Next Greater Element I - LeetCode : Swift

Next Greater Element I - LeetCode

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.

Solution

class Solution {
    func nextGreaterElement(_ num1: [Int], _ num2: [Int]) -> [Int] {
        var result: [Int] = []
        
        for i in 0..<num1.count {
            var x = num1[i]
            var resultAppended = false
            var numFound = false
            for j in 0..<num2.count {
                if (numFound && num2[j] > x){
                    resultAppended = true
                    result.append(num2[j])
                    break
                }

                if num2[j] == x {
                    numFound = true
                }
            }

            if !resultAppended {
                result.append(-1)
            }
        }

        return result
    }
}