Introduction

In this blog post, we will explore how to solve the LeetCode problem "3091" 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 positive integer k. Initially, you have an array nums = [1]. You can perform any of the following operations on the array any number of times (possibly zero): Choose any element in the array and increase its value by 1. Duplicate any element in the array and add it to the end of the array. Return the minimum number of operations required to make the sum of elements of the final array greater than or equal to k. Example 1: Input: k = 11 Output: 5 Explanation: We can do the following operations on the array nums = [1]: Increase the element by 1 three times. The resulting array is nums = [4]. Duplicate the element two times. The resulting array is nums = [4,4,4]. The sum of the final array is 4 + 4 + 4 = 12 which is greater than or equal to k = 11. The total number of operations performed is 3 + 2 = 5. Example 2: Input: k = 1 Output: 0 Explanation: The sum of the original array is already greater than or equal to 1, so no operations are needed. Constraints: 1 <= k <= 105

Explanation

Here's the approach:

• Greedy Approach: The core idea is to maximize the impact of each operation. Duplicating a larger number contributes more to the sum than duplicating a smaller number. Similarly, increasing a number that will be duplicated later is more beneficial than increasing it without duplication.
• Optimal Strategy: Increase the initial element until a certain point, then duplicate it to reach or exceed k. We need to find the optimal balance between increasing the initial element and duplicating it. Binary search can be used to find this optimal point efficiently.

• Calculate Operations: The total operations will be the sum of increment operations and duplication operations.

• Runtime & Storage Complexity: O(log k) time complexity, O(1) space complexity.

Code

    def minOperations(k: int) -> int:
    """
    Calculates the minimum number of operations to make the sum of elements in an array >= k.

    Args:
        k: The target sum.

    Returns:
        The minimum number of operations.
    """

    if k <= 1:
        return 0

    left, right = 0, k  # Possible values for the increased initial element
    ans = float('inf')

    while left <= right:
        mid = (left + right) // 2

        # Number of duplicates needed
        num_duplicates = (k + mid - 1) // mid - 1

        # Total operations
        operations = mid - 1 + num_duplicates

        ans = min(ans, operations)

        # Adjust the search range
        if num_duplicates > mid - 1 : # We need more duplicates, increase mid
            left = mid + 1
        else:
            right = mid - 1

    return ans