# Solving Leetcode Interviews in Seconds with AI: Find Occurrences of an Element in an Array


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3159" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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  

	# Explanation
	*   **Preprocessing:** Create a hash map (dictionary in Python) to store the indices of each occurrence of `x` in `nums`. The keys of the hash map will be `x`, and the values will be lists of indices.
*   **Querying:** For each query, check if `x` exists as a key in the hash map. If it does, check if the requested occurrence index (queries\[i] - 1) is within the bounds of the list of indices for `x`. If it is, return the index. Otherwise, return -1.
*   **Efficiency:** The preprocessing step allows us to answer each query in O(1) time after the initial O(n) preprocessing, where n is the length of nums.

*   **Runtime Complexity:** O(n + m), where n is the length of `nums` and m is the length of `queries`.  **Storage Complexity:** O(n) in the worst case (when all elements in `nums` are equal to `x`).

	
	# Code
	```python
	def find_indices(nums, queries, x):
    """
    Finds the index of the queries[i]th occurrence of x in the nums array.

    Args:
        nums: An integer array.
        queries: An integer array.
        x: An integer.

    Returns:
        An integer array containing the answers to all queries.
    """

    indices = {}
    for i, num in enumerate(nums):
        if num == x:
            if x not in indices:
                indices[x] = []
            indices[x].append(i)

    result = []
    for query in queries:
        if x not in indices:
            result.append(-1)
        else:
            if query <= len(indices[x]):
                result.append(indices[x][query - 1])
            else:
                result.append(-1)

    return result
	```