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feat: add solutions to lc problem: No.3012 (#2247)
No.3012.Minimize Length of Array Using Operations
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‎solution/3000-3099/3012.Minimize Length of Array Using Operations/README.md

+72-4
Original file line numberDiff line numberDiff line change
@@ -74,24 +74,92 @@ nums 的长度无法进一步减小，所以答案为 1 。
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## 解法
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### 方法一
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### 方法一：分情况讨论
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我们不妨记数组 $nums$ 的最小的元素为 $mi$。
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如果 $mi$ 只出现一次，那么我们将 $mi$ 与数组 $nums$ 的其他元素进行操作，可以将其他元素全部消去，最终剩下 $mi$ 一个元素，答案为 $1$。
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如果 $mi$ 出现多次，我们判断数组 $nums$ 中的元素是否都是 $mi$ 的倍数。如果不是，即存在至少一个元素 $x$，使得 $0 \lt x \bmod mi \lt mi$，说明我们可以通过操作，构造出一个小于 $mi$ 的元素，那么这个小于 $mi$ 的元素与其他元素进行操作，可以将其他元素全部消去，最终剩下这个小于 $mi$ 的元素，答案为 $1$；如果都是 $mi$ 的倍数，我们可以先借助 $mi$，将所有大于 $mi$ 的元素消去，最终剩下的元素都是 $mi$，个数为 $cnt$，两两配对，每两个元素进行一次操作，最终剩下 $\lceil cnt / 2 \rceil$ 个元素，答案为 $\lceil cnt / 2 \rceil$。
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时间复杂度 $O(n)$，其中 $n$ 是数组 $nums$ 的长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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```python
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class Solution:
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def minimumArrayLength(self, nums: List[int]) -> int:
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mi = min(nums)
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if any(x % mi for x in nums):
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return 1
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return (nums.count(mi) + 1) // 2
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```
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```java
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class Solution {
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public int minimumArrayLength(int[] nums) {
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int mi = Arrays.stream(nums).min().getAsInt();
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int cnt = 0;
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for (int x : nums) {
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if (x % mi != 0) {
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return 1;
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}
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if (x == mi) {
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++cnt;
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}
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}
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return (cnt + 1) / 2;
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}
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}
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```
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```cpp
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class Solution {
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public:
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int minimumArrayLength(vector<int>& nums) {
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int mi = *min_element(nums.begin(), nums.end());
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int cnt = 0;
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for (int x : nums) {
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if (x % mi) {
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return 1;
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}
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cnt += x == mi;
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}
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return (cnt + 1) / 2;
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}
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};
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```
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```go
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func minimumArrayLength(nums []int) int {
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mi := slices.Min(nums)
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cnt := 0
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for _, x := range nums {
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if x%mi != 0 {
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return 1
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}
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if x == mi {
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cnt++
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}
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}
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return (cnt + 1) / 2
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}
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```
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```ts
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function minimumArrayLength(nums: number[]): number {
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const mi = Math.min(...nums);
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let cnt = 0;
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for (const x of nums) {
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if (x % mi) {
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return 1;
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}
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if (x === mi) {
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++cnt;
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}
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}
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return (cnt + 1) >> 1;
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}
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```
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<!-- tabs:end -->

