deploy: 1ea1077

Author: yanglbme

en/lc/1955/index.html

@@ -50529,18 +50529,18 @@
       <ul class="md-nav__list">
 
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-  <a href="#solution-1" class="md-nav__link">
+  <a href="#solution-1-dynamic-programming" class="md-nav__link">
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-      Solution 1
+      Solution 1: Dynamic Programming
     </span>
   </a>
 
 </li>
 
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-  <a href="#solution-2" class="md-nav__link">
+  <a href="#solution-2-dynamic-programming-space-optimization" class="md-nav__link">
     <span class="md-ellipsis">
-      Solution 2
+      Solution 2: Dynamic Programming (Space Optimization)
     </span>
   </a>
 
@@ -87157,7 +87157,24 @@ <h2 id="description">Description</h2>
 <h2 id="solutions">Solutions</h2>
 <!-- solution:start -->
 
-<h3 id="solution-1">Solution 1</h3>
+<h3 id="solution-1-dynamic-programming">Solution 1: Dynamic Programming</h3>
+<p>We define <span class="arithmatex">\(f[i][j]\)</span> to represent the number of special subsequences ending with <span class="arithmatex">\(j\)</span> among the first <span class="arithmatex">\(i+1\)</span> elements. Initially, <span class="arithmatex">\(f[i][j]=0\)</span>, and if <span class="arithmatex">\(nums[0]=0\)</span>, then <span class="arithmatex">\(f[0][0]=1\)</span>.</p>
+<p>For <span class="arithmatex">\(i \gt 0\)</span>, we consider the value of <span class="arithmatex">\(nums[i]\)</span>:</p>
+<p>If <span class="arithmatex">\(nums[i] = 0\)</span>: If we do not choose <span class="arithmatex">\(nums[i]\)</span>, then <span class="arithmatex">\(f[i][0] = f[i-1][0]\)</span>; if we choose <span class="arithmatex">\(nums[i]\)</span>, then <span class="arithmatex">\(f[i][0]=f[i-1][0]+1\)</span>, because we can add a <span class="arithmatex">\(0\)</span> to the end of any special subsequence ending with <span class="arithmatex">\(0\)</span> to get a new special subsequence, or we can use <span class="arithmatex">\(nums[i]\)</span> alone as a special subsequence. Therefore, <span class="arithmatex">\(f[i][0] = 2 \times f[i - 1][0] + 1\)</span>. The rest of <span class="arithmatex">\(f[i][j]\)</span> is equal to <span class="arithmatex">\(f[i-1][j]\)</span>.</p>
+<p>If <span class="arithmatex">\(nums[i] = 1\)</span>: If we do not choose <span class="arithmatex">\(nums[i]\)</span>, then <span class="arithmatex">\(f[i][1] = f[i-1][1]\)</span>; if we choose <span class="arithmatex">\(nums[i]\)</span>, then <span class="arithmatex">\(f[i][1]=f[i-1][1]+f[i-1][0]\)</span>, because we can add a <span class="arithmatex">\(1\)</span> to the end of any special subsequence ending with <span class="arithmatex">\(0\)</span> or <span class="arithmatex">\(1\)</span> to get a new special subsequence. Therefore, <span class="arithmatex">\(f[i][1] = f[i-1][0] + 2 \times f[i - 1][1]\)</span>. The rest of <span class="arithmatex">\(f[i][j]\)</span> is equal to <span class="arithmatex">\(f[i-1][j]\)</span>.</p>
+<p>If <span class="arithmatex">\(nums[i] = 2\)</span>: If we do not choose <span class="arithmatex">\(nums[i]\)</span>, then <span class="arithmatex">\(f[i][2] = f[i-1][2]\)</span>; if we choose <span class="arithmatex">\(nums[i]\)</span>, then <span class="arithmatex">\(f[i][2]=f[i-1][2]+f[i-1][1]\)</span>, because we can add a <span class="arithmatex">\(2\)</span> to the end of any special subsequence ending with <span class="arithmatex">\(1\)</span> or <span class="arithmatex">\(2\)</span> to get a new special subsequence. Therefore, <span class="arithmatex">\(f[i][2] = f[i-1][1] + 2 \times f[i - 1][2]\)</span>. The rest of <span class="arithmatex">\(f[i][j]\)</span> is equal to <span class="arithmatex">\(f[i-1][j]\)</span>.</p>
+<p>In summary, we can get the following state transition equations:</p>
+<div class="arithmatex">\[
+\begin{aligned}
+f[i][0] &amp;= 2 \times f[i - 1][0] + 1, \quad nums[i] = 0 \\
+f[i][1] &amp;= f[i-1][0] + 2 \times f[i - 1][1], \quad nums[i] = 1 \\
+f[i][2] &amp;= f[i-1][1] + 2 \times f[i - 1][2], \quad nums[i] = 2 \\
+f[i][j] &amp;= f[i-1][j], \quad nums[i] \neq j
+\end{aligned}
+\]</div>
+<p>The final answer is <span class="arithmatex">\(f[n-1][2]\)</span>.</p>
+<p>The time complexity is <span class="arithmatex">\(O(n)\)</span>, and the space complexity is <span class="arithmatex">\(O(n)\)</span>. Where <span class="arithmatex">\(n\)</span> is the length of the array <span class="arithmatex">\(nums\)</span>.</p>
+<p>Similar code found with 1 license type</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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@@ -87412,7 +87429,10 @@ <h3 id="solution-1">Solution 1</h3>
 
