feat: add solutions to lc problem: No.3397

solution/3300-3399/3397.Maximum Number of Distinct Elements After Operations/README.md

@@ -66,32 +66,108 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3397.Ma
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：贪心 + 排序
+
+我们不妨对数组 $\textit{nums}$ 排序，然后从左到右考虑每个元素 $x$。
+
+对于第一个元素，我们可以贪心地将其变为 $x - k$，这样可以使得 $x$ 尽可能小，给后续的元素留下更多的空间。我们用变量 $\textit{pre}$ 当前使用到的元素的最大值，初始化为负无穷大。
+
+对于后续的元素 $x$，我们可以贪心地将其变为 $\min(x + k, \max(x - k, \textit{pre} + 1))$。这里的 $\max(x - k, \textit{pre} + 1)$ 表示我们尽可能地将 $x$ 变得更小，但不能小于 $\textit{pre} + 1$，如果存在该值，且小于 $x + k$，我们就可以将 $x$ 变为该值，不重复元素数加一，然后我们更新 $\textit{pre}$ 为该值。
+
+遍历结束，我们就得到了不重复元素的最大数量。
+
+时间复杂度 $O(n \times \log n)$，空间复杂度 $O(\log n)$。其中 $n$ 为数组 $\textit{nums}$ 的长度。
 
 <!-- tabs:start -->
 
 #### Python3
 
 ```python
-
+class Solution:
+    def maxDistinctElements(self, nums: List[int], k: int) -> int:
+        nums.sort()
+        ans = 0
+        pre = -inf
+        for x in nums:
+            cur = min(x + k, max(x - k, pre + 1))
+            if cur > pre:
+                ans += 1
+                pre = cur
+        return ans
 ```
 
 #### Java
 
 ```java
-
+class Solution {
+    public int maxDistinctElements(int[] nums, int k) {
+        Arrays.sort(nums);
+        int n = nums.length;
+        int ans = 0, pre = Integer.MIN_VALUE;
+        for (int x : nums) {
+            int cur = Math.min(x + k, Math.max(x - k, pre + 1));
+            if (cur > pre) {
+                ++ans;
+                pre = cur;
+            }
+        }
+        return ans;
+    }
+}
 ```
 
 #### C++
 
 ```cpp
-
+class Solution {
+public:
+    int maxDistinctElements(vector<int>& nums, int k) {
+        ranges::sort(nums);
+        int ans = 0, pre = INT_MIN;
+        for (int x : nums) {
+            int cur = min(x + k, max(x - k, pre + 1));
+            if (cur > pre) {
+                ++ans;
+                pre = cur;
+            }
+        }
+        return ans;
+    }
+};
 ```
 
 #### Go
 
 ```go
+func maxDistinctElements(nums []int, k int) (ans int) {
+	sort.Ints(nums)
+	pre := math.MinInt32
+	for _, x := range nums {
+		cur := min(x+k, max(x-k, pre+1))
+		if cur > pre {
+			ans++
+			pre = cur
+		}
+	}
+	return
+}
+```
 
+#### TypeScript
+
+```ts
+function maxDistinctElements(nums: number[], k: number): number {
+    nums.sort((a, b) => a - b);
+    let [ans, pre] = [0, -Infinity];
+    for (const x of nums) {
+        const cur = Math.min(x + k, Math.max(x - k, pre + 1));
+        if (cur > pre) {
+            ++ans;
+            pre = cur;
+        }
+    }
+    return ans;
+}
 ```
 
 <!-- tabs:end -->

solution/3300-3399/3397.Maximum Number of Distinct Elements After Operations/README_EN.md

@@ -64,32 +64,108 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3397.Ma
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Greedy + Sorting
+
+We can sort the array $\textit{nums}$ and then consider each element $x$ from left to right.
+
+For the first element, we can greedily change it to $x - k$, making $x$ as small as possible to leave more space for subsequent elements. We use the variable $\textit{pre}$ to track the maximum value of the elements used so far, initialized to negative infinity.
+
+For subsequent elements $x$, we can greedily change it to $\min(x + k, \max(x - k, \textit{pre} + 1))$. Here, $\max(x - k, \textit{pre} + 1)$ means we try to make $x$ as small as possible but not smaller than $\textit{pre} + 1$. If this value exists and is less than $x + k$, we can change $x$ to this value, increment the count of distinct elements, and update $\textit{pre}$ to this value.
+
+After traversing the array, we obtain the maximum number of distinct elements.
+
+The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the length of the array $\textit{nums}$.
 
