给定大小为N的数组arr []和整数K ，任务是找到需要删除的对的最小数量，以使在数组中元素对之和等于K的对不存在。
例子：
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 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 equal to K(=5), the required output is 2.
方法：可以使用两指针技术解决问题。请按照以下步骤解决此问题：
- 以升序对数组进行排序。
- 初始化两个变量，例如left = 0和right = N – 1 ，分别存储左指针和右指针的索引。
- 初始化一个变量，例如cntPairs ，以存储需要删除的对的最小数量，以使数组中的和不等于K的对不存在。
- 遍历数组并检查以下条件。
- 如果arr [left] + arr [right] == K ，则将cntPairs的值增加1并更新left + = 1和right- = 1。
- 如果arr [left] + arr [right]
- 如果arr [left] + arr [right]
- 最后，打印cntPairs的值。
下面是上述方法的实现：
C++14
// C++14 program to implement
// the above approach
#include
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 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 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
// 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
# 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#
// 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输出：
2时间复杂度： O(N)
辅助空间： O(1)