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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0401_0500.s0435_non_overlapping_intervals;

// #Medium #Array #Dynamic_Programming #Sorting #Greedy #Data_Structure_II_Day_4_Array
// #2022_07_16_Time_96_ms_(47.37%)_Space_106.6_MB_(6.15%)

import java.util.Arrays;

/**
 * 435 - Non-overlapping Intervals\.
 *
 * Medium
 *
 * 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_.
 *
 * **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 <= start_i < end_i <= 5 * 10⁴
**/
public class Solution {
    /*
     * This is sorting my starting time, the key here is that we'll want to update end time when an
     * erasure is needed: we use the smaller end time instead of the bigger one which is more likely
     * to overlap with others.
     */
    public int eraseOverlapIntervals(int[][] intervals) {
        Arrays.sort(intervals, (a, b) -> a[0] != b[0] ? a[0] - b[0] : a[1] - b[1]);
        int erasures = 0;
        int end = intervals[0][1];
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i][0] < end) {
                erasures++;
                end = Math.min(end, intervals[i][1]);
            } else {
                end = intervals[i][1];
            }
        }
        return erasures;
    }
}