给定一个大小为N的区间arr[]数组，任务是在给定数组的区间内的所有元素中找到第K个最小的元素。
例子：
Input : arr[] = {{5, 11}, {10, 15}, {12, 20}}, K =12
Output: 13
Explanation: Elements in the given array of intervals are: {5, 6, 7, 8, 9, 10, 10, 11, 11, 12, 12, 13, 13, 14, 14, 15, 15, 16, 17, 18, 19, 20}.
Therefore, the Kth(=12th) smallest element is 13.
Input: arr[] = {{5, 11}, {10, 15}, {12, 20}}, K = 7
Output:10
朴素方法：最简单的方法是生成一个新数组，该数组由区间数组中的所有元素组成。对新数组进行排序。最后，返回数组的第K个最小元素。
时间复杂度： O(X*Log(X))，其中X是区间中元素的总数。
辅助空间： O(X*log(X))
高效的方法：为了优化上述方法，想法是使用MinHeap。请按照以下步骤解决问题。
- 创建一个 MinHeap，比如说pq来存储给定数组的所有区间，以便它返回 O(1) 中剩余区间的所有元素中的最小元素。
- 从MinHeap 中弹出最小区间并检查弹出区间的最小元素是否小于弹出区间的最大元素。如果发现为真，则插入一个新的区间{minimum element of popped interval + 1, maximum element of the popped interval} 。
- 重复上述步骤K-1次。
- 最后，返回弹出间隔的最小元素。
下面是上述方法的实现：
C++14
// C++ Program to implement
// the above approach
#include
using namespace std;
// Function to get the Kth smallest
// element from an array of intervals
int KthSmallestNum(pair arr[],
int n, int k)
{
// Store all the intervals so that it
// returns the minimum element in O(1)
priority_queue,
vector >,
greater > >
pq;
// Insert all Intervals into the MinHeap
for (int i = 0; i < n; i++) {
pq.push({ arr[i].first,
arr[i].second });
}
// Stores the count of
// popped elements
int cnt = 1;
// Iterate over MinHeap
while (cnt < k) {
// Stores minimum element
// from all remaining intervals
pair interval
= pq.top();
// Remove minimum element
pq.pop();
// Check if the minimum of the current
// interval is less than the maximum
// of the current interval
if (interval.first < interval.second) {
// Insert new interval
pq.push(
{ interval.first + 1,
interval.second });
}
cnt++;
}
return pq.top().first;
}
// Driver Code
int main()
{
// Intervals given
pair arr[]
= { { 5, 11 },
{ 10, 15 },
{ 12, 20 } };
// Size of the arr
int n = sizeof(arr) / sizeof(arr[0]);
int k = 12;
cout << KthSmallestNum(arr, n, k);
} Java
// Java program to implement
// the above approach
import java.util.*;
import java.io.*;
class GFG{
// Function to get the Kth smallest
// element from an array of intervals
public static int KthSmallestNum(int arr[][], int n,
int k)
{
// Store all the intervals so that it
// returns the minimum element in O(1)
PriorityQueue pq = new PriorityQueue<>(
(a, b) -> a[0] - b[0]);
// Insert all Intervals into the MinHeap
for(int i = 0; i < n; i++)
{
pq.add(new int[]{arr[i][0],
arr[i][1]});
}
// Stores the count of
// popped elements
int cnt = 1;
// Iterate over MinHeap
while (cnt < k)
{
// Stores minimum element
// from all remaining intervals
int[] interval = pq.poll();
// Check if the minimum of the current
// interval is less than the maximum
// of the current interval
if (interval[0] < interval[1])
{
// Insert new interval
pq.add(new int[]{interval[0] + 1,
interval[1]});
}
cnt++;
}
return pq.peek()[0];
}
// Driver Code
public static void main(String args[])
{
// Intervals given
int arr[][] = { { 5, 11 },
{ 10, 15 },
{ 12, 20 } };
// Size of the arr
int n = arr.length;
int k = 12;
System.out.println(KthSmallestNum(arr, n, k));
}
}
// This code is contributed by hemanth gadarla Python3
# Python3 program to implement
# the above approach
# Function to get the Kth smallest
# element from an array of intervals
def KthSmallestNum(arr, n, k):
# Store all the intervals so that it
# returns the minimum element in O(1)
pq = []
# Insert all Intervals into the MinHeap
for i in range(n):
pq.append([arr[i][0], arr[i][1]])
# Stores the count of
# popped elements
cnt = 1
# Iterate over MinHeap
while (cnt < k):
# Stores minimum element
# from all remaining intervals
pq.sort(reverse = True)
interval = pq[0]
# Remove minimum element
pq.remove(pq[0])
# Check if the minimum of the current
# interval is less than the maximum
# of the current interval
if (interval[0] < interval[1]):
# Insert new interval
pq.append([interval[0] + 1,
interval[1]])
cnt += 1
pq.sort(reverse = True)
return pq[0][0] + 1
# Driver Code
if __name__ == '__main__':
# Intervals given
arr = [ [ 5, 11 ],
[ 10, 15 ],
[ 12, 20 ] ]
# Size of the arr
n = len(arr)
k = 12
print(KthSmallestNum(arr, n, k))
# This code is contributed by SURENDRA_GANGWARC#
// C# Program to implement
// the above approach
using System;
using System.Collections;
using System.Collections.Generic;
class GFG {
// Function to get the Kth smallest
// element from an array of intervals
static int KthSmallestNum(int[,] arr, int n, int k)
{
// Store all the intervals so that it
// returns the minimum element in O(1)
ArrayList pq = new ArrayList();
// Insert all Intervals into the MinHeap
for(int i = 0; i < n; i++)
{
pq.Add(new Tuple(arr[i,0], arr[i,1]));
}
// Stores the count of
// popped elements
int cnt = 1;
// Iterate over MinHeap
while (cnt < k)
{
// Stores minimum element
// from all remaining intervals
pq.Sort();
pq.Reverse();
Tuple interval = (Tuple)pq[0];
// Remove minimum element
pq.RemoveAt(0);
// Check if the minimum of the current
// interval is less than the maximum
// of the current interval
if (interval.Item1 < interval.Item2)
{
// Insert new interval
pq.Add(new Tuple(interval.Item1 + 1, interval.Item2));
}
cnt += 1;
}
pq.Sort();
pq.Reverse();
return ((Tuple)pq[0]).Item1 + 1;
}
// Driver code
static void Main()
{
// Intervals given
int[,] arr = { { 5, 11 },
{ 10, 15 },
{ 12, 20 } };
// Size of the arr
int n = arr.GetLength(0);
int k = 12;
Console.WriteLine(KthSmallestNum(arr, n, k));
}
}
// This code is contributed by divyeshrabadiya07 输出：
13时间复杂度： O(K*logK)
辅助空间： O(N)
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