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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

  • 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;
        }
    }
    
  • 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;
        }
    };
    
  • class Solution:
        def minimumOperations(self, nums: List[int]) -> int:
            return len({x for x in nums if x})
    
    
  • func minimumOperations(nums []int) (ans int) {
    	s := [101]bool{true}
    	for _, x := range nums {
    		if !s[x] {
    			s[x] = true
    			ans++
    		}
    	}
    	return
    }
    
  • function minimumOperations(nums: number[]): number {
        const set = new Set(nums);
        set.delete(0);
        return set.size;
    }
    
    
  • use std::collections::HashSet;
    impl Solution {
        pub fn minimum_operations(nums: Vec<i32>) -> i32 {
            let mut set = nums.iter().collect::<HashSet<&i32>>();
            set.remove(&0);
            set.len() as i32
        }
    }
    
    

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