LeetCode in Kotlin

3159. Find Occurrences of an Element in an Array

Medium

You are given an integer array nums, an integer array queries, and an integer x.

For each queries[i], you need to find the index of the queries[i]th occurrence of x in the nums array. If there are fewer than queries[i] occurrences of x, the answer should be -1 for that query.

Return an integer array answer containing the answers to all queries.

Example 1:

Input: nums = [1,3,1,7], queries = [1,3,2,4], x = 1

Output: [0,-1,2,-1]

Explanation:

• For the 1^st query, the first occurrence of 1 is at index 0.
• For the 2^nd query, there are only two occurrences of 1 in nums, so the answer is -1.
• For the 3^rd query, the second occurrence of 1 is at index 2.
• For the 4^th query, there are only two occurrences of 1 in nums, so the answer is -1.

Example 2:

Input: nums = [1,2,3], queries = [10], x = 5

Output: [-1]

Explanation:

• For the 1^st query, 5 doesn’t exist in nums, so the answer is -1.

Constraints:

• 1 <= nums.length, queries.length <= 105
• 1 <= queries[i] <= 105
• 1 <= nums[i], x <= 104

Solution

class Solution {
    fun occurrencesOfElement(nums: IntArray, queries: IntArray, x: Int): IntArray {
        val a = ArrayList<Int>()
        run {
            var i = 0
            val l = nums.size
            while (i < l) {
                if (nums[i] == x) {
                    a.add(i)
                }
                i++
            }
        }
        val l = queries.size
        val r = IntArray(l)
        for (i in 0 until l) {
            r[i] = if (queries[i] > a.size) -1 else a[queries[i] - 1]
        }
        return r
    }
}

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