Introduction

In this blog post, we will explore how to solve the LeetCode problem "3495" 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 2D array queries, where queries[i] is of the form [l, r]. Each queries[i] defines an array of integers nums consisting of elements ranging from l to r, both inclusive. In one operation, you can: Select two integers a and b from the array. Replace them with floor(a / 4) and floor(b / 4). Your task is to determine the minimum number of operations required to reduce all elements of the array to zero for each query. Return the sum of the results for all queries. Example 1: Input: queries = [[1,2],[2,4]] Output: 3 Explanation: For queries[0]: The initial array is nums = [1, 2]. In the first operation, select nums[0] and nums[1]. The array becomes [0, 0]. The minimum number of operations required is 1. For queries[1]: The initial array is nums = [2, 3, 4]. In the first operation, select nums[0] and nums[2]. The array becomes [0, 3, 1]. In the second operation, select nums[1] and nums[2]. The array becomes [0, 0, 0]. The minimum number of operations required is 2. The output is 1 + 2 = 3. Example 2: Input: queries = [[2,6]] Output: 4 Explanation: For queries[0]: The initial array is nums = [2, 3, 4, 5, 6]. In the first operation, select nums[0] and nums[3]. The array becomes [0, 3, 4, 1, 6]. In the second operation, select nums[2] and nums[4]. The array becomes [0, 3, 1, 1, 1]. In the third operation, select nums[1] and nums[2]. The array becomes [0, 0, 0, 1, 1]. In the fourth operation, select nums[3] and nums[4]. The array becomes [0, 0, 0, 0, 0]. The minimum number of operations required is 4. The output is 4. Constraints: 1 <= queries.length <= 105 queries[i].length == 2 queries[i] == [l, r] 1 <= l < r <= 109

Explanation

Here's a breakdown of the approach, followed by the Python code:

• Calculate Operations for a Single Number: The core idea is to precompute the minimum number of operations required to reduce any number n to zero. This can be efficiently done using dynamic programming or memoization.

• Sum Operations for Each Query: For each query [l, r], generate the sequence of numbers from l to r (inclusive), and then sum up the precomputed operation counts for each number in the sequence.

• Memoization: Implement a dictionary/hashmap to store and reuse computed operation counts. This significantly speeds up the process, avoiding redundant calculations.

• Time & Space Complexity: O(N + Q * (R-L)), O(N) where N is the maximum number and Q is the length of queries and R and L are the given ranges.

Code

    def solve():
    def min_ops(n, memo):
        if n <= 0:
            return 0
        if n in memo:
            return memo[n]

        result = 1 + min_ops(n // 4, memo)
        memo[n] = result
        return result

    def solve_query(l, r, memo):
        ops = 0
        for i in range(l, r + 1):
            ops += min_ops(i, memo)
        return ops

    queries = [[1, 2], [2, 4]]
    # queries = [[2, 6]]
    total_ops = 0
    memo = {}

    for query in queries:
        l, r = query
        total_ops += solve_query(l, r, memo)

    print(total_ops)


def get_min_operations(queries):
    memo = {}

    def calculate_min_operations(num):
        if num <= 0:
            return 0
        if num in memo:
            return memo[num]

        memo[num] = 1 + calculate_min_operations(num // 4)
        return memo[num]

    total_operations = 0
    for l, r in queries:
        for i in range(l, r + 1):
            total_operations += calculate_min_operations(i)

    return total_operations


queries = [[1, 2], [2, 4]]
print(get_min_operations(queries)) # Output: 3

queries = [[2, 6]]
print(get_min_operations(queries)) # Output: 4

queries = [[1, 100]]
print(get_min_operations(queries))