Leetcode 2357: Make Array Zero by Subtracting Equal Amounts

Input: The input consists of an array nums where each element is a non-negative integer.

Example: Input: nums = [1, 5, 0, 3, 5]

Constraints:

• 1 <= nums.length <= 100

• 0 <= nums[i] <= 100

Output: Return an integer representing the minimum number of operations required.

Example: Output: 3

Constraints:

• The output will always be a non-negative integer.

Goal: Sort the input array and calculate the number of operations based on distinct positive elements.

Steps:

• Sort the input array.

• Iterate over the array and count how many distinct non-zero elements exist.

• For each distinct positive element, consider an operation to reduce those elements to zero.

Goal: The size of the array is small enough (n ≤ 100) to apply sorting and iteration methods.

Steps:

• The length of nums will always be between 1 and 100.

• Each element in nums will be between 0 and 100.

Assumptions:

• All elements in nums are non-negative integers.

• At least one element in nums is positive.

• Input: Example 1: nums = [1, 5, 0, 3, 5]

• Explanation: In this case, we perform three operations: subtract 1 from all elements (resulting in [0, 4, 0, 2, 4]), subtract 2 from all elements (resulting in [0, 2, 0, 0, 2]), and finally subtract 2 again (resulting in [0, 0, 0, 0, 0]). Thus, the output is 3.

Approach: The approach is to sort the array and calculate the number of operations based on distinct positive elements in the array.

Observations:

• We can sort the array to easily track the smallest non-zero element.

• We need to perform operations for each distinct positive element in the array.

Steps:

• Sort the array of numbers.

• Iterate through the array, checking for distinct positive elements.

• For each new distinct positive number, perform an operation and increase the operation count.

Empty Inputs:

• If the array has only zeroes, the output should be 0.

Large Inputs:

• Even with the largest possible input size (100 elements), sorting and iterating through the array will be efficient enough.

Special Values:

• If all elements are zero, no operation is needed.

Constraints:

• Ensure that only positive elements are considered when counting operations.

int minimumOperations(vector<int>& nums) {
    sort(nums.begin(), nums.end());
    int mx = 0, res = 0, diff;
    for(int i = 0; i < nums.size(); i++) {
        if(mx < nums[i]) {
            mx = nums[i];
            res++;
        }
    }
    return res;
}

1 : Function Definition

int minimumOperations(vector<int>& nums) {

Defines the function `minimumOperations`, which takes a vector of integers `nums` and returns an integer representing the minimum operations needed.

2 : Sorting

    sort(nums.begin(), nums.end());

Sorts the vector `nums` in ascending order to facilitate comparison between the current element and the maximum element.

3 : Variable Initialization

    int mx = 0, res = 0, diff;

Initializes `mx` to track the maximum number encountered, `res` to count the number of operations, and `diff` which is unused in this case.

4 : Loop

    for(int i = 0; i < nums.size(); i++) {

Loops through each element of the sorted `nums` array.

5 : Condition Check

        if(mx < nums[i]) {

Checks if the current number is greater than the previously encountered maximum `mx`.

6 : Update Maximum

            mx = nums[i];

Updates `mx` to the current number `nums[i]` if it is greater.

7 : Increment Operation Count

            res++;

Increments the operation count `res` each time a new maximum is found, implying a change operation.

8 : Return Result

    return res;

Returns the total number of operations performed.

Best Case: O(n log n)

Average Case: O(n log n)

Worst Case: O(n log n)

Description: The worst-case time complexity is O(n log n) due to sorting the array.

Best Case: O(n)

Worst Case: O(n)

Description: The space complexity is O(n) due to the array being stored in memory.

Leetcode 2357: Make Array Zero by Subtracting Equal Amounts

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