You are given a non-negative integer array nums. In one operation, you must:

1. Choose a positive integer x such that x is less than or equal to the smallest non-zero element in nums.
2. Subtract x from every positive element in nums.

Return the minimum number of operations to make every element in nums equal to 0.

For example, given nums = [1, 5, 0, 3, 5], the output should be 3. Here's why:

• In the first operation, choose x = 1. Now, nums = [0, 4, 0, 2, 4]. The smallest non-zero element was 1 and we subtracted it from every positive element.
• In the second operation, choose x = 2. Now, nums = [0, 2, 0, 0, 2]. The smallest non-zero element was 2 and we subtracted it from every positive element.
• In the third operation, choose x = 2. Now, nums = [0, 0, 0, 0, 0]. The smallest non-zero element was 2 and we subtracted it from every positive element.

As another example, if nums = [0], the output should be 0 because each element in nums is already 0 so no operations are needed. Write an algorithm to efficiently calculate the minimum number of operations needed to make every element in nums equal to 0.

Question

You are given a non-negative integer array nums. In one operation, you must:

• Choose a positive integer x such that x is less than or equal to the smallest non-zero element in nums.
• Subtract x from every positive element in nums.

Return the minimum number of operations to make every element in nums equal to 0.

Example

Input: nums = [1,5,0,3,5]
Output: 3
Explanation:
In the first operation, choose x = 1. Now, nums = [0,4,0,2,4].
In the second operation, choose x = 2. Now, nums = [0,2,0,0,2].
In the third operation, choose x = 2. Now, nums = [0,0,0,0,0].

Naive Approach

The naive approach would be to simulate the operations as described in the problem statement. We repeatedly find the smallest non-zero element in the array, and then subtract that value from all positive elements. We continue this process until all elements are zero. This will definitely give us the correct answer, but it will be slow if the array contains large numbers and small differences between the elements.

def min_operations_naive(nums):
    operations = 0
    while any(nums):
        smallest_positive = float('inf')
        for num in nums:
            if num > 0 and num < smallest_positive:
                smallest_positive = num
        
        if smallest_positive == float('inf'): # no positive number
            break
        
        for i in range(len(nums)):
            if nums[i] > 0:
                nums[i] -= smallest_positive
        operations += 1
    return operations

Optimal Approach

The key observation is that the number of operations is equal to the number of distinct positive numbers in the array. We can prove this by induction. The base case is trivial. Assume the statement is true for an array with k distinct positive numbers. If we have k+1 distinct positive numbers, let x be the smallest of those numbers. After subtracting x from all positive numbers, the number x becomes zero, and all other positive numbers are reduced by x. Now we only have k distinct positive numbers to remove. Thus, by induction, we only need k + 1 steps.

def min_operations_optimal(nums):
    return len(set(num for num in nums if num > 0))

Big O Runtime Analysis

The optimal solution uses a set to keep track of distinct values. Iterating through the array takes O(n) time. Creating the set also takes O(n) time in the worst case, where n is the length of the input array nums. Therefore, the overall time complexity is O(n).

The naive solution, in the worst case, can have a runtime of O(m*n), where n is the length of the array and m is the value of the largest number in the array. This is because it would loop through the entire array m times.

Big O Space Usage Analysis

The optimal solution uses a set to store the distinct positive numbers. In the worst case, all numbers are distinct and positive, so the set will store n numbers, where n is the length of nums. Therefore, the space complexity is O(n).

The naive solution has a space complexity of O(1) because it modifies the input array in-place.

Edge Cases

1. Empty array: The problem statement specifies that 1 <= nums.length <= 100, so we don't need to handle an empty array.
2. Array with only zeros: If the array contains only zeros, the function should return 0. The provided optimal code handles this correctly because the set comprehension will be empty, resulting in len(set()) == 0.
3. Array with negative numbers: The problem statement specifies that the array contains non-negative integers, so we don't need to handle negative numbers.
4. Array with large numbers and small differences: The optimal solution handles this case efficiently because it doesn't depend on the magnitude of the numbers, only on the number of distinct positive values.