deploy: dc4d031

Author: yanglbme

en/lc/2526/index.html

@@ -60260,9 +60260,9 @@
       <ul class="md-nav__list">
 
           <li class="md-nav__item">
-  <a href="#solution-1" class="md-nav__link">
+  <a href="#solution-1-counting" class="md-nav__link">
     <span class="md-ellipsis">
-      Solution 1
+      Solution 1: Counting
     </span>
   </a>
 
@@ -80136,11 +80136,11 @@ <h2 id="description">Description</h2>
 [null, false, false, true, false]
 
 <strong>Explanation</strong>
-DataStream dataStream = new DataStream(4, 3); //value = 4, k = 3 
-dataStream.consec(4); // Only 1 integer is parsed, so returns False. 
+DataStream dataStream = new DataStream(4, 3); //value = 4, k = 3
+dataStream.consec(4); // Only 1 integer is parsed, so returns False.
 dataStream.consec(4); // Only 2 integers are parsed.
-                      // Since 2 is less than k, returns False. 
-dataStream.consec(4); // The 3 integers parsed are all equal to value, so returns True. 
+                      // Since 2 is less than k, returns False.
+dataStream.consec(4); // The 3 integers parsed are all equal to value, so returns True.
 dataStream.consec(3); // The last k integers parsed in the stream are [4,4,3].
                       // Since 3 is not equal to value, it returns False.
 </pre>
@@ -80159,7 +80159,10 @@ <h2 id="description">Description</h2>
 <h2 id="solutions">Solutions</h2>
 <!-- solution:start -->
 
-<h3 id="solution-1">Solution 1</h3>
+<h3 id="solution-1-counting">Solution 1: Counting</h3>
+<p>We can maintain a counter $\textit{cnt}$ to record the current number of consecutive integers equal to $\textit{value}$.</p>
+<p>When calling the <code>consec</code> method, if $\textit{num}$ is equal to $\textit{value}$, we increment $\textit{cnt}$ by 1; otherwise, we reset $\textit{cnt}$ to 0. Then we check whether $\textit{cnt}$ is greater than or equal to $\textit{k}$.</p>
+<p>The time complexity is $O(1)$, and the space complexity is $O(1)$.</p>
 <div class="tabbed-set tabbed-alternate" data-tabs="1:5"><input checked="checked" id="__tabbed_1_1" name="__tabbed_1" type="radio" /><input id="__tabbed_1_2" name="__tabbed_1" type="radio" /><input id="__tabbed_1_3" name="__tabbed_1" type="radio" /><input id="__tabbed_1_4" name="__tabbed_1" type="radio" /><input id="__tabbed_1_5" name="__tabbed_1" type="radio" /><div class="tabbed-labels"><label for="__tabbed_1_1">Python3</label><label for="__tabbed_1_2">Java</label><label for="__tabbed_1_3">C++</label><label for="__tabbed_1_4">Go</label><label for="__tabbed_1_5">TypeScript</label></div>
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en/lc/2528/index.html

@@ -60302,9 +60302,9 @@
       <ul class="md-nav__list">
 
           <li class="md-nav__item">
-  <a href="#solution-1" class="md-nav__link">
+  <a href="#solution-1-binary-search-difference-array-greedy" class="md-nav__link">
     <span class="md-ellipsis">
-      Solution 1
+      Solution 1: Binary Search + Difference Array + Greedy
     </span>
   </a>
 
@@ -80144,8 +80144,8 @@ <h2 id="description">Description</h2>
 <pre>
 <strong>Input:</strong> stations = [1,2,4,5,0], r = 1, k = 2
 <strong>Output:</strong> 5
-<strong>Explanation:</strong> 
-One of the optimal ways is to install both the power stations at city 1. 
+<strong>Explanation:</strong>
+One of the optimal ways is to install both the power stations at city 1.
 So stations will become [1,4,4,5,0].
 - City 0 is provided by 1 + 4 = 5 power stations.
 - City 1 is provided by 1 + 4 + 4 = 9 power stations.
@@ -80161,7 +80161,7 @@ <h2 id="description">Description</h2>
 <pre>
 <strong>Input:</strong> stations = [4,4,4,4], r = 0, k = 3
 <strong>Output:</strong> 4
-<strong>Explanation:</strong> 
+<strong>Explanation:</strong>
 It can be proved that we cannot make the minimum power of a city greater than 4.
 </pre>
 
