Non-overlapping Intervals in C++

Suppose we have a collection of intervals; we have to find the minimum number of intervals we need to remove to make the rest of the intervals non-overlapping. So if the intervals are [[1,2], [2,3], [3,4], [1,3]], then the output will be 1, as we have to remove [1,3] 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 = {{1,2},{1,2},{1,2}};
   Solution ob;
   cout << (ob.eraseOverlapIntervals(v));
}