Introduction

In this blog post, we will explore how to solve the LeetCode problem "1558" 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. You have an integer array arr of the same length with all values set to 0 initially. You also have the following modify function: You want to use the modify function to convert arr to nums using the minimum number of calls. Return the minimum number of function calls to make nums from arr. The test cases are generated so that the answer fits in a 32-bit signed integer. Example 1: Input: nums = [1,5] Output: 5 Explanation: Increment by 1 (second element): [0, 0] to get [0, 1] (1 operation). Double all the elements: [0, 1] -> [0, 2] -> [0, 4] (2 operations). Increment by 1 (both elements) [0, 4] -> [1, 4] -> [1, 5] (2 operations). Total of operations: 1 + 2 + 2 = 5. Example 2: Input: nums = [2,2] Output: 3 Explanation: Increment by 1 (both elements) [0, 0] -> [0, 1] -> [1, 1] (2 operations). Double all the elements: [1, 1] -> [2, 2] (1 operation). Total of operations: 2 + 1 = 3. Example 3: Input: nums = [4,2,5] Output: 6 Explanation: (initial)[0,0,0] -> [1,0,0] -> [1,0,1] -> [2,0,2] -> [2,1,2] -> [4,2,4] -> 4,2,5. Constraints: 1 <= nums.length <= 105 0 <= nums[i] <= 109

Explanation

Here's a solution to the problem:

• Key Idea: The core idea is to work backward from the target array nums. We prioritize doubling operations as they affect multiple elements simultaneously. Count the increment operations needed in each step when even elements become odd, prior to dividing by 2.

• Algorithm: Iteratively divide all even numbers in the array by 2 and count the number of doubling operations. Sum the difference between the original nums and the divided nums and save to count increment operations. Finally sum the doubling operations and the increment operations.

• Complexity: The runtime complexity is O(N * log(max(nums))), where N is the length of the array, and the log factor comes from the maximum number of times an element can be divided by 2. The storage complexity is O(1).

Code

    def minOperations(nums):
    """
    Calculates the minimum number of operations to transform an array of zeros to the given array nums
    using increment and double operations.

    Args:
        nums: A list of integers representing the target array.

    Returns:
        The minimum number of operations required.
    """

    ans = 0
    doubles = 0
    while True:
        increment = 0
        all_zeros = True
        can_double = True

        for i in range(len(nums)):
            if nums[i] != 0:
                all_zeros = False

            if nums[i] % 2 != 0:
                nums[i] -= 1
                increment += 1

            if nums[i] % 2 != 0 and nums[i] !=0 :
              can_double = False

        ans += increment

        if all_zeros:
            break

        if can_double:
          doubles +=1
          for i in range(len(nums)):
              nums[i] //= 2

    return ans + doubles