Count distinct pairs with given sum
Last Updated :
13 Apr, 2024
Given an array arr[] of size N and an integer K, the task is to find the count of distinct pairs in the array whose sum is equal to K.
Examples:
Input: arr[] = { 5, 6, 5, 7, 7, 8 }, K = 13
Output: 2
Explanation:
Pairs with sum K( = 13) are { (arr[0], arr[5]), (arr[1], arr[3]), (arr[1], arr[4]) }, i.e. {(5, 8), (6, 7), (6, 7)}.
Therefore, distinct pairs with sum K( = 13) are { (arr[0], arr[5]), (arr[1], arr[3]) }.
Therefore, the required output is 2.
Input: arr[] = { 2, 6, 7, 1, 8, 3 }, K = 10
Output : 2
Explanation:
Distinct pairs with sum K( = 10) are { (arr[0], arr[4]), (arr[2], arr[5]) }.
Therefore, the required output is 2.
Naive Approach: The simplest approach to solve this problem is to use Two Pointer technique. The idea is to sort the array and remove all consecutive duplicate elements from the given array. Finally, count the pairs in the given array whose sum is equal to K. Follow the steps below to solve the problem:
- Initialize a variable, say cntPairs, to store the count of distinct pairs of the array with sum K.
- Sort the array in increasing order.
- Initialize two variables, say i = 0, j = N – 1 as the index of left and right pointers to traverse the array.
- Traverse the array and check for the following conditions:
- Finally, print the value of cntPairs.
Below is the implementation of the above approach:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count distinct pairs
// in array whose sum equal to K
int cntDisPairs(int arr[], int N, int K)
{
// Stores count of distinct pairs
// whose sum equal to K
int cntPairs = 0;
// Sort the array
sort(arr, arr + N);
// Stores index of
// the left pointer
int i = 0;
// Stores index of
// the right pointer
int j = N - 1;
// Calculate count of distinct
// pairs whose sum equal to K
while (i < j) {
// If sum of current pair
// is equal to K
if (arr[i] + arr[j] == K) {
// Remove consecutive duplicate
// array elements
while (i < j && arr[i] == arr[i + 1]) {
// Update i
i++;
}
// Remove consecutive duplicate
// array elements
while (i < j && arr[j] == arr[j - 1]) {
// Update j
j--;
}
// Update cntPairs
cntPairs += 1;
// Update i
i++;
// Update j
j--;
}
// if sum of current pair
// less than K
else if (arr[i] + arr[j] < K) {
// Update i
i++;
}
else {
// Update j
j--;
}
}
return cntPairs;
}
// Driver Code
int main()
{
int arr[] = { 5, 6, 5, 7, 7, 8 };
int N = sizeof(arr) / sizeof(arr[0]);
int K = 13;
cout << cntDisPairs(arr, N, K);
}
Java
// Java program to implement
// the above approach
import java.util.*;
class GFG {
// Function to count distinct pairs
// in array whose sum equal to K
static int cntDisPairs(int arr[], int N, int K)
{
// Stores count of distinct pairs
// whose sum equal to K
int cntPairs = 0;
// Sort the array
Arrays.sort(arr);
// Stores index of
// the left pointer
int i = 0;
// Stores index of
// the right pointer
int j = N - 1;
// Calculate count of distinct
// pairs whose sum equal to K
while (i < j) {
// If sum of current pair
// is equal to K
if (arr[i] + arr[j] == K) {
// Remove consecutive duplicate
// array elements
while (i < j && arr[i] == arr[i + 1]) {
// Update i
i++;
}
// Remove consecutive duplicate
// array elements
while (i < j && arr[j] == arr[j - 1]) {
// Update j
j--;
}
// Update cntPairs
cntPairs += 1;
// Update i
i++;
// Update j
j--;
}
// if sum of current pair
// less than K
else if (arr[i] + arr[j] < K) {
// Update i
i++;
}
else {
// Update j
j--;
}
}
return cntPairs;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 5, 6, 5, 7, 7, 8 };
int N = arr.length;
int K = 13;
System.out.print(cntDisPairs(arr, N, K));
}
}
// This code is contributed by 29AjayKumar
// This code is modified by Susobhan Akhuli
Python3
