Interval Scheduling Maximization

The interval scheduling maximization (ISM) problem is to find a largest compatible set - a set of non-overlapping intervals of maximum size. The goal here is to execute as many tasks as possible.

Algorithm

Greedily select the interval with the earliest ending time.

1. Sort the intervals in ascending order of the end time. (We don't care the start time)

2. Scan through the intervals and greedily collect non-overlapping intervals.

Memorize point 1: For all the intervals whose start times are not overlapping with earlier intervals, picking the one ending first must be optimal.

Implementation

// OJ: https://leetcode.com/problems/non-overlapping-intervals/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
    int eraseOverlapIntervals(vector<vector<int>>& A) {
        sort(begin(A), end(A), [](vector<int> &a, vector<int> &b) { return a[1] < b[1]; });
        int end = INT_MIN, ans = 0; // `end` is the maximum ending time of selected intervals
        for (auto &x : A) {
            if (x[0] >= end) end = x[1]; // this interval doesn't have overlap with the previously selected interval, select it and update the `end`.
            else ++ans; // otherwise, overlapped intervals are skipped.
        }
        return ans;
    }
};

Problem

• 435. Non-overlapping Intervals (Medium)

• 452. Minimum Number of Arrows to Burst Balloons (Medium)

• 646. Maximum Length of Pair Chain (Medium) Direct Application

• 1520. Maximum Number of Non-Overlapping Substrings (Medium)

Reference

• https://en.wikipedia.org/wiki/Interval_scheduling