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Java-based LeetCode algorithm problem solutions, regularly updated

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package g1201_1300.s1288_remove_covered_intervals;

// #Medium #Array #Sorting #2022_03_11_Time_6_ms_(78.87%)_Space_47.5_MB_(6.30%)

import java.util.PriorityQueue;
import java.util.Queue;

/**
 * 1288 - Remove Covered Intervals\.
 *
 * Medium
 *
 * Given an array `intervals` where intervals[i] = [l_i, r_i] represent the interval [l_i, r_i), remove all intervals that are covered by another interval in the list.
 *
 * The interval `[a, b)` is covered by the interval `[c, d)` if and only if `c <= a` and `b <= d`.
 *
 * Return _the number of remaining intervals_.
 *
 * **Example 1:**
 *
 * **Input:** intervals = \[\[1,4],[3,6],[2,8]]
 *
 * **Output:** 2
 *
 * **Explanation:** Interval [3,6] is covered by [2,8], therefore it is removed.
 *
 * **Example 2:**
 *
 * **Input:** intervals = \[\[1,4],[2,3]]
 *
 * **Output:** 1
 *
 * **Constraints:**
 *
 * *   `1 <= intervals.length <= 1000`
 * *   `intervals[i].length == 2`
 * *   0 <= l_i < r_i <= 10⁵
 * *   All the given intervals are **unique**.
**/
public class Solution {
    public int removeCoveredIntervals(int[][] intervals) {
        Queue q = new PriorityQueue<>((a, b) -> a[0] == b[0] ? b[1] - a[1] : a[0] - b[0]);
        for (int[] interval : intervals) {
            q.offer(interval);
        }
        int[] prev = q.poll();
        int count = 0;
        while (!q.isEmpty()) {
            int[] curr = q.poll();
            if (curr[0] >= prev[0] && curr[1] <= prev[1]) {
                count++;
            } else {
                prev = curr;
            }
        }
        return intervals.length - count;
    }
}