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feat: add solutions to lc problem: No.2107
No.2107.Number of Unique Flavors After Sharing K Candies
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‎solution/2100-2199/2107.Number of Unique Flavors After Sharing K Candies/README.md

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@@ -59,27 +59,116 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：滑动窗口 + 哈希表**
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我们可以维护一个大小为 $k$ 的滑动窗口，窗口外的糖果为自己的，窗口内的 $k$ 个糖果分给妹妹和妈妈。我们可以用哈希表 $cnt$ 记录窗口外的糖果口味以及对应的数量。
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初始时，哈希表 $cnt$ 中存储的是 $candies[k]$ 到 $candies[n-1]$ 的糖果口味以及对应的数量。此时糖果口味的种类数为哈希表 $cnt$ 的大小，即 $ans = cnt.size()$。
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接下来，我们遍历 $[k,..n-1]$ 范围内的糖果，将当前糖果 $candies[i]$ 加入窗口内，同时将窗口左侧的糖果 $candies[i-k]$ 移出窗口外。然后我们更新哈希表 $cnt$，并且更新糖果口味的种类数 $ans$ 为 $max(ans, cnt.size())$。
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遍历完所有糖果后，我们即可得到最多可保留的独特口味的糖果。
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时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为糖果的数量。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def shareCandies(self, candies: List[int], k: int) -> int:
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cnt = Counter(candies[k:])
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ans = len(cnt)
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for i in range(k, len(candies)):
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cnt[candies[i]] -= 1
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cnt[candies[i - k]] += 1
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if cnt[candies[i]] == 0:
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cnt.pop(candies[i])
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ans = max(ans, len(cnt))
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int shareCandies(int[] candies, int k) {
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Map<Integer, Integer> cnt = new HashMap<>();
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int n = candies.length;
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for (int i = k; i < n; ++i) {
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cnt.merge(candies[i], 1, Integer::sum);
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}
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int ans = cnt.size();
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for (int i = k; i < candies.length; ++i) {
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if (cnt.merge(candies[i], -1, Integer::sum) == 0) {
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cnt.remove(candies[i]);
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}
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cnt.merge(candies[i - k], 1, Integer::sum);
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ans = Math.max(ans, cnt.size());
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int shareCandies(vector<int>& candies, int k) {
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unordered_map<int, int> cnt;
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int n = candies.size();
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for (int i = k; i < n; ++i) {
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++cnt[candies[i]];
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}
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int ans = cnt.size();
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for (int i = k; i < candies.size(); ++i) {
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if (--cnt[candies[i]] == 0) {
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cnt.erase(candies[i]);
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}
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++cnt[candies[i - k]];
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ans = max(ans, (int) cnt.size());
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}
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return ans;
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}
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};
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```
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### **TypeScript**
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### **Go**
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```go
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func shareCandies(candies []int, k int) (ans int) {
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cnt := map[int]int{}
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for _, c := range candies[k:] {
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cnt[c]++
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}
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ans = len(cnt)
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for i := k; i < len(candies); i++ {
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cnt[candies[i]]--
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if cnt[candies[i]] == 0 {
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delete(cnt, candies[i])
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}
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cnt[candies[i-k]]++
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ans = max(ans, len(cnt))
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}
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return
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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### **TypeScript**
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```ts
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‎solution/2100-2199/2107.Number of Unique Flavors After Sharing K Candies/README_EN.md

