Count pairs with given sum

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

Examples:

Input:  arr[] = {1, 5, 7, -1}, sum = 6
Output:  2
Explanation: Pairs with sum 6 are (1, 5) and (7, -1)

Input:  arr[] = {1, 5, 7, -1, 5}, sum = 6
Output:  3
Explanation: Pairs with sum 6 are (1, 5), (7, -1) & (1, 5)         

Input:  arr[] = {1, 1, 1, 1}, sum = 2
Output:  6
Explanation: There are 3! pairs with sum 2.

Input:  arr[] = {10, 12, 10, 15, -1, 7, 6, 5, 4, 2, 1, 1, 1}, sum = 11
Output:  9

Expected time complexity O(n)

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

C++

// 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'
int getPairsCount(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)
                count++;

    return count;
}

// 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;
    cout << "Count of pairs is "
         << getPairsCount(arr, n, sum);
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

C

// C implementation of simple method to find count of
// pairs with given sum.
#include <stdio.h>

// Returns number of pairs in arr[0..n-1] with sum equal
// to 'sum'
int getPairsCount(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)
                count++;

    return count;
}

// 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;
    printf("Count of pairs is %d",getPairsCount(arr, n, sum));
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

Java

// Java implementation of simple method to find count of
// pairs with given sum.
public class find {
    public static void main(String args[])
    {
        int[] arr = { 1, 5, 7, -1, 5 };
        int sum = 6;
        getPairsCount(arr, sum);
    }

    // Prints number of pairs in arr[0..n-1] with sum equal
    // to 'sum'
    public static void getPairsCount(int[] arr, int sum)
    {

        int count = 0; // Initialize result

        // Consider all possible pairs and check their sums
        for (int i = 0; i < arr.length; i++)
            for (int j = i + 1; j < arr.length; j++)
                if ((arr[i] + arr[j]) == sum)
                    count++;

        System.out.printf("Count of pairs is %d", count);
    }
}

// This code is contributed by Aditya Kumar (adityakumar129)

Python3

# Python3 implementation of simple method
# to find count of pairs with given sum.

# Returns number of pairs in arr[0..n-1]
# with sum equal to 'sum'


def getPairsCount(arr, n, sum):

    count = 0  # Initialize result

    # 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:
                count += 1

    return count


# Driver function
arr = [1, 5, 7, -1, 5]
n = len(arr)
sum = 6
print("Count of pairs is",
      getPairsCount(arr, n, sum))

# This code is contributed by Smitha Dinesh Semwal

C#

// C# implementation of simple
// method to find count of
// pairs with given sum.
using System;

class GFG {
    public static void getPairsCount(int[] arr, int sum)
    {

        int count = 0; // Initialize result

        // Consider all possible pairs
        // and check their sums
        for (int i = 0; i < arr.Length; i++)
            for (int j = i + 1; j < arr.Length; j++)
                if ((arr[i] + arr[j]) == sum)
                    count++;

        Console.WriteLine("Count of pairs is " + count);
    }

    // Driver Code
    static public void Main()
    {
        int[] arr = { 1, 5, 7, -1, 5 };
        int sum = 6;
        getPairsCount(arr, sum);
    }
}

// This code is contributed
// by Sach_Code

PHP

<?php
// PHP implementation of simple
// method to find count of
// pairs with given sum.

// Returns number of pairs in
// arr[0..n-1] with sum equal
// to 'sum'
function getPairsCount($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)
                $count++;

    return $count;
}

    // Driver Code
    $arr = array(1, 5, 7, -1, 5) ;
    $n = sizeof($arr);
    $sum = 6;
    echo "Count of pairs is "
         , getPairsCount($arr, $n, $sum);

// This code is contributed by nitin mittal.
?>

Javascript

<script>

// Javascript implementation of simple method to find count of
// pairs with given sum.

// Returns number of pairs in arr[0..n-1] with sum equal
// to 'sum'
function getPairsCount(arr, n, sum)
{
    let count = 0; // Initialize result

    // Consider all possible pairs and check their sums
    for (let i = 0; i < n; i++)
        for (let j = i + 1; j < n; j++)
            if (arr[i] + arr[j] == sum)
                count++;

    return count;
}

// Driver function to test the above function
    let arr = [ 1, 5, 7, -1, 5 ];
    let n = arr.length;
    let sum = 6;
    document.write("Count of pairs is "
        + getPairsCount(arr, n, sum));

// This code is contributed by Mayank Tyagi

</script>

Output

Count of pairs is 3

Time Complexity: O(n²) 
Auxiliary Space: O(1)

Efficient Approach: This approach is based on the following idea:

If the array is sorted then for each array element arr[i] we can find the number of pairs by finding all the values (sum – arr[i]) which are situated after i^th index.

