Program to find minimum number of intervals to be removed to remove overlaps in C++

Suppose we have a set of intervals; we have to find the minimum number of intervals that should be removed to make the rest of the intervals non-overlapping. So if the intervals are [[8,10],[3,5],[6,9]], then the output will be 1, as we have to remove [6,9] to make all others are non-overlapping.

To solve this, we will follow these steps −

n := size of array
if n is 0, then return 0
count := 1
sort the array based on the end time of the intervals
end := end date of the first interval
for i in range 1 to n – 1
- if start time of arr[i] >= end, then
  - end := end time of arr[i]
  - increase count by 1
return n – count

Let us see the following implementation to get better understanding −

Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
   static bool cmp(vector <int>& a, vector <int>& b){
      return a[1] < b[1];
   }
   int eraseOverlapIntervals(vector<vector<int>>& arr) {
      int n = arr.size();
      if(!n)return 0;
         int cnt = 1;
         sort(arr.begin(), arr.end(), cmp);
         int end = arr[0][1];
         for(int i = 1; i < n; i++){
            if(arr[i][0] >= end){
               end = arr[i][1];
               cnt++;
         }
      }
      return n - cnt;
   }
};
main(){
   vector<vector<int>>
   v = {{8,10},{3,5},{6,9}};
   Solution ob;
   cout << (ob.eraseOverlapIntervals(v));
}