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, 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

    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