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#
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| class Solution:
def minimumOperations(self, nums: List[int]) -> int:
return len({x for x in nums if x})
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| 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;
}
}
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| 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;
}
};
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| func minimumOperations(nums []int) (ans int) {
s := [101]bool{true}
for _, x := range nums {
if !s[x] {
s[x] = true
ans++
}
}
return
}
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| function minimumOperations(nums: number[]): number {
const set = new Set(nums);
set.delete(0);
return set.size;
}
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| 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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| 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;
}
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