# Python3 program to implement
# the above approach
# Function to count distinct pairs
# in array whose sum equal to K
def cntDisPairs(arr, N, K):
# Stores count of distinct pairs
# whose sum equal to K
cntPairs = 0
# Sort the array
arr = sorted(arr)
# Stores index of
# the left pointer
i = 0
# Stores index of
# the right pointer
j = N - 1
# Calculate count of distinct
# pairs whose sum equal to K
while (i < j):
# If sum of current pair
# is equal to K
if (arr[i] + arr[j] == K):
# Remove consecutive duplicate
# array elements
while (i < j and arr[i] == arr[i + 1]):
# Update i
i += 1
# Remove consecutive duplicate
# array elements
while (i < j and arr[j] == arr[j - 1]):
# Update j
j -= 1
# Update cntPairs
cntPairs += 1
# Update i
i += 1
# Update j
j -= 1
# If sum of current pair
# less than K
elif (arr[i] + arr[j] < K):
# Update i
i += 1
else:
# Update j
j -= 1
return cntPairs
# Driver Code
if __name__ == '__main__':
arr = [5, 6, 5, 7, 7, 8]
N = len(arr)
K = 13
print(cntDisPairs(arr, N, K))
# This code is contributed by mohit kumar 29
C#
// C# program to implement
// the above approach
using System;
class GFG {
// Function to count distinct pairs
// in array whose sum equal to K
static int cntDisPairs(int[] arr, int N, int K)
{
// Stores count of distinct pairs
// whose sum equal to K
int cntPairs = 0;
// Sort the array
Array.Sort(arr);
// Stores index of
// the left pointer
int i = 0;
// Stores index of
// the right pointer
int j = N - 1;
// Calculate count of distinct
// pairs whose sum equal to K
while (i < j) {
// If sum of current pair
// is equal to K
if (arr[i] + arr[j] == K) {
// Remove consecutive duplicate
// array elements
while (i < j && arr[i] == arr[i + 1]) {
// Update i
i++;
}
// Remove consecutive duplicate
// array elements
while (i < j && arr[j] == arr[j - 1]) {
// Update j
j--;
}
// Update cntPairs
cntPairs += 1;
// Update i
i++;
// Update j
j--;
}
// If sum of current pair
// less than K
else if (arr[i] + arr[j] < K) {
// Update i
i++;
}
else {
// Update j
j--;
}
}
return cntPairs;
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = { 5, 6, 5, 7, 7, 8 };
int N = arr.Length;
int K = 13;
Console.WriteLine(cntDisPairs(arr, N, K));
}
}
// This code is contributed by jana_sayantan
Javascript
<script>
// javascript program to implement
// the above approach
// Function to count distinct pairs
// in array whose sum equal to K
function cntDisPairs(arr,N , K)
{
// Stores count of distinct pairs
// whose sum equal to K
var cntPairs = 0;
// Sort the array
arr.sort();
// Stores index of
// the left pointer
var i = 0;
// Stores index of
// the right pointer
var j = N - 1;
// Calculate count of distinct
// pairs whose sum equal to K
while (i < j) {
// If sum of current pair
// is equal to K
if (arr[i] + arr[j] == K) {
// Remove consecutive duplicate
// array elements
while (i < j && arr[i] == arr[i + 1]) {
// Update i
i++;
}
// Remove consecutive duplicate
// array elements
while (i < j && arr[j] == arr[j - 1]) {
// Update j
j--;
}
// Update cntPairs
cntPairs += 1;
// Update i
i++;
// Update j
j--;
}
// if sum of current pair
// less than K
else if (arr[i] + arr[j] < K) {
// Update i
i++;
}
else {
// Update j
j--;
}
}
return cntPairs;
}
// Driver Code
var arr = [ 5, 6, 5, 7, 7, 8 ];
var N = arr.length;
var K = 13;
document.write(cntDisPairs(arr, N, K));
// This code contributed by shikhasingrajput
</script>
Time Complexity: O(N * log(N))
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized using hashing. Follow the steps below to solve the problem:
- Use two sets, one for numbers and one for pairs seen before.
- Traverse through the array and see if (target – arr[i]) is present in set and arr[i] is not present in seen .
- If yes, then add both arr[i] and (target – arr[i]) in seen and add one to count.
