Count pairs with given sum

Given an array of N integers, and an integer K, the task is to find the number of pairs of integers in the array whose sum is equal to K.

Examples:

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

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

Input: arr[] = {1, 1, 1, 1}, K = 2
Output:  6
Explanation: Pairs with sum 2 are (1, 1), (1, 1), (1, 1), (1, 1), (1, 1).

Input: arr[] = {10, 12, 10, 15, -1, 7, 6, 5, 4, 2, 1, 1, 1}, K = 11
Output:  9
Explanation: Pairs with sum 11 are (10, 1), (10, 1), (10, 1), (12, -1), (10, 1), (10, 1), (10, 1), (7, 4), (6, 5).

Naïve Approach: 

A simple solution is to user nested loops, one for traversal of each element and other for checking if there’s another number in the array which can be added to it to give K.

Illustration:

Consider an array arr [ ] = {1 ,5 ,7 ,1} and K = 6

• initialize count = 0
• First iteration
  • i = 0
  • j = 1
    • arr[i]+arr[j] == K
    • count = 1
  • i = 0
  • j = 2
    • arr[i] + arr[j] != K
    • count = 1
  • i = 0
  • j = 3
    • arr[i] + arr[j] != K
    • count = 1
• Second iteration
  • i = 1
  • j = 2
    • arr[i] + arr[j] != K
    • count = 1
  • i = 1
  • j = 3
    • arr[i] + arr[j] == K
    • count = 2
• Third iteration
  • i = 2
  • j = 3
    • arr[i] + arr[j] != K
    • count = 2
• Hence Output is 2 .

Follow the steps below to solve the given problem:

Follow the steps below for implementation:

• Initialize the count variable with 0 which stores the result.
• Use two nested loop and check if arr[i] + arr[j] == K, then increment the count variable.
• Return the count.

Below is the implementation of the above approach.

C++

// C++ implementation of simple method to find count of
// pairs with given sum K.
#include <bits/stdc++.h>
using namespace std;

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

    return count;
}

// Driver function to test the above function
int main()
{
    int arr[] = { 1, 5, 7, -1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int K = 6;
    cout << "Count of pairs is "
         // Function Call
         << getPairsCount(arr, n, K);
    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 K)
{
   // Initialize result
    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] == K)
                count++;

    return count;
}

// Driver function to test the above function
int main()
{
    int arr[] = { 1, 5, 7, -1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int K = 6;
    printf("Count of pairs is %d",
           // Function Call
           getPairsCount(arr, n, K));
    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 K = 6;
      // Function Call
        getPairsCount(arr, K);
    }

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

        // 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]) == K)
                    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, K):

    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] == K:
                count += 1

    return count


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

# 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

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>

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.
?>

Output

Count of pairs is 2

Time Complexity: O(n²), traversing the array for each element
Auxiliary Space: O(1)

Count pairs with given sum using Binary Search.

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

Illustration:

Given arr[ ] = {1, 5, 7, -1} and K = 6

• sort the array = {-1,1,5,7}
• initialize count = 0
• At index = 0:
  • val = K – arr[0] = 6 – (-1) = 7
  • count = count + upperBound(1, 3, 7) – lowerBound(1, 3, 7)
  • count = 1
• At index = 1:
  • val = K – arr[1] = 6 – 1 = 5
  • count = count + upperBound(2, 3, 5) – lowerBound(2, 3, 5)
  • count = 2
• At index = 2:
  • val = K – arr[2] = 6 – 5 = 1
  • count = count + upperBound(3, 3, 1) – lowerBound(3, 3, 1)
  • count = 2
• Number of pairs = 2

Follow the steps below to solve the given problem:

• 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 (K – arr[i]) using lower bound.
  • Find the index of the first element having value just greater than (K – arr[i]) using upper bound.
  • The gap between these two indices is the number of elements with value same as (K – 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 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 6;

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

Java

// Java code for the above approach
import java.io.*;
import java.util.*;

class GFG {

  // lowerBound implementation
  public static int lowerBound(int[] arr, int start,
                               int end, int key)
  {
    while (start < end) {
      int mid = start + (end - start) / 2;
      if (arr[mid] < key) {
        start = mid + 1;
      }
      else {
        end = mid;
      }
    }
    return start;
  }

  // upperBound implementation
  public static int upperBound(int[] arr, int start,
                               int end, int key)
  {
    while (start < end) {
      int mid = start + (end - start) / 2;
      if (arr[mid] <= key) {
        start = mid + 1;
      }
      else {
        end = mid;
      }
    }
    return start;
  }

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

  public static void main(String[] args)
  {
    int[] arr = { 1, 5, 7, -1};
    int n = arr.length;
    int k = 6;

    // Function call
    System.out.println("Count of pairs is "
                       + getPairsCount(arr, n, k));
  }
}

// This code is contributed by lokeshmvs21.