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### **Python3**
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```python
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class Solution:
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def shareCandies(self, candies: List[int], k: int) -> int:
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cnt = Counter(candies[k:])
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ans = len(cnt)
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for i in range(k, len(candies)):
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cnt[candies[i]] -= 1
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cnt[candies[i - k]] += 1
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if cnt[candies[i]] == 0:
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cnt.pop(candies[i])
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ans = max(ans, len(cnt))
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return ans
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```
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### **Java**
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```java
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class Solution {
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public int shareCandies(int[] candies, int k) {
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Map<Integer, Integer> cnt = new HashMap<>();
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int n = candies.length;
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for (int i = k; i < n; ++i) {
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cnt.merge(candies[i], 1, Integer::sum);
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}
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int ans = cnt.size();
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for (int i = k; i < candies.length; ++i) {
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if (cnt.merge(candies[i], -1, Integer::sum) == 0) {
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cnt.remove(candies[i]);
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}
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cnt.merge(candies[i - k], 1, Integer::sum);
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ans = Math.max(ans, cnt.size());
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int shareCandies(vector<int>& candies, int k) {
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unordered_map<int, int> cnt;
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int n = candies.size();
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for (int i = k; i < n; ++i) {
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++cnt[candies[i]];
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}
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int ans = cnt.size();
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for (int i = k; i < candies.size(); ++i) {
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if (--cnt[candies[i]] == 0) {
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cnt.erase(candies[i]);
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}
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++cnt[candies[i - k]];
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ans = max(ans, (int) cnt.size());
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func shareCandies(candies []int, k int) (ans int) {
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cnt := map[int]int{}
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for _, c := range candies[k:] {
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cnt[c]++
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}
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ans = len(cnt)
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for i := k; i < len(candies); i++ {
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cnt[candies[i]]--
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if cnt[candies[i]] == 0 {
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delete(cnt, candies[i])
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}
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cnt[candies[i-k]]++
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ans = max(ans, len(cnt))
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}
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return
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **TypeScript**

‎solution/2100-2199/2107.Number of Unique Flavors After Sharing K Candies/Solution.cpp

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class Solution {
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public:
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int shareCandies(vector<int>& candies, int k) {
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unordered_map<int, int> cnt;
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int n = candies.size();
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for (int i = k; i < n; ++i) {
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++cnt[candies[i]];
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}
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int ans = cnt.size();
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for (int i = k; i < candies.size(); ++i) {
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if (--cnt[candies[i]] == 0) {
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cnt.erase(candies[i]);
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}
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++cnt[candies[i - k]];
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ans = max(ans, (int) cnt.size());
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}
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return ans;
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}
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};

‎solution/2100-2199/2107.Number of Unique Flavors After Sharing K Candies/Solution.go

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func shareCandies(candies []int, k int) (ans int) {
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cnt := map[int]int{}
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for _, c := range candies[k:] {
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cnt[c]++
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}
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ans = len(cnt)
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for i := k; i < len(candies); i++ {
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cnt[candies[i]]--
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if cnt[candies[i]] == 0 {
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delete(cnt, candies[i])
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}
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cnt[candies[i-k]]++
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ans = max(ans, len(cnt))
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}
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return
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}

‎solution/2100-2199/2107.Number of Unique Flavors After Sharing K Candies/Solution.java

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class Solution {
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public int shareCandies(int[] candies, int k) {
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Map<Integer, Integer> cnt = new HashMap<>();
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int n = candies.length;
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for (int i = k; i < n; ++i) {
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cnt.merge(candies[i], 1, Integer::sum);
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}
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int ans = cnt.size();
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for (int i = k; i < candies.length; ++i) {
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if (cnt.merge(candies[i], -1, Integer::sum) == 0) {
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cnt.remove(candies[i]);
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}
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cnt.merge(candies[i - k], 1, Integer::sum);
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ans = Math.max(ans, cnt.size());
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}
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return ans;
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}
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}

‎solution/2100-2199/2107.Number of Unique Flavors After Sharing K Candies/Solution.py

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class Solution:
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def shareCandies(self, candies: List[int], k: int) -> int:
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cnt = Counter(candies[k:])
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ans = len(cnt)
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for i in range(k, len(candies)):
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cnt[candies[i]] -= 1
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cnt[candies[i - k]] += 1
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if cnt[candies[i]] == 0:
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cnt.pop(candies[i])
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ans = max(ans, len(cnt))
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return ans

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