 <!-- tabs:start -->
 
 #### Python3
 
 ```python
-
+class Solution:
+    def maxDistinctElements(self, nums: List[int], k: int) -> int:
+        nums.sort()
+        ans = 0
+        pre = -inf
+        for x in nums:
+            cur = min(x + k, max(x - k, pre + 1))
+            if cur > pre:
+                ans += 1
+                pre = cur
+        return ans
 ```
 
 #### Java
 
 ```java
-
+class Solution {
+    public int maxDistinctElements(int[] nums, int k) {
+        Arrays.sort(nums);
+        int n = nums.length;
+        int ans = 0, pre = Integer.MIN_VALUE;
+        for (int x : nums) {
+            int cur = Math.min(x + k, Math.max(x - k, pre + 1));
+            if (cur > pre) {
+                ++ans;
+                pre = cur;
+            }
+        }
+        return ans;
+    }
+}
 ```
 
 #### C++
 
 ```cpp
-
+class Solution {
+public:
+    int maxDistinctElements(vector<int>& nums, int k) {
+        ranges::sort(nums);
+        int ans = 0, pre = INT_MIN;
+        for (int x : nums) {
+            int cur = min(x + k, max(x - k, pre + 1));
+            if (cur > pre) {
+                ++ans;
+                pre = cur;
+            }
+        }
+        return ans;
+    }
+};
 ```
 
 #### Go
 
 ```go
+func maxDistinctElements(nums []int, k int) (ans int) {
+	sort.Ints(nums)
+	pre := math.MinInt32
+	for _, x := range nums {
+		cur := min(x+k, max(x-k, pre+1))
+		if cur > pre {
+			ans++
+			pre = cur
+		}
+	}
+	return
+}
+```
 
+#### TypeScript
+
+```ts
+function maxDistinctElements(nums: number[], k: number): number {
+    nums.sort((a, b) => a - b);
+    let [ans, pre] = [0, -Infinity];
+    for (const x of nums) {
+        const cur = Math.min(x + k, Math.max(x - k, pre + 1));
+        if (cur > pre) {
+            ++ans;
+            pre = cur;
+        }
+    }
+    return ans;
+}
 ```
 
 <!-- tabs:end -->

solution/3300-3399/3397.Maximum Number of Distinct Elements After Operations/Solution.cpp

@@ -0,0 +1,15 @@
+class Solution {
+public:
+    int maxDistinctElements(vector<int>& nums, int k) {
+        ranges::sort(nums);
+        int ans = 0, pre = INT_MIN;
+        for (int x : nums) {
+            int cur = min(x + k, max(x - k, pre + 1));
+            if (cur > pre) {
+                ++ans;
+                pre = cur;
+            }
+        }
+        return ans;
+    }
+};

solution/3300-3399/3397.Maximum Number of Distinct Elements After Operations/Solution.go

@@ -0,0 +1,12 @@
+func maxDistinctElements(nums []int, k int) (ans int) {
+	sort.Ints(nums)
+	pre := math.MinInt32
+	for _, x := range nums {
+		cur := min(x+k, max(x-k, pre+1))
+		if cur > pre {
+			ans++
+			pre = cur
+		}
+	}
+	return
+}

solution/3300-3399/3397.Maximum Number of Distinct Elements After Operations/Solution.java

@@ -0,0 +1,15 @@
+class Solution {
+    public int maxDistinctElements(int[] nums, int k) {
+        Arrays.sort(nums);
+        int n = nums.length;
+        int ans = 0, pre = Integer.MIN_VALUE;
+        for (int x : nums) {
+            int cur = Math.min(x + k, Math.max(x - k, pre + 1));
+            if (cur > pre) {
+                ++ans;
+                pre = cur;
+            }
+        }
+        return ans;
+    }
+}

solution/3300-3399/3397.Maximum Number of Distinct Elements After Operations/Solution.py

@@ -0,0 +1,11 @@
+class Solution:
+    def maxDistinctElements(self, nums: List[int], k: int) -> int:
+        nums.sort()
+        ans = 0
+        pre = -inf
+        for x in nums:
+            cur = min(x + k, max(x - k, pre + 1))
+            if cur > pre:
+                ans += 1
+                pre = cur
+        return ans

solution/3300-3399/3397.Maximum Number of Distinct Elements After Operations/Solution.ts

@@ -0,0 +1,12 @@
+function maxDistinctElements(nums: number[], k: number): number {
+    nums.sort((a, b) => a - b);
+    let [ans, pre] = [0, -Infinity];
+    for (const x of nums) {
+        const cur = Math.min(x + k, Math.max(x - k, pre + 1));
+        if (cur > pre) {
+            ++ans;
+            pre = cur;
+        }
+    }
+    return ans;
+}