@@ -80181,7 +80181,13 @@ <h2 id="description">Description</h2>
 <h2 id="solutions">Solutions</h2>
 <!-- solution:start -->
 
-<h3 id="solution-1">Solution 1</h3>
+<h3 id="solution-1-binary-search-difference-array-greedy">Solution 1: Binary Search + Difference Array + Greedy</h3>
+<p>According to the problem description, the minimum number of power stations increases as the value of $k$ increases. Therefore, we can use binary search to find the largest minimum number of power stations, ensuring that the additional power stations needed do not exceed $k$.</p>
+<p>First, we use a difference array and prefix sum to calculate the initial number of power stations in each city, recording it in the array $s$, where $s[i]$ represents the number of power stations in the $i$-th city.</p>
+<p>Next, we define the left boundary of the binary search as $0$ and the right boundary as $2^{40}$. Then, we implement a function $check(x, k)$ to determine whether the minimum number of power stations in the cities can be $x$, ensuring that the additional power stations needed do not exceed $k$.</p>
+<p>The implementation logic of the function $check(x, k)$ is as follows:</p>
+<p>Traverse each city. If the number of power stations in the current city $i$ is less than $x$, we can greedily build a power station at the rightmost possible position, $j = \min(i + r, n - 1)$, to cover as many cities as possible. During this process, we can use the difference array to add a certain value to a continuous segment. If the number of additional power stations needed exceeds $k$, then $x$ does not meet the condition, and we return <code>false</code>. Otherwise, after the traversal, return <code>true</code>.</p>
+<p>The time complexity is $O(n \times \log M)$, and the space complexity is $O(n)$. Here, $n$ is the number of cities, and $M$ is fixed at $2^{40}$.</p>
 <div class="tabbed-set tabbed-alternate" data-tabs="1:5"><input checked="checked" id="__tabbed_1_1" name="__tabbed_1" type="radio" /><input id="__tabbed_1_2" name="__tabbed_1" type="radio" /><input id="__tabbed_1_3" name="__tabbed_1" type="radio" /><input id="__tabbed_1_4" name="__tabbed_1" type="radio" /><input id="__tabbed_1_5" name="__tabbed_1" type="radio" /><div class="tabbed-labels"><label for="__tabbed_1_1">Python3</label><label for="__tabbed_1_2">Java</label><label for="__tabbed_1_3">C++</label><label for="__tabbed_1_4">Go</label><label for="__tabbed_1_5">TypeScript</label></div>
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en/search/search_index.json

@@ -84737,7 +84737,7 @@ <h2 id="_1">题目描述</h2>
 [null, false, false, true, false]
 
 <strong>解释：</strong>
-DataStream dataStream = new DataStream(4, 3); // value = 4, k = 3 
+DataStream dataStream = new DataStream(4, 3); // value = 4, k = 3
 dataStream.consec(4); // 数据流中只有 1 个整数，所以返回 False 。
 dataStream.consec(4); // 数据流中只有 2 个整数
                       // 由于 2 小于 k ，返回 False 。
@@ -84762,8 +84762,8 @@ <h2 id="_2">解法</h2>
 <!-- solution:start -->
 
