Introduction

In this blog post, we will explore how to solve the LeetCode problem "2602" 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 array nums consisting of positive integers. You are also given an integer array queries of size m. For the ith query, you want to make all of the elements of nums equal to queries[i]. You can perform the following operation on the array any number of times: Increase or decrease an element of the array by 1. Return an array answer of size m where answer[i] is the minimum number of operations to make all elements of nums equal to queries[i]. Note that after each query the array is reset to its original state. Example 1: Input: nums = [3,1,6,8], queries = [1,5] Output: [14,10] Explanation: For the first query we can do the following operations: - Decrease nums[0] 2 times, so that nums = [1,1,6,8]. - Decrease nums[2] 5 times, so that nums = [1,1,1,8]. - Decrease nums[3] 7 times, so that nums = [1,1,1,1]. So the total number of operations for the first query is 2 + 5 + 7 = 14. For the second query we can do the following operations: - Increase nums[0] 2 times, so that nums = [5,1,6,8]. - Increase nums[1] 4 times, so that nums = [5,5,6,8]. - Decrease nums[2] 1 time, so that nums = [5,5,5,8]. - Decrease nums[3] 3 times, so that nums = [5,5,5,5]. So the total number of operations for the second query is 2 + 4 + 1 + 3 = 10. Example 2: Input: nums = [2,9,6,3], queries = [10] Output: [20] Explanation: We can increase each value in the array to 10. The total number of operations will be 8 + 1 + 4 + 7 = 20. Constraints: n == nums.length m == queries.length 1 <= n, m <= 105 1 <= nums[i], queries[i] <= 109

Explanation

Here's the breakdown of the solution:

• Sorting and Binary Search: Sort the nums array to enable efficient searching. For each query, use binary search to find the insertion point of the query value in the sorted nums array. This allows us to efficiently determine the number of elements smaller and larger than the query.
• Prefix Sum Optimization: Calculate the prefix sum of the sorted nums array. This enables us to compute the sum of elements less than and greater than the query value in O(1) time using the insertion point obtained from binary search.

• Calculating Operations: Using the prefix sum and the insertion point, calculate the total number of operations required to make all elements equal to the query value. This involves calculating the sum of differences between the query value and elements smaller than it, and the sum of differences between elements larger than the query value and the query value.

• Runtime Complexity: O(n log n + m log n), where n is the length of nums and m is the length of queries. Sorting nums takes O(n log n), and each query requires a binary search (O(log n)).

• Storage Complexity: O(n) for the sorted nums array and the prefix sum array.

Code

    def minOperations(nums, queries):
    nums.sort()
    n = len(nums)
    prefix_sum = [0] * (n + 1)
    for i in range(n):
        prefix_sum[i + 1] = prefix_sum[i] + nums[i]

    def calculate_operations(query):
        l, r = 0, n - 1
        index = n
        while l <= r:
            mid = (l + r) // 2
            if nums[mid] >= query:
                index = mid
                r = mid - 1
            else:
                l = mid + 1

        less_sum = prefix_sum[index]
        greater_sum = prefix_sum[n] - prefix_sum[index]

        less_ops = query * index - less_sum
        greater_ops = greater_sum - query * (n - index)

        return less_ops + greater_ops

    result = []
    for query in queries:
        result.append(calculate_operations(query))
    return result