Follow the steps below to implement this approach:

• Sort the array arr[] in increasing order.
• Loop from i = 0 to N-1.
  • Find the index of the first element having value same or just greater than (sum – arr[i]) using lower bound.
  • Find the index of the first element having value just greater than (sum – arr[i]) using upper bound.
  • The gap between these two indices is the number of elements with value same as (sum – arr[i]).
  • Add this with the final count of pairs.
• Return the final count after the iteration is over.

Below is the implementation of the above approach.

C++

// C++ code to implement the approach

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

// Function to find the count of pairs
int getPairsCount(int arr[], int n, int k)
{
    sort(arr, arr + n);
    int x = 0, c = 0, y, z;
    for (int i = 0; i < n - 1; i++) {
        x = k - arr[i];

        // Lower bound from i+1
        int y = lower_bound(arr + i + 1,
                            arr + n, x) - arr;

        // Upper bound from i+1
        int z = upper_bound(arr + i + 1,
                            arr + n, x) - arr;
        c = c + z - y;
    }
    return c;
}

// Driver code
int main()
{
    int arr[] = { 1, 5, 7, -1, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 6;

    // Function call
    cout << "Count of pairs is "
         << getPairsCount(arr, n, k);
    return 0;
}

Output

Count of pairs is 3

Time Complexity: O(n * log(n) )
Auxiliary Space: O(1)

This approach is contributed by tene2902

Most Efficient Approach: 
A better solution is possible in O(n) time. Below is the Algorithm – 

1. Create a map to store frequency of each number in the array. (Single traversal is required)
2. In the next traversal, for every element check if it can be combined with any other element (other than itself!) to give the desired sum. Increment the counter accordingly.
3. After completion of second traversal, we’d have twice the required value stored in counter because every pair is counted two times. Hence divide count by 2 and return.

Below is the implementation of above idea : 

C++

// 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'
int getPairsCount(int arr[], int n, int sum)
{
    unordered_map<int, int> m;

    // Store counts of all elements in map m
    for (int i = 0; i < n; i++)
        m[arr[i]]++;

    int twice_count = 0;

    // iterate through each element and increment the
    // count (Notice that every pair is counted twice)
    for (int i = 0; i < n; i++) {
        twice_count += m[sum - arr[i]];

        // if (arr[i], arr[i]) pair satisfies the condition,
        // then we need to ensure that the count is
        // decreased by one such that the (arr[i], arr[i])
        // pair is not considered
        if (sum - arr[i] == arr[i])
            twice_count--;
    }

    // return the half of twice_count
    return twice_count / 2;
}

// 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;
    cout << "Count of pairs is "
         << getPairsCount(arr, n, sum);
    return 0;
}

Java

/* Java implementation of simple method to find count of
pairs with given sum*/

import java.util.HashMap;

class Test {
    static int arr[] = new int[] { 1, 5, 7, -1, 5 };

    // Returns number of pairs in arr[0..n-1] with sum equal
    // to 'sum'
    static int getPairsCount(int n, int sum)
    {
        HashMap<Integer, Integer> hm = new HashMap<>();

        // Store counts of all elements in map hm
        for (int i = 0; i < n; i++) {

            // initializing value to 0, if key not found
            if (!hm.containsKey(arr[i]))
                hm.put(arr[i], 0);

            hm.put(arr[i], hm.get(arr[i]) + 1);
        }
        int twice_count = 0;

        // iterate through each element and increment the
        // count (Notice that every pair is counted twice)
        for (int i = 0; i < n; i++) {
            if (hm.get(sum - arr[i]) != null)
                twice_count += hm.get(sum - arr[i]);

            // if (arr[i], arr[i]) pair satisfies the
            // condition, then we need to ensure that the
            // count is decreased by one such that the
            // (arr[i], arr[i]) pair is not considered
            if (sum - arr[i] == arr[i])
                twice_count--;
        }

