notes/OJ notes/pages/Leetcode Non-Overlapping-Intervals.md

Leetcode Non-Overlapping-Intervals

2022-09-03 15:34

Algorithms:

#algorithm #greedy

Data structures:

#DS #array

Difficulty:

#coding_problem #difficulty-medium

Additional tags:

# #

Revisions:

N/A

Links:

• Link to problem

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Problem

Given an array of intervals intervals where intervals[i] = [starti, endi], return the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.

Examples

Example 1:

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.

Example 2:

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.

Example 3:

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 <= 105
• intervals[i].length == 2
• -5 * 104 <= starti < endi <= 5 * 104

Thoughts

[!summary] This is a #greedy problem, similar to Leetcode Merge-Intervals

Key concept:

first sort the intervals.

pick the intervals with smallest end, which will allow us to make more room for following ones.

#tip: Use & to reference vectors in comp.function to save time.

Solution

class Solution {
  static bool comp(vector<int> &a, vector<int> &b) { return a[1] < b[1]; }

public:
  int eraseOverlapIntervals(vector<vector<int>> &intervals) {

    sort(intervals.begin(), intervals.end(), comp);

    // Using this var to skip overlapping intervals.
    auto before = intervals[0];

    int ans = 0;

    for (int i = 1, size = intervals.size(); i < size; i++) {
      if (intervals[i][0] < before[1]) {
        ans++;
      } else {
        before = intervals[i];
      }
    }

    return ans;
  }
};