 <!-- solution:start -->
 
-<h3 id="solution-2">Solution 2</h3>
+<h3 id="solution-2-dynamic-programming-space-optimization">Solution 2: Dynamic Programming (Space Optimization)</h3>
+<p>We notice that in the above state transition equations, the value of <span class="arithmatex">\(f[i][j]\)</span> is only related to <span class="arithmatex">\(f[i-1][j]\)</span>. Therefore, we can remove the first dimension and optimize the space complexity to <span class="arithmatex">\(O(1)\)</span>.</p>
+<p>We can use an array <span class="arithmatex">\(f\)</span> of length 3 to represent the number of special subsequences ending with 0, 1, and 2, respectively. For each element in the array, we update the array <span class="arithmatex">\(f\)</span> according to the value of the current element.</p>
+<p>The time complexity is <span class="arithmatex">\(O(n)\)</span>, and the space complexity is <span class="arithmatex">\(O(1)\)</span>. Where <span class="arithmatex">\(n\)</span> is the length of the array <span class="arithmatex">\(nums\)</span>.</p>
 <div class="tabbed-set tabbed-alternate" data-tabs="2:5"><input checked="checked" id="__tabbed_2_1" name="__tabbed_2" type="radio" /><input id="__tabbed_2_2" name="__tabbed_2" type="radio" /><input id="__tabbed_2_3" name="__tabbed_2" type="radio" /><input id="__tabbed_2_4" name="__tabbed_2" type="radio" /><input id="__tabbed_2_5" name="__tabbed_2" type="radio" /><div class="tabbed-labels"><label for="__tabbed_2_1">Python3</label><label for="__tabbed_2_2">Java</label><label for="__tabbed_2_3">C++</label><label for="__tabbed_2_4">Go</label><label for="__tabbed_2_5">TypeScript</label></div>
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en/search/search_index.json

@@ -50580,7 +50580,7 @@
           <li class="md-nav__item">
   <a href="#_4" class="md-nav__link">
     <span class="md-ellipsis">
-      方法二
+      方法二：动态规划（空间优化）
     </span>
   </a>
 
