LeetCode in Kotlin

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:

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:

Constraints:

Follow up: Could you find an O(nums1.length + nums2.length) solution?

Solution

class Solution {
    fun nextGreaterElement(nums1: IntArray, nums2: IntArray): IntArray {
        val indexMap: MutableMap<Int, Int> = HashMap()
        for (i in nums2.indices) {
            indexMap[nums2[i]] = i
        }
        for (i in nums1.indices) {
            val num = nums1[i]
            var index = indexMap[num]!!
            if (index == nums2.size - 1) {
                nums1[i] = -1
            } else {
                var found = false
                while (index < nums2.size) {
                    if (nums2[index] > num) {
                        nums1[i] = nums2[index]
                        found = true
                        break
                    }
                    index++
                }
                if (!found) {
                    nums1[i] = -1
                }
            }
        }
        return nums1
    }
}
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