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Commit 0603c3f

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feat: add solutions to lc problem: No.3046 (doocs#2375)
No.3046.Split the Array
1 parent 8d7a71a commit 0603c3f

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7 files changed

+149
-8
lines changed

7 files changed

+149
-8
lines changed

‎solution/3000-3099/3046.Split the Array/README.md‎

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## 解法
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### 方法一
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### 方法一:计数
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根据题意,我们需要将数组分成两部分,每部分的元素都是互不相同的。因此,我们可以统计数组中每个元素的出现次数,如果某个元素出现的次数大于等于 3ドル$ 次,那么就无法满足题意。否则,我们可以将数组分成两部分。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 是数组的长度。
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<!-- tabs:start -->
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```python
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class Solution:
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def isPossibleToSplit(self, nums: List[int]) -> bool:
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return max(Counter(nums).values()) < 3
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```
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```java
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class Solution {
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public boolean isPossibleToSplit(int[] nums) {
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int[] cnt = new int[101];
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for (int x : nums) {
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if (++cnt[x] >= 3) {
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return false;
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}
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}
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return true;
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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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bool isPossibleToSplit(vector<int>& nums) {
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int cnt[101]{};
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for (int x : nums) {
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if (++cnt[x] >= 3) {
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return false;
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}
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}
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return true;
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}
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};
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```
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```go
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func isPossibleToSplit(nums []int) bool {
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cnt := [101]int{}
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for _, x := range nums {
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cnt[x]++
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if cnt[x] >= 3 {
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return false
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}
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}
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return true
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}
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```
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```ts
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function isPossibleToSplit(nums: number[]): boolean {
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const cnt: number[] = Array(101).fill(0);
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for (const x of nums) {
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if (++cnt[x] >= 3) {
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return false;
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}
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}
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return true;
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}
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```
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<!-- tabs:end -->

‎solution/3000-3099/3046.Split the Array/README_EN.md‎

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## Solutions
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### Solution 1
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### Solution 1: Counting
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According to the problem, we need to divide the array into two parts, and the elements in each part are all distinct. Therefore, we can count the occurrence of each element in the array. If an element appears three or more times, it cannot satisfy the problem's requirements. Otherwise, we can divide the array into two parts.
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The time complexity is $O(n),ドル and the space complexity is $O(n)$. Where $n$ is the length of the array.
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<!-- tabs:start -->
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```python
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class Solution:
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def isPossibleToSplit(self, nums: List[int]) -> bool:
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return max(Counter(nums).values()) < 3
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```
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```java
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class Solution {
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public boolean isPossibleToSplit(int[] nums) {
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int[] cnt = new int[101];
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for (int x : nums) {
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if (++cnt[x] >= 3) {
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return false;
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}
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}
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return true;
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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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bool isPossibleToSplit(vector<int>& nums) {
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int cnt[101]{};
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for (int x : nums) {
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if (++cnt[x] >= 3) {
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return false;
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}
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}
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return true;
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}
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};
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```
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```go
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func isPossibleToSplit(nums []int) bool {
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cnt := [101]int{}
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for _, x := range nums {
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cnt[x]++
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if cnt[x] >= 3 {
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return false
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}
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}
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return true
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}
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```
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```ts
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function isPossibleToSplit(nums: number[]): boolean {
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const cnt: number[] = Array(101).fill(0);
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for (const x of nums) {
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if (++cnt[x] >= 3) {
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return false;
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}
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}
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return true;
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}
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```
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<!-- tabs:end -->
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class Solution {
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public:
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bool isPossibleToSplit(vector<int>& nums) {
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int cnt[101]{};
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for (int x : nums) {
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if (++cnt[x] >= 3) {
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return false;
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}
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}
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return true;
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}
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};
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func isPossibleToSplit(nums []int) bool {
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cnt := [101]int{}
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for _, x := range nums {
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cnt[x]++
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if cnt[x] >= 3 {
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return false
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}
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}
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return true
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}
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class Solution {
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public boolean isPossibleToSplit(int[] nums) {
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int[] cnt = new int[101];
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for (int x : nums) {
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if (++cnt[x] >= 3) {
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return false;
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}
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}
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return true;
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}
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}
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class Solution:
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def isPossibleToSplit(self, nums: List[int]) -> bool:
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return max(Counter(nums).values()) < 3
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function isPossibleToSplit(nums: number[]): boolean {
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const cnt: number[] = Array(101).fill(0);
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for (const x of nums) {
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if (++cnt[x] >= 3) {
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return false;
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}
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}
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return true;
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}

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