Introduction

In this blog post, we will explore how to solve the LeetCode problem "3488" 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 a circular array nums and an array queries. For each query i, you have to find the following: The minimum distance between the element at index queries[i] and any other index j in the circular array, where nums[j] == nums[queries[i]]. If no such index exists, the answer for that query should be -1. Return an array answer of the same size as queries, where answer[i] represents the result for query i. Example 1: Input: nums = [1,3,1,4,1,3,2], queries = [0,3,5] Output: [2,-1,3] Explanation: Query 0: The element at queries[0] = 0 is nums[0] = 1. The nearest index with the same value is 2, and the distance between them is 2. Query 1: The element at queries[1] = 3 is nums[3] = 4. No other index contains 4, so the result is -1. Query 2: The element at queries[2] = 5 is nums[5] = 3. The nearest index with the same value is 1, and the distance between them is 3 (following the circular path: 5 -> 6 -> 0 -> 1). Example 2: Input: nums = [1,2,3,4], queries = [0,1,2,3] Output: [-1,-1,-1,-1] Explanation: Each value in nums is unique, so no index shares the same value as the queried element. This results in -1 for all queries. Constraints: 1 <= queries.length <= nums.length <= 105 1 <= nums[i] <= 106 0 <= queries[i] < nums.length

Explanation

Here's the breakdown of the solution:

• Index Mapping: Create a dictionary (hash map) to store the indices of each number in nums. This allows for quick lookups of all indices that have the same value as the query element.
• Circular Distance Calculation: For each query, iterate through the indices of the same value (obtained from the dictionary). Calculate the circular distance between the query index and each of the other indices.

• Minimum Distance: Find the minimum of these circular distances. If no other index with the same value exists, return -1.

• Runtime Complexity: O(N + Q * K), where N is the length of nums, Q is the length of queries, and K is the average number of occurrences of each element in nums. In the worst case, K could be N, leading to O(N^2) time complexity, but on average, K will be much smaller.

• Storage Complexity: O(N) to store the index map.

Code

    def circular_array_queries(nums, queries):
    """
    Finds the minimum circular distance between the element at each query index
    and any other index with the same value in the circular array nums.

    Args:
        nums: A list of integers representing the circular array.
        queries: A list of integers representing the query indices.

    Returns:
        A list of integers representing the minimum distances for each query.
        Returns -1 if no other index with the same value exists.
    """

    index_map = {}
    n = len(nums)

    # Create a dictionary to store indices of each number
    for i in range(n):
        if nums[i] not in index_map:
            index_map[nums[i]] = []
        index_map[nums[i]].append(i)

    results = []
    for query_index in queries:
        num = nums[query_index]
        min_distance = float('inf')
        found = False

        if num not in index_map:
            results.append(-1)
            continue

        indices = index_map[num]
        if len(indices) == 1:
            results.append(-1)
            continue


        for index in indices:
            if index != query_index:
                distance = abs(index - query_index)
                circular_distance = min(distance, n - distance)
                min_distance = min(min_distance, circular_distance)
                found = True

        if found:
            results.append(min_distance)
        else:
            results.append(-1)

    return results