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Commit c90ca0e

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feat: add solutions to lc problem: No.3375 (doocs#4339)
No.3375.Minimum Operations to Make Array Values Equal to K
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‎solution/3300-3399/3375.Minimum Operations to Make Array Values Equal to K/README.md‎

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<!-- solution:start -->
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### 方法一
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### 方法一:哈希表
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根据题目描述,我们每次可以选择当前数组中的次大值作为合法整数 $h,ドル将所有大于 $h$ 的数都变为 $h,ドル这样可以使得操作次数最少。另外,由于操作会使得数字变小,因此,如果当前数组中存在小于 $k$ 的数,那么我们就无法将所有数都变为 $k,ドル直接返回 -1 即可。
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我们遍历数组 $\textit{nums},ドル对于当前的数 $x,ドル如果 $x < k,ドル直接返回 -1;否则,我们将 $x$ 加入哈希表中,并且更新当前数组中的最小值 $\textit{mi}$。最后,我们返回哈希表的大小减去 1(如果 $\textit{mi} = k,ドル则需要减去 1)。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 是数组 $\textit{nums}$ 的长度。
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<!-- tabs:start -->
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn min_operations(nums: Vec<i32>, k: i32) -> i32 {
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use std::collections::HashSet;
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let mut s = HashSet::new();
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let mut mi = i32::MAX;
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for &x in &nums {
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if x < k {
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return -1;
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}
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s.insert(x);
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mi = mi.min(x);
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}
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(s.len() as i32) - if mi == k { 1 } else { 0 }
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/3300-3399/3375.Minimum Operations to Make Array Values Equal to K/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Hash Table
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According to the problem description, we can choose the second largest value in the current array as the valid integer $h$ each time, and change all numbers greater than $h$ to $h$. This minimizes the number of operations. Additionally, since the operation reduces the numbers, if there are numbers in the current array smaller than $k,ドル we cannot make all numbers equal to $k,ドル so we directly return -1.
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We iterate through the array $\textit{nums}$. For the current number $x,ドル if $x < k,ドル we directly return -1. Otherwise, we add $x$ to the hash table and update the minimum value $\textit{mi}$ in the current array. Finally, we return the size of the hash table minus 1 (if $\textit{mi} = k,ドル we need to subtract 1).
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Time complexity is $O(n),ドル and space complexity is $O(n),ドル where $n$ is the length of the array $\textit{nums}$.
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<!-- tabs:start -->
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn min_operations(nums: Vec<i32>, k: i32) -> i32 {
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use std::collections::HashSet;
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let mut s = HashSet::new();
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let mut mi = i32::MAX;
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for &x in &nums {
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if x < k {
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return -1;
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}
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s.insert(x);
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mi = mi.min(x);
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}
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(s.len() as i32) - if mi == k { 1 } else { 0 }
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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impl Solution {
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pub fn min_operations(nums: Vec<i32>, k: i32) -> i32 {
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use std::collections::HashSet;
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let mut s = HashSet::new();
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let mut mi = i32::MAX;
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for &x in &nums {
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if x < k {
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return -1;
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}
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s.insert(x);
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mi = mi.min(x);
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}
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(s.len() as i32) - if mi == k { 1 } else { 0 }
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}
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}

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