给定大小为N的间隔arr []数组，任务是在给定数组的间隔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
天真的方法：最简单的方法是生成一个新数组，其中包含间隔数组中的所有元素。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是间隔中元素的总数。
辅助空间： O(X * log(X))
高效的方法：为了优化上述方法，我们的想法是使用MinHeap。请按照以下步骤解决问题。
- 创建一个MinHeap，说pq来存储给定数组的所有间隔，以便它返回O(1)中其余间隔的所有元素中的最小元素。
- 从MinHeap中弹出最小间隔，并检查弹出间隔的最小元素是否小于弹出间隔的最大元素。如果发现是真的，则插入一个新的间隔{弹出间隔的最小元素+ 1，弹出间隔的最大元素} 。
- 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)