# Minimum removals required to make any interval equal to the union of the given Set

• Last Updated : 30 Aug, 2022

Given a set S of size N (1 â‰¤ N â‰¤ 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:

1. Traverse the given set of intervals.
2. 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.
3. Find the minimum value of all N – Count (since N – Count would give the number of intervals deleted)values for each interval.
4. Print the minimum value obtained as the required answer.

Below is the implementation of above approach :

## C++

 // C++ implementation 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 Setint 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 Codeint 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 implementation of the above approachimport java.util.*;class GFG{ // Function to count minimum number of removals// required to make an interval equal to the// union of the given Setstatic 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 Codepublic 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

 # 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 Setdef 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 Codeif __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#

 // C# implementation of the above approachusing System; public class GFG{ // Function to count minimum number of removals// required to make an interval equal to the// union of the given Setstatic 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 Codepublic 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

## Javascript



Output:

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

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