3159. Find Occurrences of an Element in an Array

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3159. Find Occurrences of an Element in an Array

Description

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:

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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:

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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 <= 10^5
  • 1 <= queries[i] <= 10^5
  • 1 <= nums[i], x <= 10^4

Hints/Notes

  • Biweekly Contest 131

Solution

Language: C++

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class Solution {
public:
    vector<int> occurrencesOfElement(vector<int>& nums, vector<int>& queries,
                                     int x) {
        vector<int> occurances;
        for (int i = 0; i < nums.size(); i++) {
            if (nums[i] == x) {
                occurances.push_back(i);
            }
        }
        vector<int> res;
        for (auto query : queries) {
            if (query - 1 >= occurances.size()) {
                res.push_back(-1);
            } else {
                res.push_back(occurances[query - 1]);
            }
        }
        return res;
    }
};