Introduction

In this blog post, we will explore how to solve the LeetCode problem "1906" 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

The minimum absolute difference of an array a is defined as the minimum value of |a[i] - a[j]|, where 0 <= i < j < a.length and a[i] != a[j]. If all elements of a are the same, the minimum absolute difference is -1. For example, the minimum absolute difference of the array [5,2,3,7,2] is |2 - 3| = 1. Note that it is not 0 because a[i] and a[j] must be different. You are given an integer array nums and the array queries where queries[i] = [li, ri]. For each query i, compute the minimum absolute difference of the subarray nums[li...ri] containing the elements of nums between the 0-based indices li and ri (inclusive). Return an array ans where ans[i] is the answer to the ith query. A subarray is a contiguous sequence of elements in an array. The value of |x| is defined as: x if x >= 0. -x if x < 0. Example 1: Input: nums = [1,3,4,8], queries = [[0,1],[1,2],[2,3],[0,3]] Output: [2,1,4,1] Explanation: The queries are processed as follows: - queries[0] = [0,1]: The subarray is [1,3] and the minimum absolute difference is |1-3| = 2. - queries[1] = [1,2]: The subarray is [3,4] and the minimum absolute difference is |3-4| = 1. - queries[2] = [2,3]: The subarray is [4,8] and the minimum absolute difference is |4-8| = 4. - queries[3] = [0,3]: The subarray is [1,3,4,8] and the minimum absolute difference is |3-4| = 1. Example 2: Input: nums = [4,5,2,2,7,10], queries = [[2,3],[0,2],[0,5],[3,5]] Output: [-1,1,1,3] Explanation: The queries are processed as follows: - queries[0] = [2,3]: The subarray is [2,2] and the minimum absolute difference is -1 because all the elements are the same. - queries[1] = [0,2]: The subarray is [4,5,2] and the minimum absolute difference is |4-5| = 1. - queries[2] = [0,5]: The subarray is [4,5,2,2,7,10] and the minimum absolute difference is |4-5| = 1. - queries[3] = [3,5]: The subarray is [2,7,10] and the minimum absolute difference is |7-10| = 3. Constraints: 2 <= nums.length <= 105 1 <= nums[i] <= 100 1 <= queries.length <= 2 * 104 0 <= li < ri < nums.length

Explanation

Here's a breakdown of the solution:

• Leverage Constraints: The constraint that 1 <= nums[i] <= 100 is crucial. It allows us to use a counting-based approach for each query.
• Frequency Counting: For each query, we'll create a frequency count array (or dictionary) for the subarray specified by the query.

• Efficient Min-Diff Calculation: Iterate through the frequency count to find the minimum absolute difference, only comparing elements that actually exist in the subarray.

• Runtime Complexity: O(q n), where q is the number of queries and n is the length of the subarray (ri - li + 1) for each query. In the worst-case scenario each query iterates across the entire array nums, but given the constraints on nums[i] the inner loop for calculating the diff is O(1). Therefore O(qn).

• Storage Complexity: O(1) because the frequency count array is fixed at size 101.

Code

    def minDifference(nums, queries):
    ans = []
    for l, r in queries:
        sub_array = nums[l:r+1]
        counts = {}
        for x in sub_array:
            counts[x] = counts.get(x, 0) + 1

        unique_nums = sorted(counts.keys())

        if len(unique_nums) < 2:
            ans.append(-1)
        else:
            min_diff = float('inf')
            for i in range(len(unique_nums) - 1):
                min_diff = min(min_diff, unique_nums[i+1] - unique_nums[i])
            ans.append(min_diff)
    return ans