Count maximum possible pairs from an array having sum K
Last Updated :
17 Nov, 2021
Given an array arr[] consisting of N integers and an integer K, the task is to find the maximum number of pairs having a sum K possible from the given array.
Note: Every array element can be part of a single pair.
Examples:
Input: arr[] = {1, 2, 3, 4}, K = 5
Output: 2
Explanation: Pairs with sum K from the array are (1, 4), and (2, 3).
Input: arr[] = {3, 1, 3, 4, 3}, K = 6
Output: 1
Explanation: Pair with sum K from the array is (3, 3).
Two-Pointer Approach: The idea is to use the Two Pointer Technique. Follow the steps below to solve the problem:
- Initialize the variable ans as 0 to store the maximum number of pairs with the sum K.
- Sort the array arr[] in increasing order.
- Initialize two index variables L as 0 and R as (N - 1) to find the candidate elements in the sorted array.
- Iterate until L is less than R and do the following:
- Check if the sum of arr[L] and arr[R] is K or not. If found to be true, then increment ans and L by 1 and decrement R by 1.
- If the sum of arr[L] and arr[R] is less than K, then increment L by 1.
- Otherwise, decrement R by 1.
- After the above steps, print the value of ans as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count the maximum number
// of pairs from given array with sum K
void maxPairs(int nums[], int n, int k)
{
// Sort array in increasing order
sort(nums, nums + n);
// Stores the final result
int result = 0;
// Initialize the left and right pointers
int start = 0, end = n - 1;
// Traverse array until start < end
while (start < end) {
if (nums[start] + nums[end] > k)
// Decrement right by 1
end--;
else if (nums[start] + nums[end] < k)
// Increment left by 1
start++;
// Increment result and left
// pointer by 1 and decrement
// right pointer by 1
else
{
start++;
end--;
result++;
}
}
// Print the result
cout << result << endl;;
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 3, 4 };
int n = sizeof(arr)/sizeof(arr[0]);
int K = 5;
// Function Call
maxPairs(arr, n, K);
return 0;
}
// This code is contributed by AnkThon
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to count the maximum number
// of pairs from given array with sum K
public static void maxPairs(int[] nums, int k)
{
// Sort array in increasing order
Arrays.sort(nums);
// Stores the final result
int result = 0;
// Initialize the left and right pointers
int start = 0, end = nums.length - 1;
// Traverse array until start < end
while (start < end) {
if (nums[start] + nums[end] > k)
// Decrement right by 1
end--;
else if (nums[start] + nums[end] < k)
// Increment left by 1
start++;
// Increment result and left
// pointer by 1 and decrement
// right pointer by 1
else {
start++;
end--;
result++;
}
}
// Print the result
System.out.println(result);
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 1, 2, 3, 4 };
int K = 5;
// Function Call
maxPairs(arr, K);
}
}
# Python3 program for the above approach
# Function to count the maximum number
# of pairs from given array with sum K
def maxPairs(nums, k):
# Sort array in increasing order
nums = sorted(nums)
# Stores the final result
result = 0
# Initialize the left and right pointers
start, end = 0, len(nums) - 1
# Traverse array until start < end
while (start < end):
if (nums[start] + nums[end] > k):
# Decrement right by 1
end -= 1
elif (nums[start] + nums[end] < k):
# Increment left by 1
start += 1
# Increment result and left
# pointer by 1 and decrement
# right pointer by 1
else:
start += 1
end -= 1
result += 1
# Print the result
print(result)
# Driver Code
if __name__ == '__main__':
arr = [ 1, 2, 3, 4 ]
K = 5
# Function Call
maxPairs(arr, K)
# This code is contributed by mohit kumar 29
// C# program for the above approach
using System;
class GFG{
// Function to count the maximum number
// of pairs from given array with sum K
public static void maxPairs(int[] nums, int k)
{
// Sort array in increasing order
Array.Sort(nums);
// Stores the final result
int result = 0;
// Initialize the left and right pointers
int start = 0, end = nums.Length - 1;
// Traverse array until start < end
while (start < end) {
if (nums[start] + nums[end] > k)
// Decrement right by 1
end--;
else if (nums[start] + nums[end] < k)
// Increment left by 1
start++;
// Increment result and left
// pointer by 1 and decrement
// right pointer by 1
else
{
start++;
end--;
result++;
}
}
// Print the result
Console.Write(result);
}
// Driver Code
public static void Main()
{
int[] arr = { 1, 2, 3, 4 };
int K = 5;
// Function Call
maxPairs(arr, K);
}
}
// This code is contributed by susmitakundugoaldanga
<script>
// JavaScript program for above approach
// Function to count the maximum number
// of pairs from given array with sum K
function maxPairs(nums, k)
{
// Sort array in increasing order
nums.sort();
// Stores the final result
let result = 0;
// Initialize the left and right pointers
let start = 0, end = nums.length - 1;
// Traverse array until start < end
while (start < end) {
if (nums[start] + nums[end] > k)
// Decrement right by 1
end--;
else if (nums[start] + nums[end] < k)
// Increment left by 1
start++;
// Increment result and left
// pointer by 1 and decrement
// right pointer by 1
else {
start++;
end--;
result++;
}
}
// Print the result
document.write(result);
}
// Driver Code
let arr = [ 1, 2, 3, 4 ];
let K = 5;
// Function Call
maxPairs(arr, K);
</script>
Output:
2
Time Complexity: O(N*log N)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to use hashing. Follow the steps below to solve the problem:
- Initialize a variable, say ans, to store the maximum number of pairs with the sum K.
