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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0001_0100.s0057_insert_interval;

// #Medium #Array #Level_2_Day_17_Interval #2023_08_11_Time_0_ms_(100.00%)_Space_43.7_MB_(95.60%)

import java.util.Arrays;

/**
 * 57 - Insert Interval\.
 *
 * Medium
 *
 * You are given an array of non-overlapping intervals `intervals` where intervals[i] = [start_i, end_i] represent the start and the end of the i^th interval and `intervals` is sorted in ascending order by start_i. You are also given an interval `newInterval = [start, end]` that represents the start and end of another interval.
 *
 * Insert `newInterval` into `intervals` such that `intervals` is still sorted in ascending order by start_i and `intervals` still does not have any overlapping intervals (merge overlapping intervals if necessary).
 *
 * Return `intervals` _after the insertion_.
 *
 * **Example 1:**
 *
 * **Input:** intervals = \[\[1,3],[6,9]], newInterval = [2,5]
 *
 * **Output:** [[1,5],[6,9]] 
 *
 * **Example 2:**
 *
 * **Input:** intervals = \[\[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
 *
 * **Output:** [[1,2],[3,10],[12,16]]
 *
 * **Explanation:** Because the new interval `[4,8]` overlaps with `[3,5],[6,7],[8,10]`.
 *
 * **Example 3:**
 *
 * **Input:** intervals = [], newInterval = [5,7]
 *
 * **Output:** [[5,7]] 
 *
 * **Example 4:**
 *
 * **Input:** intervals = \[\[1,5]], newInterval = [2,3]
 *
 * **Output:** [[1,5]] 
 *
 * **Example 5:**
 *
 * **Input:** intervals = \[\[1,5]], newInterval = [2,7]
 *
 * **Output:** [[1,7]] 
 *
 * **Constraints:**
 *
 * *   0 <= intervals.length <= 10⁴
 * *   `intervals[i].length == 2`
 * *   0 <= start_i <= end_i <= 10⁵
 * *   `intervals` is sorted by start_i in **ascending** order.
 * *   `newInterval.length == 2`
 * *   0 <= start <= end <= 10⁵
**/
public class Solution {
    public int[][] insert(int[][] intervals, int[] newInterval) {
        int n = intervals.length;
        int l = 0;
        int r = n - 1;
        while (l < n && newInterval[0] > intervals[l][1]) {
            l++;
        }
        while (r >= 0 && newInterval[1] < intervals[r][0]) {
            r--;
        }
        int[][] res = new int[l + n - r][2];
        for (int i = 0; i < l; i++) {
            res[i] = Arrays.copyOf(intervals[i], intervals[i].length);
        }
        res[l][0] = Math.min(newInterval[0], l == n ? newInterval[0] : intervals[l][0]);
        res[l][1] = Math.max(newInterval[1], r == -1 ? newInterval[1] : intervals[r][1]);
        for (int i = l + 1, j = r + 1; j < n; i++, j++) {
            res[i] = intervals[j];
        }
        return res;
    }
}