给定数组中的第 K 个最小对和
给定一个大小为N的整数数组arr[]和一个整数K ，任务是找到给定数组的第 K 个最小对和。
例子：
Input: arr[] = {1, 5, 6, 3, 2}, K = 3
Output: 5
Explanation: The sum of all the pairs are: 1+5 = 6, 1+6 = 7, 1+3 = 4, 1+2 = 3, 5+6 = 11, 5+3 = 8, 5+2 = 7, 6+3 = 9, 6+2 = 8, 2+3 = 5. The 3rd smallest among these is 5.
Input: arr[] = {2, 4, 5, 6}, K = 6
Output: 11
天真的方法：这是一种贪婪的方法。我们将获得所有可能对的总和并将它们存储在一个数组中。然后我们将按升序对该数组进行排序，第 K 个值是所需的答案。
时间复杂度： O (N 2 * log(N 2 ))
辅助空间： O(N 2 )
高效方法：这种方法的概念基于max-heap 。我们将实现一个大小为 K 的最大堆。请按照以下步骤操作：
- 开始从i = 0到i = N-2迭代数组。
- 对于每个 i从j = i+1迭代到j = N-1 。
- 获取这个 (i, j) 对的总和并将其插入到最大堆中。
- 如果堆已满，则将堆的顶部元素与此总和进行比较。
- 如果顶部元素的值更大，则将其替换为新的 sum 。
- 当迭代超过最大堆的最顶部元素时，是第K 个最小的对 sum 。
下面是上述方法的实现：
C++
#include
using namespace std;
// Function to get the Kth smallest pair sum
int kthPairSum(vector& arr, int K)
{
int n = arr.size();
// Priority queue to implement max-heap
priority_queue pq;
// Loop to calculate all possible pair sum
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
// Variable to store the sum
int temp = arr[i] + arr[j];
// If the heap is full
if (pq.size() == K) {
// If the top element is greater
if (pq.top() > temp) {
pq.pop();
pq.push(temp);
}
}
else
pq.push(temp);
}
}
// Return the Kth smallest value
return pq.top();
}
// Driver code
int main()
{
vector arr = { 1, 5, 6, 3, 2 };
int K = 3;
cout << kthPairSum(arr, K);
return 0;
} Java
// Java code for the above approach
import java.util.*;
class GFG {
// Function to get the Kth smallest pair sum
static int kthPairSum(int[] arr, int K)
{
int n = arr.length;
// Priority queue to implement max-heap
// Creating empty priority queue
PriorityQueue pq
= new PriorityQueue(
Collections.reverseOrder());
// Loop to calculate all possible pair sum
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
// Variable to store the sum
int temp = arr[i] + arr[j];
// If the heap is full
if (pq.size() == K) {
// If the top element is greater
if (pq.peek() > temp) {
pq.poll();
pq.add(temp);
}
}
else
pq.add(temp);
}
}
// Return the Kth smallest value
return pq.peek();
}
// Driver code
public static void main(String[] args)
{
int[] arr = { 1, 5, 6, 3, 2 };
int K = 3;
System.out.println(kthPairSum(arr, K));
}
}
// This code is contributed by Potta Lokesh Python3
from queue import PriorityQueue
# Function to get the Kth smallest pair sum
def kthPairSum(arr, K):
n = len(arr);
# Priority queue to implement max-heap
pq = PriorityQueue()
# Loop to calculate all possible pair sum
for i in range(n - 1):
for j in range(i + 1, n):
# Variable to store the sum
temp = arr[i] + arr[j];
# If the heap is full
if (pq.qsize() == K):
# If the top element is greater
if (pq.queue[-1] > temp):
pq.get();
pq.put(temp);
else:
pq.put(temp);
# Return the Kth smallest value
return pq.queue[0];
# Driver code
arr = [ 1, 5, 6, 3, 2 ];
K = 3;
print(kthPairSum(arr, K));
# This code is contributed by saurabh_jaiswal.C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to get the Kth smallest pair sum
static int kthPairSum(int[] arr, int K)
{
int n = arr.Length;
// Priority queue to implement max-heap
// Creating empty priority queue
List pq = new List();
// Loop to calculate all possible pair sum
for (int i = 0; i < n - 1; i++) {
for (int j = i + 1; j < n; j++) {
// Variable to store the sum
int temp = arr[i] + arr[j];
// If the heap is full
if (pq.Count == K) {
// If the top element is greater
if (pq[0] > temp) {
pq.Remove(0);
pq.Add(temp);
}
}
else
pq.Add(temp);
}
}
// Return the Kth smallest value
return pq[0]-1;
}
// Driver Code
public static void Main()
{
int[] arr = { 1, 5, 6, 3, 2 };
int K = 3;
Console.Write(kthPairSum(arr, K));
}
}
// This code is contributed by avijitmondal1998. 输出
5时间复杂度： O(N 2 * logK)
辅助空间： O(K)