Given a set S of size N (1 ? N ? 1e5) consisting of intervals, the task is to find the minimum intervals required to be removed from the set such that any one of the remaining intervals becomes equal to the union of this set.
Examples:
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.
Approach : The problem can be solved based on the following observation:
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.
Follow the steps below to solve the problem:
- Traverse the given set of intervals.
- For every interval in the Set, find all the intervals which have their left boundary greater than or equal to its left boundary as well as have their right boundary less than or equal to its right boundary. Store the count of such intervals in a variable, say Count.
- Find the minimum value of all N – Count (since N – Count would give the number of intervals deleted)values for each interval.
- Print the minimum value obtained as the required answer.
Below is the implementation of above approach :
// C++ implementation of the above approach #include <bits/stdc++.h> 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<pair< int , int > >& 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<pair< int , int > > 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 implementation 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 29AjayKumar |
# Python3 implementation 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 29 |
// C# implementation 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 |
<script> // Javascript program to implement // the above approach // Function to count minimum number of removals // required to make an interval equal to the // union of the given Set function findMinDeletions(v, n)
{ // Stores the minimum number of removals
let minDel = Number.MAX_VALUE;
// Traverse the Set
for (let i = 0; i < n; i++)
{
// Left Boundary
let L = v[i][0];
// Right Boundary
let R = v[i][1];
// Stores count of intervals
// lying within current interval
let Count = 0;
// Traverse over all remaining intervals
for (let 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 let v = [[ 1, 3 ],
[ 4, 12 ],
[ 5, 8 ],
[ 13, 20 ]];
let N = v.length;
// Returns the minimum removals
document.write(findMinDeletions(v, N));
// This code is contributed by souravghosh0416.
</script> |
2
Time Complexity: O(N2), since two nested loops are used.
Auxiliary space: O(1), since no extra array is used so the space taken by the algorithm is constant