fix: update

Author: yanglbme

solution/3100-3199/3134.Find the Median of the Uniqueness Array/README.md

@@ -84,7 +84,21 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：二分查找 + 双指针
+
+我们记数组 $\textit{nums}$ 的长度为 $n$，那么唯一性数组的长度为 $m = \frac{(1 + n) \times n}{2}$，而唯一性数组的中位数就是这 $m$ 个数中的第 $\frac{m + 1}{2}$ 小的数字。
+
+考虑唯一性数组中，有多少个数小于等于 $x$。随着 $x$ 的增大，只会有越来越多的数小于等于 $x$。这存在着单调性，因此，我们可以二分枚举 $x$，找到第一个 $x$，满足唯一性数组中小于等于 $x$ 的数的个数大于等于 $\frac{m + 1}{2}$，这个 $x$ 就是唯一性数组的中位数。
+
+我们定义二分查找的左边界 $l = 0$，右边界 $r = n$，然后进行二分查找，对于每个 $\textit{mid}$，我们检查唯一性数组中小于等于 $\textit{mid}$ 的数的个数是否大于等于 $\frac{m + 1}{2}$。我们通过函数 $\text{check}(mx)$ 来实现这一点。
+
+函数 $\text{check}(mx)$ 的实现思路如下：
+
+由于子数组越长，不同元素的个数越多，因此，我们可以利用双指针维护一个滑动窗口，使得窗口中的子数组的不同元素的个数不超过 $mx$。具体地，我们维护一个哈希表 $\textit{cnt}$，$\textit{cnt}[x]$ 表示窗口中元素 $x$ 的个数。我们使用两个指针 $l$ 和 $r$，其中 $l$ 表示窗口的左边界，而 $r$ 表示窗口的右边界。初始时 $l = r = 0$。
+
+我们枚举 $r$，对于每个 $r$，我们将 $\textit{nums}[r]$ 加入窗口中，并更新 $\textit{cnt}[\textit{nums}[r]]$。如果窗口中的不同元素的个数超过了 $mx$，我们需要将 $l$ 右移，直到窗口中的不同元素的个数不超过 $mx$。此时，右端点为 $r$，而左端点为 $[l,..r]$ 的子数组都是满足条件的，一共有 $r - l + 1$ 个子数组。我们将这个数量累加到 $k$ 中，如果 $k$ 大于等于 $\frac{m + 1}{2}$，那么说明唯一性数组中小于等于 $\textit{mid}$ 的数的个数大于等于 $\frac{m + 1}{2}$，我们返回 $\text{true}$，否则返回 $\text{false}$。
+
+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/3100-3199/3134.Find the Median of the Uniqueness Array/README_EN.md

@@ -80,7 +80,21 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Binary Search + Two Pointers
+
+Let the length of the array $\textit{nums}$ be $n$. The length of the uniqueness array is $m = \frac{(1 + n) \times n}{2}$, and the median of the uniqueness array is the $\frac{m + 1}{2}$-th smallest number among these $m$ numbers.
+
+Consider how many numbers in the uniqueness array are less than or equal to $x$. As $x$ increases, there will be more and more numbers less than or equal to $x$. This property is monotonic, so we can use binary search to enumerate $x$ and find the first $x$ such that the number of elements in the uniqueness array less than or equal to $x$ is greater than or equal to $\frac{m + 1}{2}$. This $x$ is the median of the uniqueness array.
+
+We define the left boundary of the binary search as $l = 0$ and the right boundary as $r = n$. Then we perform binary search. For each $\textit{mid}$, we check whether the number of elements in the uniqueness array less than or equal to $\textit{mid}$ is greater than or equal to $\frac{m + 1}{2}$. We achieve this through the function $\text{check}(mx)$.
+
+The implementation idea of the function $\text{check}(mx)$ is as follows:
+
+Since the longer the subarray, the more different elements it contains, we can use two pointers to maintain a sliding window such that the number of different elements in the window does not exceed $mx$. Specifically, we maintain a hash table $\textit{cnt}$, where $\textit{cnt}[x]$ represents the number of occurrences of element $x$ in the window. We use two pointers $l$ and $r$, where $l$ represents the left boundary of the window and $r$ represents the right boundary. Initially, $l = r = 0$.
+
+We enumerate $r$. For each $r$, we add $\textit{nums}[r]$ to the window and update $\textit{cnt}[\textit{nums}[r]]$. If the number of different elements in the window exceeds $mx$, we need to move $l$ to the right until the number of different elements in the window does not exceed $mx$. At this point, the subarrays with the right endpoint $r$ and left endpoints in the range $[l, .., r]$ all meet the condition, and there are $r - l + 1$ such subarrays. We accumulate this count into $k$. If $k$ is greater than or equal to $\frac{m + 1}{2}$, it means that the number of elements in the uniqueness array less than or equal to $\textit{mid}$ is greater than or equal to $\frac{m + 1}{2}$, and we return $\text{true}$; otherwise, we return $\text{false}$.
+
+The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{nums}$.
 
 <!-- tabs:start -->