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Leetcode 2357. Make Array Zero by Subtracting Equal Amounts

Problem Statement

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

1. Choose a non-negative integer x that is present in the array.
2. Subtract x from every positive element of the array.

Return the minimum number of operations required to make every element of the array equal to 0.

Clarifying Questions

1. What constraints are given for the array size?
   • There are no specific size constraints mentioned, but it can be inferred it fits typical problem constraints (e.g., within a few thousand elements).
2. Is the array guaranteed to be non-empty?
   • Yes, the array is guaranteed to be non-empty.
3. Can the array have duplicate elements?
   • Yes, the array can have duplicate elements.
4. Will every element in the array be a non-negative integer?
   • Yes, every element is a non-negative integer, as mentioned in the problem statement.

Strategy

The idea is based on the observation that each unique positive number in the array must be used at least once as x in the operations to eventually reduce all elements to zero.

Key Insights:

• Each unique non-zero value needs to be subtracted at least once to ensure all elements become zero.
• Thus, the minimum number of operations is equivalent to the number of unique non-zero elements in the array.

Step-by-Step Plan

1. Use a set to keep track of unique non-zero elements in the array.
2. The size of the set will give the number of unique non-zero elements.
3. Return the size of the set as the result.

Time Complexity

• O(n) where n is the number of elements in the array. This is because we iterate through the array once to populate the set.
• Space complexity is also O(n) in the worst case where all elements are unique and non-zero.

Code

#include <vector>
#include <unordered_set>

int minimumOperations(std::vector<int>& nums) {
    std::unordered_set<int> uniqueValues;
    
    for (int num : nums) {
        // Insert the number into the set if it's not zero
        if (num != 0) {
            uniqueValues.insert(num);
        }
    }
    
    // The size of the set represents the number of different values subtracted.
    return uniqueValues.size();
}

Explanation

• We use an unordered_set to keep track of unique non-zero elements.
• We iterate through the array, adding each non-zero element to the set.
• The size of the set at the end of iteration gives us the number of unique values, which is our answer.

This approach ensures we count each unique value only once, giving us the minimum number of required operations.

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