2357. Make Array Zero by Subtracting Equal Amounts - Detailed Explanation

Problem Statement

Description:
You are given an integer array nums. In one operation, you can choose a positive integer x and subtract x from every nonzero element in nums. Your goal is to make every element in the array equal to zero using the minimum number of operations.

Operation Details:

• In one operation, the same positive integer x is subtracted from each nonzero element in the array.
• Zeros remain unchanged.

Constraints:

• 1 ≤ nums.length ≤ 100
• 0 ≤ nums[i] ≤ 100

Example 1:

• Input: nums = [1, 5, 0, 3, 5]
• Output: 3
• Explanation:
  1. Operation 1: Choose x = 1 (the smallest nonzero value).
     • New array: [0, 4, 0, 2, 4]
  2. Operation 2: Now, the smallest nonzero value is 2. Choose x = 2.
     • New array: [0, 2, 0, 0, 2]
  3. Operation 3: Finally, choose x = 2 again (since 2 is the only distinct nonzero value left).
     • New array: [0, 0, 0, 0, 0]
       Total operations = 3.

Example 2:

• Input: nums = [0, 0, 0]
• Output: 0
• Explanation:
  All elements are already zero, so no operations are needed.

Example 3:

• Input: nums = [2, 2, 2]
• Output: 1
• Explanation:
  One operation with x = 2 subtracts 2 from each element, resulting in [0, 0, 0].

Hints

• Hint 1: Realize that the optimal strategy is to always subtract the smallest nonzero number from all nonzero elements. This operation will reduce at least one distinct number in the array to zero.
• Hint 2: Notice that the order in which you subtract doesn’t matter—the number of operations is directly related to the number of distinct nonzero elements present in the array.
• Hint 3: Consider that after subtracting the smallest nonzero element, some numbers might become zero, but the remaining positive numbers still retain their relative differences. This observation suggests that the final answer is simply the count of distinct positive numbers in the array.

Approaches

Brute Force (Simulation) Approach

Idea:
Simulate the process by repeatedly finding the smallest nonzero element, subtracting it from all nonzero elements, and counting each operation until every element becomes zero.

Steps:

1. While there is any nonzero element in nums:
   • Find the minimum nonzero element, say m.
   • Subtract m from every element in the array that is nonzero.
   • Count this as one operation.
2. Continue until all elements are zero.

Complexity:

• Time Complexity: O(n²) in the worst case (each operation involves scanning and updating the array).
• Space Complexity: O(1) extra space (if done in place).

Visual Example (using Example 1):
For nums = [1, 5, 0, 3, 5]:

• First Operation: Subtract 1 → [0, 4, 0, 2, 4]
• Second Operation: Subtract 2 → [0, 2, 0, 0, 2]
• Third Operation: Subtract 2 → [0, 0, 0, 0, 0]

Optimal Approach: Count Distinct Nonzero Elements

Idea:
A key observation is that subtracting the minimum nonzero element from all nonzero numbers reduces one distinct number from the array. In other words, no matter how you perform the operations, you need one operation for every distinct positive number in the array.

Steps:

1. Traverse the array and add every nonzero element to a set (to automatically filter out duplicates).
2. The size of this set represents the minimum number of operations required.
3. Return the size of the set.

Complexity:

• Time Complexity: O(n) – one pass through the array.
• Space Complexity: O(n) – in the worst case, if all elements are distinct.

Why It Works:

• Each operation effectively “removes” one unique nonzero value from the array.
• Zeros are ignored because they are already at the target value.

Detailed Walkthrough (Optimal Approach)

Consider the array: nums = [1, 5, 0, 3, 5]

1. Initialization:
   • Create an empty set to store distinct nonzero elements.
2. Iteration:
   • Look at each element in nums:
     • For 1: Since 1 > 0, add 1 to the set → set becomes {1}.
     • For 5: 5 > 0, add 5 → set becomes {1, 5}.
     • For 0: Skip (already zero).
     • For 3: 3 > 0, add 3 → set becomes {1, 3, 5}.
     • For 5: Already in the set, so skip.
3. Result:
   • The set has 3 elements: {1, 3, 5}.
   • Therefore, the minimum number of operations is 3.

Code Implementation

Python Code