NextGreaterElementI496.java

/**
 * You are given two arrays (without duplicates) nums1 and nums2 where nums1’s
 * elements are subset of nums2. Find all the next greater numbers for nums1's
 * elements in the corresponding places of nums2.
 * 
 * The Next Greater Number of a number x in nums1 is the first greater number
 * to its right in nums2. If it does not exist, output -1 for this number.
 * 
 * Example 1:
 * Input: nums1 = [4,1,2], nums2 = [1,3,4,2].
 * Output: [-1,3,-1]
 * Explanation:
 *     For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1.
 *     For number 1 in the first array, the next greater number for it in the second array is 3.
 *     For number 2 in the first array, there is no next greater number for it in the second array, so output -1.
 * 
 * Example 2:
 * Input: nums1 = [2,4], nums2 = [1,2,3,4].
 * Output: [3,-1]
 * Explanation:
 *     For number 2 in the first array, the next greater number for it in the second array is 3.
 *     For number 4 in the first array, there is no next greater number for it in the second array, so output -1.
 * 
 * Note:
 * All elements in nums1 and nums2 are unique.
 * The length of both nums1 and nums2 would not exceed 1000.
 */

public class NextGreaterElementI496 {
    public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        if (nums1 == null || nums1.length == 0) return new int[0];
        int[] res = new int[nums1.length];
        Map<Integer, Integer> map = new HashMap<>();
        Stack<Integer> stack = new Stack<>();
        for (int i=0; i<nums2.length; i++) {
            if (stack.isEmpty() || stack.peek() >= nums2[i]) {
                stack.push(nums2[i]);
                continue;
            }
            
            while (!stack.isEmpty() && stack.peek() < nums2[i]) {
                map.put(stack.pop(), nums2[i]);
            }
            stack.push(nums2[i]);
        }
        
        for (int i=0; i<nums1.length; i++) {
            res[i] = map.getOrDefault(nums1[i], -1);
        } 
        return res;
    }
}