        // return the half of twice_count
        return twice_count / 2;
    }

    // Driver method to test the above function
    public static void main(String[] args)
    {

        int sum = 6;
        System.out.println(
            "Count of pairs is "
            + getPairsCount(arr.length, sum));
    }
}
// This code is contributed by Gaurav Miglani

Python3

# Python 3 implementation of simple method
# to find count of pairs with given sum.
import sys

# Returns number of pairs in arr[0..n-1]
# with sum equal to 'sum'


def getPairsCount(arr, n, sum):

    m = [0] * 1000

    # Store counts of all elements in map m
    for i in range(0, n):
        m[arr[i]] += 1

    twice_count = 0

    # Iterate through each element and increment
    # the count (Notice that every pair is
    # counted twice)
    for i in range(0, n):

        twice_count += m[sum - arr[i]]

        # if (arr[i], arr[i]) pair satisfies the
        # condition, then we need to ensure that
        # the count is  decreased by one such
        # that the (arr[i], arr[i]) pair is not
        # considered
        if (sum - arr[i] == arr[i]):
            twice_count -= 1

    # return the half of twice_count
    return int(twice_count / 2)


# Driver function
arr = [1, 5, 7, -1, 5]
n = len(arr)
sum = 6

print("Count of pairs is", getPairsCount(arr,
                                         n, sum))

# This code is contributed by
# Smitha Dinesh Semwal

C#

// C# implementation of simple method to
// find count of pairs with given sum
using System;
using System.Collections.Generic;

class GFG {
    public static int[] arr = new int[] { 1, 5, 7, -1, 5 };

    // Returns number of pairs in arr[0..n-1]
    // with sum equal to 'sum'
    public static int getPairsCount(int n, int sum)
    {
        Dictionary<int, int> hm
            = new Dictionary<int, int>();

        // Store counts of all elements
        // in map hm
        for (int i = 0; i < n; i++) {

            // initializing value to 0,
            // if key not found
            if (!hm.ContainsKey(arr[i])) {
                hm[arr[i]] = 0;
            }

            hm[arr[i]] = hm[arr[i]] + 1;
        }
        int twice_count = 0;

        // iterate through each element and
        // increment the count (Notice that
        // every pair is counted twice)
        for (int i = 0; i < n; i++) {
            if (hm[sum - arr[i]] != 0) {
                twice_count += hm[sum - arr[i]];
            }

            // if (arr[i], arr[i]) pair satisfies
            // the condition, then we need to ensure
            // that the count is decreased by one
            // such that the (arr[i], arr[i])
            // pair is not considered
            if (sum - arr[i] == arr[i]) {
                twice_count--;
            }
        }

        // return the half of twice_count
        return twice_count / 2;
    }

    // Driver Code
    public static void Main(string[] args)
    {
        int sum = 6;
        Console.WriteLine("Count of pairs is "
                          + getPairsCount(arr.Length, sum));
    }
}

// This code is contributed by Shrikant13

Javascript

<script>
/* javascript implementation of simple method to find count of
pairs with given sum*/
    var arr = [ 1, 5, 7, -1, 5 ];

    // Returns number of pairs in arr[0..n-1] with sum equal
    // to 'sum'
    function getPairsCount(n , sum) {
        var hm = new Map();

        // Store counts of all elements in map hm
        for (var i = 0; i < n; i++) {

            // initializing value to 0, if key not found
            if (!hm.has(arr[i]))
                hm.set(arr[i], 0);

            hm.set(arr[i], hm.get(arr[i]) + 1);
        }
        var twice_count = 0;

        // iterate through each element and increment the
        // count (Notice that every pair is counted twice)
        for (i = 0; i < n; i++) {
            if (hm.get(sum - arr[i]) != null)
                twice_count += hm.get(sum - arr[i]);

            // if (arr[i], arr[i]) pair satisfies the
            // condition, then we need to ensure that the
            // count is decreased by one such that the
            // (arr[i], arr[i]) pair is not considered
            if (sum - arr[i] == arr[i])
                twice_count--;
        }

        // return the half of twice_count
        return twice_count / 2;
    }

    // Driver method to test the above function
        var sum = 6;
        document.write("Count of pairs is " + getPairsCount(arr.length, sum));

// This code is contributed by umadevi9616
</script>

Output

Count of pairs is 3

Time Complexity: O(n), to iterate over the array
Auxiliary Space: O(n), to make a map of size n

This article is contributed by Ashutosh Kumar. 

