Understanding the Problem

We're given a positive integer k and an initial array nums = [1]. Our goal is to find the minimum number of operations needed to make the sum of the elements in nums greater than or equal to k. The allowed operations are incrementing an element and duplicating an element.

Naive Approach (Brute Force)

A brute force approach would involve exploring all possible combinations of incrementing and duplicating elements until the sum of the array is greater than or equal to k. However, this would lead to exponential time complexity and is highly inefficient.

Optimal Approach (Greedy)

A more efficient approach is a greedy algorithm. The idea is to prioritize operations based on their impact on the sum and number of elements. Here's the breakdown:

1. Maximize the Value: Increase the initial element to a value that helps reach k faster. This will reduce the duplication steps needed.
2. Efficient Duplication: After maximizing the value, duplicate it as many times as necessary to reach the target sum k.

Algorithm

1. Calculate the target value each element must have: target = ceil(k / size) where size is number of elements. We start with a size of 1.
2. Compute increment operations needed: increments = target - 1.
3. Compute duplication operations: duplicates = size - 1.

Edge Cases

• k = 1: No operations are needed.

Code Implementation

import math

def min_operations(k: int) -> int:
    if k == 1:
        return 0

    operations = 0
    num = 1
    sum_val = 1
    count = 1

    while sum_val < k:
        increase_ops = k - sum_val
        duplicate_ops = math.ceil((k / num) - count)

        if increase_ops < duplicate_ops:
            diff = k - sum_val
            operations += diff
            sum_val += diff
        else:
            operations += 1
            sum_val += num
            num = sum_val/count
            count+=1

    return operations

def min_operations_optimal(k: int) -> int:
    if k == 1:
        return 0

    operations = 0
    num = 1
    sum_val = 1
    count = 1

    while sum_val < k:
        diff = (k + count - 1) // count - num
        if diff > 0:
            operations += diff
            sum_val += diff * count
            num += diff
        else:
            operations += 1
            sum_val += num
            count += 1


    return operations

Example

For k = 11:

1. Initialize operations = 0, num = 1, sum_val = 1, count = 1.
2. The while loop continues since sum_val (1) < k (11). We increase the initial number by the target value: target = 4. increase operations is 3.
3. Now, operations = 3, sum_val = 4, duplicate two times to reach the target: duplicate operations are 2.
4. Now the sum_val = 12 and operations = 5.

Time and Space Complexity

• Time Complexity: O(sqrt(k)).
• Space Complexity: O(1).