给定大小为N ( 1≤N≤1e5 )的集合S ，该集合S由间隔组成，任务是找到需要从集合中删除的最小间隔，以使其余间隔中的任何一个都等于该集合的并集。
例子：
Input: S = {[1, 3], [4, 12], [5, 8], [13, 20]}
Output: 2
Explanation:
Removing the intervals [1, 3] and [13, 20] modifies the set to { [4, 12], [5, 8]}. The interval [4, 12] is the union of the set.
Input : S = {[1, 2], [1, 10], [4, 8], [3, 7]}
Output: 0
Explanation:
Union of the given set = {[1, 10]}, which is already present in the set. Therefore, no removals required.
方法：该问题可以基于以下观察得到解决：
Observation: To make any interval equal to the union of the set, the set must contain an interval [L, R] such that all the remaining intervals have their left boundary greater than equal to L and right boundary less than equal to R.
请按照以下步骤解决问题：
- 遍历给定的时间间隔集。
- 对于集合中的每个间隔，找到其左边界大于或等于其左边界以及右边界小于或等于其右边界的所有间隔。将此类间隔的计数存储在变量中，例如Count 。
- 找到每个间隔的所有N – Count (因为N – Count会给出删除的间隔数)值的最小值。
- 打印获得的最小值作为所需答案。
下面是上述方法的实现：
C++
// C++ implementaion of the above approach
#include
using namespace std;
// Function to count minimum number of removals
// required to make an interval equal to the
// union of the given Set
int findMinDeletions(vector >& v,
int n)
{
// Stores the minimum number of removals
int minDel = INT_MAX;
// Traverse the Set
for (int i = 0; i < n; i++) {
// Left Boundary
int L = v[i].first;
// Right Boundary
int R = v[i].second;
// Stores count of intervals
// lying within current interval
int Count = 0;
// Traverse over all remaining intervals
for (int j = 0; j < n; j++) {
// Check if interval lies within
// the current interval
if (v[j].first >= L && v[j].second <= R) {
// Increase count
Count += 1;
}
}
// Update minimum removals required
minDel = min(minDel, n - Count);
}
return minDel;
}
// Driver Code
int main()
{
vector > v;
v.push_back({ 1, 3 });
v.push_back({ 4, 12 });
v.push_back({ 5, 8 });
v.push_back({ 13, 20 });
int N = v.size();
// Returns the minimum removals
cout << findMinDeletions(v, N);
return 0;
} Java
// Java implementaion of the above approach
import java.util.*;
class GFG{
// Function to count minimum number of removals
// required to make an interval equal to the
// union of the given Set
static int findMinDeletions(int [][]v,
int n)
{
// Stores the minimum number of removals
int minDel = Integer.MAX_VALUE;
// Traverse the Set
for (int i = 0; i < n; i++)
{
// Left Boundary
int L = v[i][0];
// Right Boundary
int R = v[i][1];
// Stores count of intervals
// lying within current interval
int Count = 0;
// Traverse over all remaining intervals
for (int j = 0; j < n; j++)
{
// Check if interval lies within
// the current interval
if (v[j][0] >= L && v[j][1] <= R)
{
// Increase count
Count += 1;
}
}
// Update minimum removals required
minDel = Math.min(minDel, n - Count);
}
return minDel;
}
// Driver Code
public static void main(String[] args)
{
int [][]v = {{ 1, 3 },
{ 4, 12 },
{ 5, 8 },
{ 13, 20 }};
int N = v.length;
// Returns the minimum removals
System.out.print(findMinDeletions(v, N));
}
}
// This code is contributed by 29AjayKumarPython3
# Python3 implementaion of the above approach
# Function to count minimum number of removals
# required to make an interval equal to the
# union of the given Set
def findMinDeletions(v, n):
# Stores the minimum number of removals
minDel = 10**18
# Traverse the Set
for i in range(n):
# Left Boundary
L = v[i][0]
# Right Boundary
R = v[i][1]
# Stores count of intervals
# lying within current interval
Count = 0
# Traverse over all remaining intervals
for j in range(n):
# Check if interval lies within
# the current interval
if (v[j][1] >= L and v[j][0] <= R):
# Increase count
Count += 1
# Update minimum removals required
minDel = min(minDel, n - Count)
return minDel
# Driver Code
if __name__ == '__main__':
v = []
v.append([1, 3])
v.append([4, 12])
v.append([5, 8])
v.append([13, 2])
N = len(v)
# Returns the minimum removals
print (findMinDeletions(v, N))
# This code is contributed by mohit kumar 29C#
// C# implementaion of the above approach
using System;
public class GFG{
// Function to count minimum number of removals
// required to make an interval equal to the
// union of the given Set
static int findMinDeletions(int [,]v,
int n)
{
// Stores the minimum number of removals
int minDel = int.MaxValue;
// Traverse the Set
for (int i = 0; i < n; i++)
{
// Left Boundary
int L = v[i,0];
// Right Boundary
int R = v[i,1];
// Stores count of intervals
// lying within current interval
int Count = 0;
// Traverse over all remaining intervals
for (int j = 0; j < n; j++)
{
// Check if interval lies within
// the current interval
if (v[j,0] >= L && v[j,1] <= R)
{
// Increase count
Count += 1;
}
}
// Update minimum removals required
minDel = Math.Min(minDel, n - Count);
}
return minDel;
}
// Driver Code
public static void Main(String[] args)
{
int [,]v = {{ 1, 3 },
{ 4, 12 },
{ 5, 8 },
{ 13, 20 }};
int N = v.GetLength(0);
// Returns the minimum removals
Console.Write(findMinDeletions(v, N));
}
}
// This code is contributed by 29AjayKumar输出：
2时间复杂度： O(N 2 )
辅助空间： O(1)