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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 each operation, you must:

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

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

Example:

  1. Input: nums = [1,5,0,3,5] Output: 3

  2. Input: nums = [0] Output: 0

Clarifying Questions

  1. What is the length range of the array nums?
    • The length of array nums could be anywhere from 0 to 10^5.
  2. What values can the elements of nums take?
    • Elements of nums are non-negative integers and can range from 0 to 10^9.
  3. What should be returned if nums is already all zeroes?
    • If nums is already all zeroes, the number of operations required is 0.
  4. Can nums contain duplicate numbers?
    • Yes, nums can contain duplicate numbers.

Strategy

To solve the problem, we need to make every element in the array nums equal to 0 using the least number of operations described.

One efficient approach is to understand that each unique positive number in the array nums must eventually be reduced to zero. Hence, the number of operations required to make the entire array zero is equivalent to the number of unique positive integers in the array.

Key Steps:

  1. Create a set of all unique positive integers from nums.
  2. The size of this set will be the number of operations required.

Code

import java.util.HashSet;
import java.util.Set;

public class MakeArrayZero {
    public int minimumOperations(int[] nums) {
        Set<Integer> uniquePositives = new HashSet<>();
        
        for (int num : nums) {
            if (num > 0) {
                uniquePositives.add(num);
            }
        }
        
        // The count of unique positive numbers is the number of operations required.
        return uniquePositives.size();
    }

    public static void main(String[] args) {
        MakeArrayZero maZero = new MakeArrayZero();
        
        int[] nums1 = {1, 5, 0, 3, 5};
        System.out.println(maZero.minimumOperations(nums1)); // Output: 3
        
        int[] nums2 = {0};
        System.out.println(maZero.minimumOperations(nums2)); // Output: 0
    }
}

Time Complexity

The total time complexity of this approach is O(n) where n is the number of elements in the array nums.

Hence, the overall time complexity is O(n), and this approach is efficient given the constraints of the problem.

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