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Java-based LeetCode algorithm problem solutions, regularly updated
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package g0401_0500.s0496_next_greater_element_i;

// #Easy #Array #Hash_Table #Stack #Monotonic_Stack #Programming_Skills_I_Day_5_Function
// #2022_07_21_Time_4_ms_(81.18%)_Space_43.7_MB_(77.46%)

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

/**
 * 496 - Next Greater Element I\.
 *
 * Easy
 *
 * The **next greater element** of some element `x` in an array is the **first greater** element that is **to the right** of `x` in the same array.
 *
 * You are given two **distinct 0-indexed** integer arrays `nums1` and `nums2`, where `nums1` is a subset of `nums2`.
 *
 * For each `0 <= i < nums1.length`, find the index `j` such that `nums1[i] == nums2[j]` and determine the **next greater element** of `nums2[j]` in `nums2`. If there is no next greater element, then the answer for this query is `-1`.
 *
 * Return _an array_ `ans` _of length_ `nums1.length` _such that_ `ans[i]` _is the **next greater element** as described above._
 *
 * **Example 1:**
 *
 * **Input:** nums1 = [4,1,2], nums2 = [1,3,4,2]
 *
 * **Output:** [-1,3,-1]
 *
 * **Explanation:** 
 *
 * The next greater element for each value of nums1 is as follows: 
 *
 * - 4 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1. 
 *
 * - 1 is underlined in nums2 = [1,3,4,2]. The next greater element is 3. 
 *
 * - 2 is underlined in nums2 = [1,3,4,2]. There is no next greater element, so the answer is -1.
 *
 * **Example 2:**
 *
 * **Input:** nums1 = [2,4], nums2 = [1,2,3,4]
 *
 * **Output:** [3,-1]
 *
 * **Explanation:** 
 *
 * The next greater element for each value of nums1 is as follows: 
 *
 * - 2 is underlined in nums2 = [1,2,3,4]. The next greater element is 3. 
 *
 * - 4 is underlined in nums2 = [1,2,3,4]. There is no next greater element, so the answer is -1.
 *
 * **Constraints:**
 *
 * *   `1 <= nums1.length <= nums2.length <= 1000`
 * *   0 <= nums1[i], nums2[i] <= 10⁴
 * *   All integers in `nums1` and `nums2` are **unique**.
 * *   All the integers of `nums1` also appear in `nums2`.
 *
 * **Follow up:** Could you find an `O(nums1.length + nums2.length)` solution?
**/
public class Solution {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        Map indexMap = new HashMap<>();
        for (int i = 0; i < nums2.length; i++) {
            indexMap.put(nums2[i], i);
        }
        for (int i = 0; i < nums1.length; i++) {
            int num = nums1[i];
            int index = indexMap.get(num);
            if (index == nums2.length - 1) {
                nums1[i] = -1;
            } else {
                boolean found = false;
                while (index < nums2.length) {
                    if (nums2[index] > num) {
                        nums1[i] = nums2[index];
                        found = true;
                        break;
                    }
                    index++;
                }
                if (!found) {
                    nums1[i] = -1;
                }
            }
        }
        return nums1;
    }
}