‎solution/3000-3099/3012.Minimize Length of Array Using Operations/README_EN.md

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@@ -70,24 +70,92 @@ It can be shown that 1 is the minimum achievable length.</pre>
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## Solutions
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### Solution 1
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### Solution 1: Case Discussion
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Let's denote the smallest element in the array $nums$ as $mi$.
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If $mi$ appears only once, we can perform operations with $mi$ and the other elements in the array $nums$ to eliminate all other elements, leaving only $mi$. The answer is $1$.
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If $mi$ appears multiple times, we need to check whether all elements in the array $nums$ are multiples of $mi$. If not, there exists at least one element $x$ such that $0 < x \bmod mi < mi$. This means we can construct an element smaller than $mi$ through operations. This smaller element can eliminate all other elements through operations, leaving only this smaller element. The answer is $1$. If all elements are multiples of $mi$, we can first use $mi$ to eliminate all elements larger than $mi$. The remaining elements are all $mi$, with a count of $cnt$. Pair them up, and perform an operation for each pair. Finally, there will be $\lceil cnt / 2 \rceil$ elements left, so the answer is $\lceil cnt / 2 \rceil$.
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The time complexity is $O(n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.
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<!-- tabs:start -->
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```python
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class Solution:
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def minimumArrayLength(self, nums: List[int]) -> int:
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mi = min(nums)
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if any(x % mi for x in nums):
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return 1
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return (nums.count(mi) + 1) // 2
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```
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```java
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class Solution {
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public int minimumArrayLength(int[] nums) {
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int mi = Arrays.stream(nums).min().getAsInt();
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int cnt = 0;
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for (int x : nums) {
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if (x % mi != 0) {
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return 1;
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}
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if (x == mi) {
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++cnt;
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}
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}
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return (cnt + 1) / 2;
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}
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}
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```
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```cpp
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class Solution {
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public:
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int minimumArrayLength(vector<int>& nums) {
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int mi = *min_element(nums.begin(), nums.end());
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int cnt = 0;
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for (int x : nums) {
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if (x % mi) {
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return 1;
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}
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cnt += x == mi;
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}
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return (cnt + 1) / 2;
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}
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};
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```
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```go
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func minimumArrayLength(nums []int) int {
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mi := slices.Min(nums)
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cnt := 0
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for _, x := range nums {
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if x%mi != 0 {
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return 1
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}
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if x == mi {
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cnt++
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}
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}
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return (cnt + 1) / 2
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}
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```
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```ts
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function minimumArrayLength(nums: number[]): number {
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const mi = Math.min(...nums);
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let cnt = 0;
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for (const x of nums) {
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if (x % mi) {
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return 1;
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}
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if (x === mi) {
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++cnt;
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}
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}
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return (cnt + 1) >> 1;
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}
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```
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<!-- tabs:end -->

‎solution/3000-3099/3012.Minimize Length of Array Using Operations/Solution.cpp

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class Solution {
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public:
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int minimumArrayLength(vector<int>& nums) {
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int mi = *min_element(nums.begin(), nums.end());
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int cnt = 0;
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for (int x : nums) {
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if (x % mi) {
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return 1;
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}
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cnt += x == mi;
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}
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return (cnt + 1) / 2;
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}
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};

‎solution/3000-3099/3012.Minimize Length of Array Using Operations/Solution.go

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func minimumArrayLength(nums []int) int {
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mi := slices.Min(nums)
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cnt := 0
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for _, x := range nums {
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if x%mi != 0 {
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return 1
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}
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if x == mi {
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cnt++
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}
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}
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return (cnt + 1) / 2
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}

‎solution/3000-3099/3012.Minimize Length of Array Using Operations/Solution.java

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class Solution {
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public int minimumArrayLength(int[] nums) {
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int mi = Arrays.stream(nums).min().getAsInt();
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int cnt = 0;
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for (int x : nums) {
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if (x % mi != 0) {
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return 1;
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}
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if (x == mi) {
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++cnt;
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}
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}
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return (cnt + 1) / 2;
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}
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}

‎solution/3000-3099/3012.Minimize Length of Array Using Operations/Solution.py

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class Solution:
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def minimumArrayLength(self, nums: List[int]) -> int:
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mi = min(nums)
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if any(x % mi for x in nums):
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return 1
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return (nums.count(mi) + 1) // 2

‎solution/3000-3099/3012.Minimize Length of Array Using Operations/Solution.ts

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function minimumArrayLength(nums: number[]): number {
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const mi = Math.min(...nums);
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let cnt = 0;
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for (const x of nums) {
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if (x % mi) {
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return 1;
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}
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if (x === mi) {
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++cnt;
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}
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}
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return (cnt + 1) >> 1;
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}

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