logseq_notes/pages/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_problems #difficulty_medium

  Additional tags:

  #leetcode

  Revisions:

  N/A

Links:

• Link to problem

  ---

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;
}
};