496. Next Greater Element I #

Problem #

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.

Problem Summary #

This problem is also an easy one. The problem gives 2 arrays A and B. For each element in array A, you need to find in array B a number greater than the element in array A, while keeping the order of elements in B unchanged. If found, output this value; if not found, output -1.

Solution Approach #

Easy problem; just implement it according to the problem statement.

Code #

package leetcode

func nextGreaterElement(nums1 []int, nums2 []int) []int {
	if len(nums1) == 0 || len(nums2) == 0 {
		return []int{}
	}
	res, reocrd := []int{}, map[int]int{}
	for i, v := range nums2 {
		reocrd[v] = i
	}
	for i := 0; i < len(nums1); i++ {
		flag := false
		for j := reocrd[nums1[i]]; j < len(nums2); j++ {
			if nums2[j] > nums1[i] {
				res = append(res, nums2[j])
				flag = true
				break
			}
		}
		if flag == false {
			res = append(res, -1)
		}
	}
	return res
}