To keep the maximum number of non-overlapping intervals, we need to remove the minimum number of overlapping ones. This is a classic greedy scenario.We sort intervals by their end times and always pick the next interval that starts after the last selected one ends. The rest need to be removed.

Sort all intervals based on their end time in ascending order.
Initialize a variable to store the end time of the last non-overlapping interval (e.g., set to the end of the first interval).
Initialize a counter to track how many intervals need to be removed.
Traverse the remaining intervals:
If the start of the current interval is greater than or equal to the end time of the last added interval, update the end time.
Else, it overlaps increment the removal counter.
Return the counter at the end as the minimum number of intervals to remove.

Code

#include <bits/stdc++.h>
using namespace std;

class Solution {
public: 
    // Function to find the minimum number of intervals to remove to make all intervals non-overlapping
    int eraseOverlapIntervals(vector<vector<int>>& intervals) {
        // Sort intervals based on their end time (greedy strategy)
        sort(intervals.begin(), intervals.end(), [](auto& a, auto& b) {
            return a[1] < b[1];
        });

        // Count of intervals to remove
        int count = 0;

        // Store end time of the last non-overlapping interval
        int prevEnd = intervals[0][1];

        // Iterate through intervals starting from the second
 