deploy: da21eee

Author: yanglbme

en/lc/1110/index.html

@@ -55125,9 +55125,9 @@
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-  <a href="#solution-1" class="md-nav__link">
+  <a href="#solution-1-simulation" class="md-nav__link">
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-      Solution 1
+      Solution 1: Simulation
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@@ -77364,7 +77364,9 @@ <h2 id="description">Description</h2>
 <h2 id="solutions">Solutions</h2>
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-<h3 id="solution-1">Solution 1</h3>
+<h3 id="solution-1-simulation">Solution 1: Simulation</h3>
+<p>According to the problem statement, we can simulate the entire process, and the remaining number will be the answer. In implementation, we do not need to create an additional array; we can directly operate on the original array.</p>
+<p>The time complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$. The space complexity is $O(1)$.</p>
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en/lc/2293/index.html

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       <ul class="md-nav__list">
 
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-  <a href="#solution-1" class="md-nav__link">
+  <a href="#solution-1-greedy-sorting" class="md-nav__link">
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-      Solution 1
+      Solution 1: Greedy + Sorting
     </span>
   </a>
 
@@ -77379,7 +77379,9 @@ <h2 id="description">Description</h2>
 <h2 id="solutions">Solutions</h2>
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-<h3 id="solution-1">Solution 1</h3>
+<h3 id="solution-1-greedy-sorting">Solution 1: Greedy + Sorting</h3>
+<p>The problem requires dividing into subsequences, not subarrays, so the elements in a subsequence can be non-continuous. We can sort the array $\textit{nums}$. Assuming the first element of the current subsequence is $a$, the difference between the maximum and minimum values in the subsequence will not exceed $k$. Therefore, we can iterate through the array $\textit{nums}$. If the difference between the current element $b$ and $a$ is greater than $k$, then update $a$ to $b$ and increase the number of subsequences by 1. After the iteration, we can obtain the minimum number of subsequences, noting that the initial number of subsequences is $1$.</p>
+<p>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 $\textit{nums}$.</p>
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