Python3

# Python code to implement the approach
import bisect

# Function to find the count of pairs


def getPairsCount(arr, n, k):
    arr.sort()
    x, c = 0, 0
    for i in range(n-1):
        x = k-arr[i]

        # Lower bound from i+1
        y = bisect.bisect_left(arr, x, i+1, n)

        # Upper bound from i+1
        z = bisect.bisect(arr, x, i+1, n)
        c = c+z-y
    return c


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

# Function call
print("Count of pairs is", getPairsCount(arr, n, k))

# This code is contributed by Pushpesh Raj

C#

// C# code for the above approach

using System;
using System.Collections;

public class GFG {

    // lowerBound implementation
    public static int lowerBound(int[] arr, int start,
                                 int end, int key)
    {
        while (start < end) {
            int mid = start + (end - start) / 2;
            if (arr[mid] < key) {
                start = mid + 1;
            }
            else {
                end = mid;
            }
        }
        return start;
    }

    // upperBound implementation
    public static int upperBound(int[] arr, int start,
                                 int end, int key)
    {
        while (start < end) {
            int mid = start + (end - start) / 2;
            if (arr[mid] <= key) {
                start = mid + 1;
            }
            else {
                end = mid;
            }
        }
        return start;
    }

    // Function to find the count of pairs
    public static int getPairsCount(int[] arr, int n, int k)
    {
        Array.Sort(arr);
        int c = 0;
        for (int i = 0; i < n - 1; i++) {
            int x = k - arr[i];
            int y = lowerBound(arr, i + 1, n, x);
            int z = upperBound(arr, i + 1, n, x);
            c = c + z - y;
        }
        return c;
    }

    static public void Main()
    {

        // Code
        int[] arr = { 1, 5, 7, -1};
        int n = arr.Length;
        int k = 6;

        // Function call
        Console.WriteLine("Count of pairs is "
                          + getPairsCount(arr, n, k));
    }
}

// This code is contributed by lokesh.

Javascript

// JavaScript code for the above approach

// lowerBound implementation
function lowerBound(arr, start, end, key) {
  while (start < end) {
    var mid = start + Math.floor((end - start) / 2);
    if (arr[mid] < key) {
      start = mid + 1;
    } else {
      end = mid;
    }
  }
  return start;
}

// upperBound implementation
function upperBound(arr, start, end, key) {
  while (start < end) {
    var mid = start + Math.floor((end - start) / 2);
    if (arr[mid] <= key) {
      start = mid + 1;
    } else {
      end = mid;
    }
  }
  return start;
}

// Function to find the count of pairs
function getPairsCount(arr, n, k) {
  arr.sort((a, b) => a - b);
  var c = 0;
  for (var i = 0; i < n - 1; i++) {
    var x = k - arr[i];
    var y = lowerBound(arr, i + 1, n, x);
    var z = upperBound(arr, i + 1, n, x);
    c = c + z - y;
  }
  return c;
}

var arr = [1, 5, 7, -1];
var n = arr.length;
var k = 6;

// Function call
console.log("Count of pairs is " + getPairsCount(arr, n, k));

Output

Count of pairs is 2

Time Complexity: O(n * log(n) ), applying binary search on each element
Auxiliary Space: O(1)

Count pairs with given sum using Hashing

• Check the frequency of K – arr[i] in the arr
• This can be achieved using Hashing.

Illustration:

Given arr[] = {1, 5, 7, -1} and K = 6

• initialize count = 0
• At index = 0:
  • freq[K – arr[0]] = freq[6 – 1] = freq[5] = 0
  • count = 0
  • freq[arr[0]] = freq[1] = 1
• At index = 1:
  • freq[K – arr[1]] = freq[6 – 5] = freq[1] = 1
  • count = 1
  • freq[arr[1]] = freq[5] = 1
• At index = 2:
  • freq[K – arr[2]] = freq[6 – 7] = freq[-1] = 0
  • count = 1
  • freq[arr[2]] = freq[7] = 1
• At index = 3:
  • freq[K – arr[3]] = freq[6 – (-1)] = freq[7] = 1
  • count = 2
  • freq[arr[3]] = freq[-1] = 1

• Number of pairs  = 2

Follow the steps below to solve the given problem: 

• Create a map to store the frequency of each number in the array.
• Check if (K – arr[i]) is present in the map, if present then increment the count variable by its frequency.
• After traversal is over, return the count.

Below is the implementation of the 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 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 code
int main()
{
    int arr[] = { 1, 5, 7, -1};
    int n = sizeof(arr) / sizeof(arr[0]);
    int sum = 6;

    // Function Call
    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};
        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]
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 };
        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 ];
        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 2

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