Given an array arr[] of size N and an integer K, the task is to find the minimum count of pairs required to be removed such that no pair exists in the array whose sum of elements is equal to K.
Examples:
Input: arr[] = { 3, 1, 3, 4, 3 }, K = 6
Output: 1
Explanation:
Removing the pair (arr[0], arr[2]) modifies arr[] to arr[] = { 1, 4, 3 }
Since no pair exists in arr[] whose sum of elements is equal to K(=6), the required output is 1.Input: arr = { 1, 2, 3, 4 }, K = 5
Output: 2
Explanation:
Removing the pair (arr[0], arr[3]) modifies arr[] to arr[] = { 2, 3 }
Removing the pair (arr[0], arr[1]) modifies arr[] to arr[] = { }
Since no pair exists in arr[] whose sum of elements is equal to K(=5), the required output is 2.
Approach: The problem can be solved using the two-pointer technique. Follow the steps below to solve this problem:
- Sort the array in ascending order.
- Initialize two variables, say left = 0 and right = N – 1 to store the index of left and right pointers respectively.
- Initialize a variable, say cntPairs, to store the minimum count of pairs required to be removed such that no pair exists in the array whose sum is equal to K.
-
Traverse the array and check the following conditions.
- If arr[left] + arr[right] == K, then increment the value of cntPairs by 1 and update left += 1 and right -= 1.
- If arr[left] + arr[right] < K, then update left += 1.
- If arr[left] + arr[right] < K, then update right -= 1.
- Finally, print the value of cntPairs.
Below is the implementation of the above approach:
// C++14 program to implement // the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the maximum count of pairs // required to be removed such that no pairs // exist whose sum equal to K int maxcntPairsSumKRemoved(vector< int > arr, int k)
{ // Stores maximum count of pairs required
// to be removed such that no pairs
// exist whose sum equal to K
int cntPairs = 0;
// Base Case
if (arr.size() <= 1)
return cntPairs;
// Sort the array
sort(arr.begin(), arr.end());
// Stores index of
// left pointer
int left = 0;
// Stores index of
// right pointer
int right = arr.size() - 1;
while (left < right)
{
// Stores sum of left
// and right pointer
int s = arr[left] + arr[right];
// If s equal to k
if (s == k)
{
// Update cntPairs
cntPairs += 1;
// Update left
left += 1;
// Update right
right -= 1;
}
// If s > k
else if (s > k)
// Update right
right -= 1;
else
// Update left
left += 1;
}
// Return the cntPairs
return cntPairs;
} // Driver Code int main()
{ vector< int > arr = { 1, 2, 3, 4 };
int K = 5;
// Function call
cout << (maxcntPairsSumKRemoved(arr, K));
return 0;
} // This code is contributed by mohit kumar 29 |
// Java program for the above approach import java.util.*;
class GFG{
// Function to find the maximum count of pairs // required to be removed such that no pairs // exist whose sum equal to K static int maxcntPairsSumKRemoved( int [] arr, int k)
{ // Stores maximum count of pairs required
// to be removed such that no pairs
// exist whose sum equal to K
int cntPairs = 0 ;
// Base Case
if (arr.length <= 1 )
return cntPairs;
// Sort the array
Arrays.sort(arr);
// Stores index of
// left pointer
int left = 0 ;
// Stores index of
// right pointer
int right = arr.length - 1 ;
while (left < right)
{
// Stores sum of left
// and right pointer
int s = arr[left] + arr[right];
// If s equal to k
if (s == k)
{
// Update cntPairs
cntPairs += 1 ;
// Update left
left += 1 ;
// Update right
right -= 1 ;
}
// If s > k
else if (s > k)
// Update right
right -= 1 ;
else
// Update left
left += 1 ;
}
// Return the cntPairs
return cntPairs;
} // Driver Code public static void main (String[] args)
{ int [] arr = { 1 , 2 , 3 , 4 };
int K = 5 ;
// Function call
System.out.println (maxcntPairsSumKRemoved(arr, K));
} } |
# Python3 program to implement # the above approach # Function to find the maximum count of pairs # required to be removed such that no pairs # exist whose sum equal to K def maxcntPairsSumKRemoved(arr, k):
# Stores maximum count of pairs required
# to be removed such that no pairs
# exist whose sum equal to K
cntPairs = 0
# Base Case
if not arr or len (arr) = = 1 :
return cntPairs
# Sort the array
arr.sort()
# Stores index of
# left pointer
left = 0
# Stores index of
# right pointer
right = len (arr) - 1
while left < right:
# Stores sum of left
# and right pointer
s = arr[left] + arr[right]
# If s equal to k
if s = = k:
# Update cntPairs
cntPairs + = 1
# Update left
left + = 1
# Update right
right - = 1
# If s > k
elif s > k:
# Update right
right - = 1
else :
# Update left
left + = 1
# Return the cntPairs
return cntPairs
# Driver Code if __name__ = = "__main__" :
arr = [ 1 , 2 , 3 , 4 ]
K = 5
# Function call
print (maxcntPairsSumKRemoved(arr, K))
|
// C# program for the above approach using System;
class GFG{
// Function to find the maximum count of pairs // required to be removed such that no pairs // exist whose sum equal to K static int maxcntPairsSumKRemoved( int [] arr, int k)
{ // Stores maximum count of pairs required
// to be removed such that no pairs
// exist whose sum equal to K
int cntPairs = 0;
// Base Case
if (arr.Length <= 1)
return cntPairs;
// Sort the array
Array.Sort(arr);
// Stores index of
// left pointer
int left = 0;
// Stores index of
// right pointer
int right = arr.Length - 1;
while (left < right)
{
// Stores sum of left
// and right pointer
int s = arr[left] + arr[right];
// If s equal to k
if (s == k)
{
// Update cntPairs
cntPairs += 1;
// Update left
left += 1;
// Update right
right -= 1;
}
// If s > k
else if (s > k)
// Update right
right -= 1;
else
// Update left
left += 1;
}
// Return the cntPairs
return cntPairs;
} // Driver Code public static void Main(String[] args)
{ int [] arr = { 1, 2, 3, 4 };
int K = 5;
// Function call
Console.WriteLine (maxcntPairsSumKRemoved(arr, K));
} } // This code is contributed by 29AjayKumar |
<script> //Javascript program to implement // the above approach // Function to find the maximum count of pairs // required to be removed such that no pairs // exist whose sum equal to K function maxcntPairsSumKRemoved( arr, k)
{ // Stores maximum count of pairs required
// to be removed such that no pairs
// exist whose sum equal to K
var cntPairs = 0;
// Base Case
if (arr.length <= 1)
return cntPairs;
// Sort the array
arr.sort();
// Stores index of
// left pointer
var left = 0;
// Stores index of
// right pointer
var right = arr.length - 1;
while (left < right)
{
// Stores sum of left
// and right pointer
var s = arr[left] + arr[right];
// If s equal to k
if (s == k)
{
// Update cntPairs
cntPairs += 1;
// Update left
left += 1;
// Update right
right -= 1;
}
// If s > k
else if (s > k)
// Update right
right -= 1;
else
// Update left
left += 1;
}
// Return the cntPairs
return cntPairs;
} var arr = [ 1,2,3,4];
var K = 5;
// Function call document.write(maxcntPairsSumKRemoved(arr, K)); //This code is contributed by SoumikMondal </script> |
2
Time Complexity: O(N*logN) as it sorts the given array
Auxiliary Space: O(1)