Minimum Array Sum

Naive Solution

A brute-force solution would involve trying all possible combinations of applying operation 1 and operation 2 to each element in the nums array. This means exploring every subset of operations and calculating the resulting sum. This approach is highly inefficient because it leads to exponential time complexity due to the vast search space.

Optimal Solution

We can use a greedy approach combined with sorting. The idea is to prioritize applying operations that yield the greatest reduction in the sum of the array elements at each step.

1. Sort the array: Sort the nums array in descending order. This helps us focus on the largest elements first, which are more likely to benefit significantly from the operations.
2. Apply Operation 2: Iterate through the sorted array, applying operation 2 (subtracting k) to elements greater than or equal to k until op2 becomes 0.
3. Apply Operation 1: After applying operation 2, iterate through the array again, applying operation 1 (dividing by 2 and rounding up) until op1 becomes 0.
4. Calculate the Sum: Finally, calculate and return the sum of the modified array.

Edge Cases

• Empty array: If nums is empty, return 0.
• k = 0: If k is 0, operation 2 has no effect, so we only focus on operation 1.
• op1 = 0 and op2 = 0: If both operation counts are 0, simply return the sum of the original array.
• op1 or op2 > nums.length: We should not apply the operations more than once on each element. Thus, limit it to nums.length.

Code Implementation (Python)

import math

def min_sum(nums, k, op1, op2):
    n = len(nums)
    nums = sorted(nums, reverse=True)
    
    # Apply Operation 2
    for i in range(min(n, op2)):
        if nums[i] >= k:
            nums[i] -= k
        else:
            break

    # Apply Operation 1
    nums = sorted(nums, reverse=True) #sort again after op2
    for i in range(min(n, op1)):
        nums[i] = math.ceil(nums[i] / 2)

    return sum(nums)

# Example Usage
nums = [2, 8, 3, 19, 3]
k = 3
op1 = 1
op2 = 1
print(min_sum(nums, k, op1, op2))  # Output: 23

nums = [2, 4, 3]
k = 3
op1 = 2
op2 = 1
print(min_sum(nums, k, op1, op2))  # Output: 3

Time and Space Complexity

• Time Complexity: O(n log n) due to sorting the array. The iterations for applying operations are O(n), but since sorting dominates, the overall time complexity is O(n log n).
• Space Complexity: O(1) (or O(n) depending on the sorting algorithm used) since we modify the array in-place. Some sorting algorithms may require additional space, but in many common implementations, the space complexity is constant.