More efficient solution in one loop:-

C++

// 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'
int getPairsCount(int arr[], int n, int k)
{
    unordered_map<int, int> m;
    int count = 0;
    for (int i = 0; i < n; i++) {
        if (m.find(k - arr[i]) != m.end()) {
            count += m[k - arr[i]];
        }
        m[arr[i]]++;
    }
    return count;
}

// 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;
    cout << "Count of pairs is "
         << getPairsCount(arr, n, sum);
    return 0;
}

Java

// Java implementation of simple method to find count of
// pairs with given sum.
import java.util.*;

class GFG{

// Returns number of pairs in arr[0..n-1] with sum equal
// to 'sum'
static int getPairsCount(int arr[], int n, int k)
{
    HashMap<Integer,Integer> m = new HashMap<>();
    int count = 0;
    for (int i = 0; i < n; i++) {
        if (m.containsKey(k - arr[i])) {
            count += m.get(k - arr[i]);
        }
        if(m.containsKey(arr[i])){
            m.put(arr[i], m.get(arr[i])+1);
        }
        else{
            m.put(arr[i], 1);
        }
    }
    return count;
}

// Driver function to test the above function
public static void main(String[] args)
{
    int arr[] = { 1, 5, 7, -1, 5};
    int n = arr.length;
    int sum = 6;
    System.out.print("Count of pairs is "
         + getPairsCount(arr, n, sum));
}
}

// This code is contributed by umadevi9616

Python3

# Python implementation of simple method to find count of
# pairs with given sum.

# Returns number of pairs in arr[0..n-1] with sum equal to 'sum'
def getPairsCount(arr, n, sum):
  unordered_map = {}
  count = 0
  for i in range(n):
    if sum - arr[i] in unordered_map:
      count += unordered_map[sum - arr[i]]
    if arr[i] in unordered_map:
      unordered_map[arr[i]] += 1
    else:
      unordered_map[arr[i]] = 1
  return count

# Driver code
arr = [1, 5, 7, -1, 5]
n = len(arr)
sum = 6
print('Count of pairs is', getPairsCount(arr, n, sum))

# This code is contributed by Manish Thapa

C#

// C# implementation of simple method to find count of
// pairs with given sum.
using System;
using System.Collections.Generic;

public class GFG {

    // Returns number of pairs in arr[0..n-1] with sum equal
    // to 'sum'
    static int getPairsCount(int []arr, int n, int k) {
        Dictionary<int, int> m = new Dictionary<int, int>();
        int count = 0;
        for (int i = 0; i < n; i++) {
            if (m.ContainsKey(k - arr[i])) {
                count += m[k - arr[i]];
            }
            if (m.ContainsKey(arr[i])) {
                m[arr[i]] = m[arr[i]] + 1;
            } else {
                m.Add(arr[i], 1);
            }
        }
        return count;
    }

    // Driver function to test the above function
    public static void Main(String[] args) {
        int []arr = { 1, 5, 7, -1, 5 };
        int n = arr.Length;
        int sum = 6;
        Console.Write("Count of pairs is " + getPairsCount(arr, n, sum));
    }
}

// This code is contributed by umadevi9616

Javascript

<script>
// javascript implementation of simple method to find count of
// pairs with given sum.

    // Returns number of pairs in arr[0..n-1] with sum equal
    // to 'sum'
    function getPairsCount(arr , n , k) {
        var m = new Map();
        var count = 0;
        for (var i = 0; i < n; i++) {
            if (m.has(k - arr[i])) {
                count += m.get(k - arr[i]);
            }
            if (m.has(arr[i])) {
                m.set(arr[i], m.get(arr[i]) + 1);
            } else {
                m.set(arr[i], 1);
            }
        }
        return count;
    }

    // Driver function to test the above function
        var arr = [ 1, 5, 7, -1, 5 ];
        var n = arr.length;
        var sum = 6;
        document.write("Count of pairs is " + getPairsCount(arr, n, sum));

// This code is contributed by umadevi9616
</script>

Output

Count of pairs is 3

Time Complexity: O(n), to iterate over the array
Auxiliary Space: O(n), to make a map of size n

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