@@ -91959,20 +91959,19 @@ <h3 id="_3">方法一：动态规划</h3>
 <p>我们定义 <span class="arithmatex">\(f[i][j]\)</span> 表示前 <span class="arithmatex">\(i+1\)</span> 个元素中，以 <span class="arithmatex">\(j\)</span> 结尾的特殊子序列的个数。初始时 <span class="arithmatex">\(f[i][j]=0\)</span>，如果 <span class="arithmatex">\(nums[0]=0\)</span>，则 <span class="arithmatex">\(f[0][0]=1\)</span>。</p>
 <p>对于 <span class="arithmatex">\(i \gt 0\)</span>，我们考虑 <span class="arithmatex">\(nums[i]\)</span> 的值：</p>
 <p>如果 <span class="arithmatex">\(nums[i] = 0\)</span>：如果我们不选择 <span class="arithmatex">\(nums[i]\)</span>，则 <span class="arithmatex">\(f[i][0] = f[i-1][0]\)</span>；如果我们选择 <span class="arithmatex">\(nums[i]\)</span>，那么 <span class="arithmatex">\(f[i][0]=f[i-1][0]+1\)</span>，因为我们可以在任何一个以 <span class="arithmatex">\(0\)</span> 结尾的特殊子序列后面加上一个 <span class="arithmatex">\(0\)</span> 得到一个新的特殊子序列，也可以将 <span class="arithmatex">\(nums[i]\)</span> 单独作为一个特殊子序列。因此 <span class="arithmatex">\(f[i][0] = 2 \times f[i - 1][0] + 1\)</span>。其余的 <span class="arithmatex">\(f[i][j]\)</span> 与 <span class="arithmatex">\(f[i-1][j]\)</span> 相等。</p>
-<p>如果 <span class="arithmatex">\(nums[i] = 1\)</span>：如果我们不选择 <span class="arithmatex">\(nums[i]\)</span>，则 <span class="arithmatex">\(f[i][1] = f[i-1][1]\)</span>；如果我们选择 <span class="arithmatex">\(nums[i]\)</span>，那么 <span class="arithmatex">\(f[i][1]=f[i-1][1]+f[i-1][0]\)</span>，因为我们可以在任何一个以 <span class="arithmatex">\(0\)</span> 或 <span class="arithmatex">\(1\)</span> 结尾的特殊子序列后面加上一个 <span class="arithmatex">\(1\)</span> 得到一个新的特殊子序列。因此 <span class="arithmatex">\(f[i][1] = f[i-1][1] + 2 \times f[i - 1][0]\)</span>。其余的 <span class="arithmatex">\(f[i][j]\)</span> 与 <span class="arithmatex">\(f[i-1][j]\)</span> 相等。</p>
-<p>如果 <span class="arithmatex">\(nums[i] = 2\)</span>：如果我们不选择 <span class="arithmatex">\(nums[i]\)</span>，则 <span class="arithmatex">\(f[i][2] = f[i-1][2]\)</span>；如果我们选择 <span class="arithmatex">\(nums[i]\)</span>，那么 <span class="arithmatex">\(f[i][2]=f[i-1][2]+f[i-1][1]\)</span>，因为我们可以在任何一个以 <span class="arithmatex">\(1\)</span> 或 <span class="arithmatex">\(2\)</span> 结尾的特殊子序列后面加上一个 <span class="arithmatex">\(2\)</span> 得到一个新的特殊子序列。因此 <span class="arithmatex">\(f[i][2] = f[i-1][2] + 2 \times f[i - 1][1]\)</span>。其余的 <span class="arithmatex">\(f[i][j]\)</span> 与 <span class="arithmatex">\(f[i-1][j]\)</span> 相等。</p>
+<p>如果 <span class="arithmatex">\(nums[i] = 1\)</span>：如果我们不选择 <span class="arithmatex">\(nums[i]\)</span>，则 <span class="arithmatex">\(f[i][1] = f[i-1][1]\)</span>；如果我们选择 <span class="arithmatex">\(nums[i]\)</span>，那么 <span class="arithmatex">\(f[i][1]=f[i-1][1]+f[i-1][0]\)</span>，因为我们可以在任何一个以 <span class="arithmatex">\(0\)</span> 或 <span class="arithmatex">\(1\)</span> 结尾的特殊子序列后面加上一个 <span class="arithmatex">\(1\)</span> 得到一个新的特殊子序列。因此 <span class="arithmatex">\(f[i][1] = f[i-1][0] + 2 \times f[i - 1][1]\)</span>。其余的 <span class="arithmatex">\(f[i][j]\)</span> 与 <span class="arithmatex">\(f[i-1][j]\)</span> 相等。</p>