 <h3 id="_3">方法一：计数</h3>
-<p>维护一个计数器 $cnt$，记录当前连续整数为 <code>value</code> 的个数。</p>
-<p>当 <code>num</code> 与 <code>value</code> 相等时，$cnt$ 自增 1，否则 $cnt$ 重置为 0。然后判断 $cnt$ 是否大于等于 <code>k</code> 即可。</p>
+<p>我们可以维护一个计数器 $\textit{cnt}$，记录当前连续整数为 $\textit{value}$ 的个数。</p>
+<p>调用 <code>consec</code> 方法时，如果 $\textit{num}$ 与 $\textit{value}$ 相等，我们将 $\textit{cnt}$ 自增 1，否则将 $\textit{cnt}$ 重置为 0。然后判断 $\textit{cnt}$ 是否大于等于 $\textit{k}$ 即可。</p>
 <p>时间复杂度 $O(1)$，空间复杂度 $O(1)$。</p>
 <div class="tabbed-set tabbed-alternate" data-tabs="1:5"><input checked="checked" id="__tabbed_1_1" name="__tabbed_1" type="radio" /><input id="__tabbed_1_2" name="__tabbed_1" type="radio" /><input id="__tabbed_1_3" name="__tabbed_1" type="radio" /><input id="__tabbed_1_4" name="__tabbed_1" type="radio" /><input id="__tabbed_1_5" name="__tabbed_1" type="radio" /><div class="tabbed-labels"><label for="__tabbed_1_1">Python3</label><label for="__tabbed_1_2">Java</label><label for="__tabbed_1_3">C++</label><label for="__tabbed_1_4">Go</label><label for="__tabbed_1_5">TypeScript</label></div>
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lc/2526/index.html

@@ -84788,7 +84788,7 @@ <h3 id="_3">方法一：二分查找 + 差分数组 + 贪心</h3>
 <p>我们先利用差分数组以及前缀和算出初始时每座城市的供电站数目，记录在数组 $s$ 中，其中 $s[i]$ 表示第 $i$ 座城市的供电站数目。</p>
 <p>接下来，我们定义二分查找的左边界为 $0$，右边界为 $2^{40}$。然后实现一个 $check(x, k)$ 函数，用于判断是否城市供电站数目的最小值是否可以为 $x$，使得额外建造的供电站不超过 $k$ 座。</p>
 <p>函数 $check(x, k)$ 的实现逻辑是：</p>
-<p>遍历每座城市，如果当前城市 $i$ 的供电站数目小于 $x$，此时我们可以贪心地在尽可能右边的位置上建造供电站，位置 $j = min(i + r, n - 1)$，这样可以使得供电站覆盖尽可能多的城市。过程中我们可以借助差分数组，给一段连续的位置加上某个值。如果需要额外建造的供电站数量超过 $k$，那么 $x$ 不满足条件，返回 <code>false</code>。否则遍历结束后，返回 <code>true</code>。</p>
+<p>遍历每座城市，如果当前城市 $i$ 的供电站数目小于 $x$，此时我们可以贪心地在尽可能右边的位置上建造供电站，位置 $j = \min(i + r, n - 1)$，这样可以使得供电站覆盖尽可能多的城市。过程中我们可以借助差分数组，给一段连续的位置加上某个值。如果需要额外建造的供电站数量超过 $k$，那么 $x$ 不满足条件，返回 <code>false</code>。否则遍历结束后，返回 <code>true</code>。</p>
 <p>时间复杂度 $O(n \times \log M)$，空间复杂度 $O(n)$。其中 $n$ 为城市数量，而 $M$ 我们固定取 $2^{40}$。</p>
 <div class="tabbed-set tabbed-alternate" data-tabs="1:5"><input checked="checked" id="__tabbed_1_1" name="__tabbed_1" type="radio" /><input id="__tabbed_1_2" name="__tabbed_1" type="radio" /><input id="__tabbed_1_3" name="__tabbed_1" type="radio" /><input id="__tabbed_1_4" name="__tabbed_1" type="radio" /><input id="__tabbed_1_5" name="__tabbed_1" type="radio" /><div class="tabbed-labels"><label for="__tabbed_1_1">Python3</label><label for="__tabbed_1_2">Java</label><label for="__tabbed_1_3">C++</label><label for="__tabbed_1_4">Go</label><label for="__tabbed_1_5">TypeScript</label></div>
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