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LeetCode 3159: Find Occurrences of an Element in an Array Solution

Master LeetCode problem 3159 (Find Occurrences of an Element in an Array), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3159. Find Occurrences of an Element in an Array

Medium
ArrayHash Table

Problem Explanation

Explanation:

To solve this problem, we can iterate through the nums array and maintain a hashmap to store the indices of occurrences of element x. Then, for each query, we can check the hashmap to find the corresponding index of the queries[i]th occurrence of x.

  1. Iterate through the nums array and store the indices of occurrences of x in a hashmap.
  2. For each query in the queries array, check if there are enough occurrences of x and return the corresponding index from the hashmap.
  3. If there are fewer occurrences of x, return -1 for that query.
  4. Return the array of answers.

Time Complexity:

  • O(N) where N is the length of the nums array, as we iterate through the array to build the hashmap and for each query.

Space Complexity:

  • O(N) to store the hashmap of indices.

:

Solution Code

class Solution {
    public int[] findOccurrences(int[] nums, int[] queries, int x) {
        Map<Integer, List<Integer>> indexMap = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == x) {
                indexMap.computeIfAbsent(nums[i], k -> new ArrayList<>()).add(i);
            }
        }
        
        int[] answer = new int[queries.length];
        for (int i = 0; i < queries.length; i++) {
            List<Integer> indices = indexMap.getOrDefault(x, new ArrayList<>());
            if (queries[i] <= indices.size()) {
                answer[i] = indices.get(queries[i] - 1);
            } else {
                answer[i] = -1;
            }
        }
        
        return answer;
    }
}

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