Problem

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

Pre analysis

will keep a dictionary of all the indexes of x and will return the value from the dictionary for each query