feat: add solutions to lc problem: No.3346 (#4794)

Author: yanglbme

solution/3300-3399/3346.Maximum Frequency of an Element After Performing Operations I/README.md

@@ -87,7 +87,30 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：差分
+
+根据题目描述，对于数组 $\textit{nums}$ 中的每个元素 $x$，我们可以将其变为 $[x-k, x+k]$ 范围内的任意整数。我们希望通过对 $\textit{nums}$ 中的若干元素进行操作，使得某个整数在数组中出现的次数最多。
+
+题目可以转化为，将每个元素 $x$ 对应的区间 $[x-k, x+k]$ 的所有元素进行合并，找出合并后区间中包含最多原始元素的整数。这可以通过差分数组来实现。
+
+我们使用一个字典 $d$ 来记录差分数组。对于每个元素 $x$，我们在差分数组中执行以下操作：
+
+-   在位置 $x-k$ 处加 $1$，表示从这个位置开始，有一个新的区间开始。
+-   在位置 $x+k+1$ 处减 $1$，表示从这个位置开始，有一个区间结束。
+-   在位置 $x$ 处加 $0$，确保位置 $x$ 存在于差分数组中，方便后续计算。
+
+同时，我们还需要记录每个元素在原始数组中出现的次数，使用字典 $cnt$ 来实现。
+
+接下来，我们对差分数组进行前缀和计算，得到每个位置上有多少个区间覆盖。对于每个位置 $x$，我们计算其覆盖的区间数 $s$。接下来分类讨论：
+
+-   如果 $x$ 在原始数组中出现过，对于 $x$ 自身的操作，没有意义，因此会有 $s - cnt[x]$ 个其他的元素可以通过操作变为 $x$，但最多只能操作 $\textit{numOperations}$ 次，所以该位置的最大频率为 $\textit{cnt}[x] + \min(s - \textit{cnt}[x], \textit{numOperations})$。
+-   如果 $x$ 在原始数组中没有出现过，那么最多只能通过操作 $\textit{numOperations}$ 次将其他元素变为 $x$，因此该位置的最大频率为 $\min(s, \textit{numOperations})$。
+
+综合以上两种情况，我们可以统一表示为 $\min(s, \textit{cnt}[x] + \textit{numOperations})$。
+
+最后，我们遍历所有位置，计算出每个位置的最大频率，并取其中的最大值作为答案。
+
+时间复杂度 $O(n \times \log n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $\textit{nums}$ 的长度。
 
 <!-- tabs:start -->
 

solution/3300-3399/3346.Maximum Frequency of an Element After Performing Operations I/README_EN.md

@@ -83,7 +83,30 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Difference Array
+
+According to the problem description, for each element $x$ in the array $\textit{nums}$, we can change it to any integer within the range $[x-k, x+k]$. We want to perform operations on some elements in $\textit{nums}$ to maximize the frequency of a certain integer in the array.
+
+The problem can be transformed into merging all elements in the interval $[x-k, x+k]$ corresponding to each element $x$, and finding the integer that contains the most original elements in the merged intervals. This can be implemented using a difference array.
+
+We use a dictionary $d$ to record the difference array. For each element $x$, we perform the following operations on the difference array:
+
+-   Add $1$ at position $x-k$, indicating that a new interval starts from this position.
+-   Subtract $1$ at position $x+k+1$, indicating that an interval ends from this position.
+-   Add $0$ at position $x$, ensuring that position $x$ exists in the difference array for subsequent calculations.
+
+At the same time, we need to record the number of occurrences of each element in the original array, using a dictionary $cnt$ to implement this.
+
+Next, we perform prefix sum calculation on the difference array to get how many intervals cover each position. For each position $x$, we calculate the number of intervals covering it as $s$. Then we discuss by cases:
+
+-   If $x$ appears in the original array, operations on $x$ itself are meaningless. Therefore, there are $s - cnt[x]$ other elements that can be changed to $x$ through operations, but at most $\textit{numOperations}$ operations can be performed. So the maximum frequency at this position is $\textit{cnt}[x] + \min(s - \textit{cnt}[x], \textit{numOperations})$.
+-   If $x$ does not appear in the original array, then at most $\textit{numOperations}$ operations can be performed to change other elements to $x$. Therefore, the maximum frequency at this position is $\min(s, \textit{numOperations})$.
+
+Combining the above two cases, we can uniformly express it as $\min(s, \textit{cnt}[x] + \textit{numOperations})$.
+
+Finally, we traverse all positions, calculate the maximum frequency at each position, and take the maximum value among them as the answer.
+
+The time complexity is $O(n \times \log n)$ and the space complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$.
 