+<p>如果 <span class="arithmatex">\(nums[i] = 2\)</span>：如果我们不选择 <span class="arithmatex">\(nums[i]\)</span>，则 <span class="arithmatex">\(f[i][2] = f[i-1][2]\)</span>；如果我们选择 <span class="arithmatex">\(nums[i]\)</span>，那么 <span class="arithmatex">\(f[i][2]=f[i-1][2]+f[i-1][1]\)</span>，因为我们可以在任何一个以 <span class="arithmatex">\(1\)</span> 或 <span class="arithmatex">\(2\)</span> 结尾的特殊子序列后面加上一个 <span class="arithmatex">\(2\)</span> 得到一个新的特殊子序列。因此 <span class="arithmatex">\(f[i][2] = f[i-1][1] + 2 \times f[i - 1][2]\)</span>。其余的 <span class="arithmatex">\(f[i][j]\)</span> 与 <span class="arithmatex">\(f[i-1][j]\)</span> 相等。</p>
 <p>综上，我们可以得到如下的状态转移方程：</p>
 <div class="arithmatex">\[
 \begin{aligned}
 f[i][0] &amp;= 2 \times f[i - 1][0] + 1, \quad nums[i] = 0 \\
-f[i][1] &amp;= f[i-1][1] + 2 \times f[i - 1][0], \quad nums[i] = 1 \\
-f[i][2] &amp;= f[i-1][2] + 2 \times f[i - 1][1], \quad nums[i] = 2 \\
+f[i][1] &amp;= f[i-1][0] + 2 \times f[i - 1][1], \quad nums[i] = 1 \\
+f[i][2] &amp;= f[i-1][1] + 2 \times f[i - 1][2], \quad nums[i] = 2 \\
 f[i][j] &amp;= f[i-1][j], \quad nums[i] \neq j
 \end{aligned}
 \]</div>
 <p>最终的答案即为 <span class="arithmatex">\(f[n-1][2]\)</span>。</p>
 <p>时间复杂度 <span class="arithmatex">\(O(n)\)</span>，空间复杂度 <span class="arithmatex">\(O(n)\)</span>。其中 <span class="arithmatex">\(n\)</span> 是数组 <span class="arithmatex">\(nums\)</span> 的长度。</p>
-<p>我们注意到，上述的状态转移方程中，<span class="arithmatex">\(f[i][j]\)</span> 的值仅与 <span class="arithmatex">\(f[i-1][j]\)</span> 有关，因此我们可以去掉第一维，将空间复杂度优化到 <span class="arithmatex">\(O(1)\)</span>。</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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@@ -92227,7 +92226,8 @@ <h3 id="_3">方法一：动态规划</h3>
 
 <!-- solution:start -->
 
-<h3 id="_4">方法二</h3>
+<h3 id="_4">方法二：动态规划（空间优化）</h3>
+<p>我们注意到，上述的状态转移方程中，<span class="arithmatex">\(f[i][j]\)</span> 的值仅与 <span class="arithmatex">\(f[i-1][j]\)</span> 有关，因此我们可以去掉第一维，将空间复杂度优化到 <span class="arithmatex">\(O(1)\)</span>。</p>
 <div class="tabbed-set tabbed-alternate" data-tabs="2:5"><input checked="checked" id="__tabbed_2_1" name="__tabbed_2" type="radio" /><input id="__tabbed_2_2" name="__tabbed_2" type="radio" /><input id="__tabbed_2_3" name="__tabbed_2" type="radio" /><input id="__tabbed_2_4" name="__tabbed_2" type="radio" /><input id="__tabbed_2_5" name="__tabbed_2" type="radio" /><div class="tabbed-labels"><label for="__tabbed_2_1">Python3</label><label for="__tabbed_2_2">Java</label><label for="__tabbed_2_3">C++</label><label for="__tabbed_2_4">Go</label><label for="__tabbed_2_5">TypeScript</label></div>
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