- Initialize a hash table, say S, to store the frequency of elements in arr[].
- Traverse the array arr[] using a variable, say i, and perform the following steps:
- If the frequency of (K - arr[i]) is positive, then increment ans by 1 and decrement the frequency of (K - arr[i]) by 1.
- Otherwise, insert arr[i] with frequency 1 in the Hash Table.
- After completing the above steps, print the value of ans as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
#include <string.h>
using namespace std;
// Function to find the maximum number
// of pairs with a sum K such that
// same element can't be used twice
void maxPairs(vector<int> nums, int k)
{
// Initialize a hashm
map<int, int> m;
// Store the final result
int result = 0;
// Iterate over the array nums[]
for(auto i : nums)
{
// Decrement its frequency
// in m and increment
// the result by 1
if (m.find(i) != m.end() && m[i] > 0)
{
m[i] = m[i] - 1;
result++;
}
// Increment its frequency by 1
// if it is already present in m.
// Otherwise, set its frequency to 1
else
{
m[k - i] = m[k - i] + 1;
}
}
// Print the result
cout << result;
}
// Driver Code
int main()
{
vector<int> arr = { 1, 2, 3, 4 };
int K = 5;
// Function Call
maxPairs(arr, K);
}
// This code is contributed by grand_master
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to find the maximum number
// of pairs with a sum K such that
// same element can't be used twice
public static void maxPairs(
int[] nums, int k)
{
// Initialize a hashmap
Map<Integer, Integer> map
= new HashMap<>();
// Store the final result
int result = 0;
// Iterate over the array nums[]
for (int i : nums) {
// Decrement its frequency
// in map and increment
// the result by 1
if (map.containsKey(i) &&
map.get(i) > 0)
{
map.put(i, map.get(i) - 1);
result++;
}
// Increment its frequency by 1
// if it is already present in map.
// Otherwise, set its frequency to 1
else
{
map.put(k - i,
map.getOrDefault(k - i, 0) + 1);
}
}
// Print the result
System.out.println(result);
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 1, 2, 3, 4 };
int K = 5;
// Function Call
maxPairs(arr, K);
}
}
# Python3 program for the above approach
# Function to find the maximum number
# of pairs with a sum K such that
# same element can't be used twice
def maxPairs(nums, k) :
# Initialize a hashm
m = {}
# Store the final result
result = 0
# Iterate over the array nums[]
for i in nums :
# Decrement its frequency
# in m and increment
# the result by 1
if ((i in m) and m[i] > 0) :
m[i] = m[i] - 1
result += 1
# Increment its frequency by 1
# if it is already present in m.
# Otherwise, set its frequency to 1
else :
if k - i in m :
m[k - i] += 1
else :
m[k - i] = 1
# Print the result
print(result)
# Driver code
arr = [ 1, 2, 3, 4 ]
K = 5
# Function Call
maxPairs(arr, K)
# This code is contributed by divyesh072019
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to find the maximum number
// of pairs with a sum K such that
// same element can't be used twice
public static void maxPairs(
int[] nums, int k)
{
// Initialize a hashmap
Dictionary<int, int> map
= new Dictionary<int, int>();
// Store the readonly result
int result = 0;
// Iterate over the array nums[]
foreach (int i in nums)
{
// Decrement its frequency
// in map and increment
// the result by 1
if (map.ContainsKey(i) &&
map[i] > 0)
{
map[i] = map[i] - 1;
result++;
}
// Increment its frequency by 1
// if it is already present in map.
// Otherwise, set its frequency to 1
else
{
if (!map.ContainsKey(k - i))
map.Add(k - i, 1);
else
map[i] = map[i] + 1;
}
}
// Print the result
Console.WriteLine(result);
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = {1, 2, 3, 4};
int K = 5;
// Function Call
maxPairs(arr, K);
}
}
// This code is contributed by 29AjayKumar
<script>
// Javascript program for the above approach
// Function to find the maximum number
// of pairs with a sum K such that
// same element can't be used twice
function maxPairs(nums, k)
{
// Initialize a hashm
var m = new Map();
// Store the final result
var result = 0;
// Iterate over the array nums[]
nums.forEach(i => {
// Decrement its frequency
// in m and increment
// the result by 1
if (m.has(i) && m.get(i) > 0)
{
m.set(i, m.get(i)-1);
result++;
}
// Increment its frequency by 1
// if it is already present in m.
// Otherwise, set its frequency to 1
else
{
if(m.has(k-i))
m.set(k-i, m.get(k-i)+1)
else
m.set(k-i, 1)
}
});
// Print the result
document.write( result);
}
// Driver Code
var arr = [1, 2, 3, 4];
var K = 5;
// Function Call
maxPairs(arr, K);
</script>
Time Complexity: O(N)
Auxiliary Space: O(N)
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