2357. Make Array Zero by Subtracting Equal Amounts 

Description

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 1:

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].

Example 2:

Input: nums = [0]
Output: 0
Explanation: Each element in nums is already 0 so no operations are needed.

Constraints:

1 <= nums.length <= 100
0 <= nums[i] <= 100

Solutions

Solution 1: Hash Table or Array

We observe that in each operation, all identical nonzero elements in the array $\textit{nums}$ can be reduced to $0$. Therefore, we only need to count the number of distinct nonzero elements in $\textit{nums}$, which is the minimum number of operations required. To count the distinct nonzero elements, we can use a hash table or an array.

The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$.

Python3

class Solution:
    def minimumOperations(self, nums: List[int]) -> int:
        return len({x for x in nums if x})

Java

class Solution {
    public int minimumOperations(int[] nums) {
        boolean[] s = new boolean[101];
        s[0] = true;
        int ans = 0;
        for (int x : nums) {
            if (!s[x]) {
                ++ans;
                s[x] = true;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int minimumOperations(vector<int>& nums) {
        bool s[101]{};
        s[0] = true;
        int ans = 0;
        for (int& x : nums) {
            if (!s[x]) {
                ++ans;
                s[x] = true;
            }
        }
        return ans;
    }
};

Go

func minimumOperations(nums []int) (ans int) {
	s := [101]bool{true}
	for _, x := range nums {
		if !s[x] {
			s[x] = true
			ans++
		}
	}
	return
}

TypeScript

function minimumOperations(nums: number[]): number {
    const s = new Set(nums);
    s.delete(0);
    return s.size;
}

Rust

use std::collections::HashSet;
impl Solution {
    pub fn minimum_operations(nums: Vec<i32>) -> i32 {
        let mut s = nums.iter().collect::<HashSet<&i32>>();
        s.remove(&0);
        s.len() as i32
    }
}

C

int minimumOperations(int* nums, int numsSize) {
    int vis[101] = {0};
    vis[0] = 1;
    int ans = 0;
    for (int i = 0; i < numsSize; i++) {
        if (vis[nums[i]]) {
            continue;
        }
        vis[nums[i]] = 1;
        ans++;
    }
    return ans;
}