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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0401_0500.s0436_find_right_interval;

// #Medium #Array #Sorting #Binary_Search #Binary_Search_II_Day_11
// #2022_07_16_Time_20_ms_(81.51%)_Space_59.1_MB_(5.83%)

import java.util.HashMap;
import java.util.Map;

/**
 * 436 - Find Right Interval\.
 *
 * Medium
 *
 * You are given an array of `intervals`, where intervals[i] = [start_i, end_i] and each start_i is **unique**.
 *
 * The **right interval** for an interval `i` is an interval `j` such that start_j >= end_i and start_j is **minimized**.
 *
 * Return _an array of **right interval** indices for each interval `i`_. If no **right interval** exists for interval `i`, then put `-1` at index `i`.
 *
 * **Example 1:**
 *
 * **Input:** intervals = \[\[1,2]]
 *
 * **Output:** [-1]
 *
 * **Explanation:** There is only one interval in the collection, so it outputs -1. 
 *
 * **Example 2:**
 *
 * **Input:** intervals = \[\[3,4],[2,3],[1,2]]
 *
 * **Output:** [-1,0,1]
 *
 * **Explanation:**
 *
 * There is no right interval for [3,4].
 *
 * The right interval for [2,3] is [3,4] since start₀ = 3 is the smallest start that is >= end₁ = 3.
 *
 * The right interval for [1,2] is [2,3] since start₁ = 2 is the smallest start that is >= end₂ = 2.
 *
 * **Example 3:**
 *
 * **Input:** intervals = \[\[1,4],[2,3],[3,4]]
 *
 * **Output:** [-1,2,-1]
 *
 * **Explanation:**
 *
 * There is no right interval for [1,4] and [3,4].
 *
 * The right interval for [2,3] is [3,4] since start₂ = 3 is the smallest start that is >= end₁ = 3.
 *
 * **Constraints:**
 *
 * *   1 <= intervals.length <= 2 * 10⁴
 * *   `intervals[i].length == 2`
 * *   -10⁶ <= start_i <= end_i <= 10⁶
 * *   The start point of each interval is **unique**.
**/
public class Solution {
    private int[] findminmax(int[][] num) {
        int min = num[0][0];
        int max = num[0][0];
        for (int i = 1; i < num.length; i++) {
            min = Math.min(min, num[i][0]);
            max = Math.max(max, num[i][0]);
        }
        return new int[] {min, max};
    }

    public int[] findRightInterval(int[][] intervals) {
        if (intervals.length <= 1) {
            return new int[] {-1};
        }
        int n = intervals.length;
        int[] result = new int[n];
        Map map = new HashMap<>();
        for (int i = 0; i < n; i++) {
            map.put(intervals[i][0], i);
        }
        int[] minmax = findminmax(intervals);
        for (int i = minmax[1] - 1; i >= minmax[0] + 1; i--) {
            map.computeIfAbsent(i, k -> map.get(k + 1));
        }
        for (int i = 0; i < n; i++) {
            result[i] = map.getOrDefault(intervals[i][1], -1);
        }
        return result;
    }
}