Below is the implementation of the above approach:
C++
// C++ program to implement
#include <bits/stdc++.h>
using namespace std;
int cntDisPairs(vector<int> arr, int target)
{
unordered_set<int> set;
unordered_set<int> seen;
int count = 0;
for (int num : arr) {
if (set.find(target - num) != set.end()
&& seen.find(num) == seen.end()) {
count++;
seen.insert(num);
seen.insert(target - num);
}
set.insert(num);
}
return count;
}
int main()
{
vector<int> arr = { 5, 6, 5, 7, 7, 8 };
int K = 13;
cout << cntDisPairs(arr, K);
}
Java
// Java program to implement
// the above approach
import java.util.*;
class GFG {
// Function to count distinct pairs
// in array whose sum equal to K
static int cntDisPairs(int arr[], int N, int K)
{
// Stores count of distinct pairs
// whose sum equal to K
int cntPairs = 0;
// Store frequency of each distinct
// element of the array
HashMap<Integer, Integer> cntFre
= new HashMap<Integer, Integer>();
for (int i = 0; i < N; i++) {
// Update frequency
// of arr[i]
if (cntFre.containsKey(arr[i]))
cntFre.put(arr[i], cntFre.get(arr[i]) + 1);
else
cntFre.put(arr[i], 1);
}
// Traverse the map
for (Map.Entry<Integer, Integer> it :
cntFre.entrySet()) {
// Stores key value
// of the map
int i = it.getKey();
// If i is the half of K
if (2 * i == K) {
// If frequency of i
// greater than 1
if (cntFre.get(i) > 1)
cntPairs += 2;
}
else {
if (cntFre.containsKey(K - i)) {
// Update cntPairs
cntPairs += 1;
}
}
}
// Update cntPairs
cntPairs = cntPairs / 2;
return cntPairs;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 5, 6, 5, 7, 7, 8 };
int N = arr.length;
int K = 13;
System.out.print(cntDisPairs(arr, N, K));
}
}
// This code is contributed by shikhasingrajput
Python3
# Python3 program to implement
# the above approach
# Function to count distinct pairs
# in array whose sum equal to K
def cntDisPairs(arr, N, K):
# Stores count of distinct pairs
# whose sum equal to K
cntPairs = 0
# Store frequency of each distinct
# element of the array
cntFre = {}
for i in arr:
# Update frequency
# of arr[i]
if i in cntFre:
cntFre[i] += 1
else:
cntFre[i] = 1
# Traverse the map
for key, value in cntFre.items():
# Stores key value
# of the map
i = key
# If i is the half of K
if (2 * i == K):
# If frequency of i
# greater than 1
if (cntFre[i] > 1):
cntPairs += 2
else:
if (cntFre[K - i]):
# Update cntPairs
cntPairs += 1
# Update cntPairs
cntPairs = cntPairs / 2
return cntPairs
# Driver Code
arr = [5, 6, 5, 7, 7, 8]
N = len(arr)
K = 13
print(int(cntDisPairs(arr, N, K)))
# This code is contributed by Dharanendra L V
C#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG {
// Function to count distinct pairs
// in array whose sum equal to K
static int cntDisPairs(int[] arr, int N, int K)
{
// Stores count of distinct pairs
// whose sum equal to K
int cntPairs = 0;
// Store frequency of each distinct
// element of the array
Dictionary<int, int> cntFre
= new Dictionary<int, int>();
for (int i = 0; i < N; i++) {
// Update frequency
// of arr[i]
if (cntFre.ContainsKey(arr[i]))
cntFre[arr[i]] = cntFre[arr[i]] + 1;
else
cntFre.Add(arr[i], 1);
}
// Traverse the map
foreach(KeyValuePair<int, int> it in cntFre)
{
// Stores key value
// of the map
int i = it.Key;
// If i is the half of K
if (2 * i == K) {
// If frequency of i
// greater than 1
if (cntFre[i] > 1)
cntPairs += 2;
}
else {
if (cntFre.ContainsKey(K - i)) {
// Update cntPairs
cntPairs += 1;
}
}
}
// Update cntPairs
cntPairs = cntPairs / 2;
return cntPairs;
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = { 5, 6, 5, 7, 7, 8 };
int N = arr.Length;
int K = 13;
Console.Write(cntDisPairs(arr, N, K));
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// javascript program to implement
// the above approach
// Function to count distinct pairs
// in array whose sum equal to K
function cntDisPairs(arr , N , K) {
// Stores count of distinct pairs
// whose sum equal to K
var cntPairs = 0;
// Store frequency of each distinct
// element of the array
var cntFre =new Map();
for (i = 0; i < N; i++) {
// Update frequency
// of arr[i]
if (cntFre.has(arr[i]))
cntFre.set(arr[i], cntFre.get(arr[i]) + 1);
else
cntFre.set(arr[i], 1);
}
// Traverse the map
for (var it of cntFre) {
// Stores key value
// of the map
var i = it[0];
// If i is the half of K
if (2 * i == K) {
// If frequency of i
// greater than 1
if (cntFre[i] > 1)
cntPairs += 2;
}
else {
if (cntFre.has(K - i)) {
// Update cntPairs
cntPairs += 1;
}
}
}
// Update cntPairs
cntPairs = cntPairs / 2;
return cntPairs;
}
// Driver Code
var arr = [ 5, 6, 5, 7, 7, 8 ];
var N = arr.length;
var K = 13;
document.write(cntDisPairs(arr, N, K));
// This code is contributed by umadevi9616
</script>
Time Complexity: O(N)
Auxiliary Space: O(N)
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