 <!-- tabs:start -->
 

solution/3300-3399/3347.Maximum Frequency of an Element After Performing Operations II/README.md

@@ -87,7 +87,30 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：差分
+
+根据题目描述，对于数组 $\textit{nums}$ 中的每个元素 $x$，我们可以将其变为 $[x-k, x+k]$ 范围内的任意整数。我们希望通过对 $\textit{nums}$ 中的若干元素进行操作，使得某个整数在数组中出现的次数最多。
+
+题目可以转化为，将每个元素 $x$ 对应的区间 $[x-k, x+k]$ 的所有元素进行合并，找出合并后区间中包含最多原始元素的整数。这可以通过差分数组来实现。
+
+我们使用一个字典 $d$ 来记录差分数组。对于每个元素 $x$，我们在差分数组中执行以下操作：
+
+-   在位置 $x-k$ 处加 $1$，表示从这个位置开始，有一个新的区间开始。
+-   在位置 $x+k+1$ 处减 $1$，表示从这个位置开始，有一个区间结束。
+-   在位置 $x$ 处加 $0$，确保位置 $x$ 存在于差分数组中，方便后续计算。
+
+同时，我们还需要记录每个元素在原始数组中出现的次数，使用字典 $cnt$ 来实现。
+
+接下来，我们对差分数组进行前缀和计算，得到每个位置上有多少个区间覆盖。对于每个位置 $x$，我们计算其覆盖的区间数 $s$。接下来分类讨论：
+
+-   如果 $x$ 在原始数组中出现过，对于 $x$ 自身的操作，没有意义，因此会有 $s - cnt[x]$ 个其他的元素可以通过操作变为 $x$，但最多只能操作 $\textit{numOperations}$ 次，所以该位置的最大频率为 $\textit{cnt}[x] + \min(s - \textit{cnt}[x], \textit{numOperations})$。
+-   如果 $x$ 在原始数组中没有出现过，那么最多只能通过操作 $\textit{numOperations}$ 次将其他元素变为 $x$，因此该位置的最大频率为 $\min(s, \textit{numOperations})$。
+
+综合以上两种情况，我们可以统一表示为 $\min(s, \textit{cnt}[x] + \textit{numOperations})$。
+
+最后，我们遍历所有位置，计算出每个位置的最大频率，并取其中的最大值作为答案。
+
+时间复杂度 $O(n \times \log n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $\textit{nums}$ 的长度。
 
 <!-- tabs:start -->
 

solution/3300-3399/3347.Maximum Frequency of an Element After Performing Operations II/README_EN.md

@@ -83,7 +83,30 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Difference Array
+
+According to the problem description, for each element $x$ in the array $\textit{nums}$, we can change it to any integer within the range $[x-k, x+k]$. We want to perform operations on some elements in $\textit{nums}$ to maximize the frequency of a certain integer in the array.
+
+The problem can be transformed into merging all elements in the interval $[x-k, x+k]$ corresponding to each element $x$, and finding the integer that contains the most original elements in the merged intervals. This can be implemented using a difference array.
+
+We use a dictionary $d$ to record the difference array. For each element $x$, we perform the following operations on the difference array:
+
+-   Add $1$ at position $x-k$, indicating that a new interval starts from this position.
+-   Subtract $1$ at position $x+k+1$, indicating that an interval ends from this position.
+-   Add $0$ at position $x$, ensuring that position $x$ exists in the difference array for subsequent calculations.
+
+At the same time, we need to record the number of occurrences of each element in the original array, using a dictionary $cnt$ to implement this.
+
+Next, we perform prefix sum calculation on the difference array to get how many intervals cover each position. For each position $x$, we calculate the number of intervals covering it as $s$. Then we discuss by cases:
+
+-   If $x$ appears in the original array, operations on $x$ itself are meaningless. Therefore, there are $s - cnt[x]$ other elements that can be changed to $x$ through operations, but at most $\textit{numOperations}$ operations can be performed. So the maximum frequency at this position is $\textit{cnt}[x] + \min(s - \textit{cnt}[x], \textit{numOperations})$.
+-   If $x$ does not appear in the original array, then at most $\textit{numOperations}$ operations can be performed to change other elements to $x$. Therefore, the maximum frequency at this position is $\min(s, \textit{numOperations})$.
+
+Combining the above two cases, we can uniformly express it as $\min(s, \textit{cnt}[x] + \textit{numOperations})$.
+
+Finally, we traverse all positions, calculate the maximum frequency at each position, and take the maximum value among them as the answer.
+
+The time complexity is $O(n \times \log n)$ and the space complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$.
 
 <!-- tabs:start -->