text: /**
   * 2.3 区间问题1
   * 435. Non-overlapping Intervals(Hard)
   * 
   * Given an array of intervals intervals where intervals[i] = [start_i, end_i], 
   * return the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.
   * 
   * Input: intervals = [[1,2],[2,3],[3,4],[1,3]], Output: 1
   * Explanation: [1,3] can be removed and the rest of the intervals are non-overlapping.
   * 
   * Input: intervals = [[1,2],[1,2],[1,2]], Output: 2
   * Explanation: You need to remove two [1,2] to make the rest of the intervals non-overlapping.
   * 
   * Input: intervals = [[1,2],[2,3]], Output: 0
   * Explanation: You don't need to remove any of the intervals since they're already non-overlapping.
   * 
   * Constraints:
   *  * 1 <= intervals.length <= 10^5
   *  * intervals[i].length == 2
   *  * -5 * 10^4 <= starti < endi <= 5 * 10^4
   */
  
  #include <bits/stdc++.h>
  using namespace std;
  
  int eraseOverlapIntervals(vector<vector<int>>& intervals) {
      if (intervals.empty()) return 0;
      int n = intervals.size();
      sort(intervals.begin(), intervals.end(), [](vector<int> &a, vector<int> &b) {
          return a[1] < b[1];
      });
  
      int removed = 0, prev = intervals[0][1];
      for (int i = 1; i < n; ++i) {
          if (intervals[i][0] < prev) { // find an overlap
              ++removed;
          } else {
              prev = intervals[i][1];
          }
      }
  
      return removed;
  }