Princi Singh

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Output :

The pair is : (-2, 8)
The pair is : (2, 4)


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