deploy: b1fe6a2

Author: yanglbme

en/lc/1526/index.html

@@ -39034,9 +39034,9 @@
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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
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@@ -77567,7 +77567,16 @@ <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 $f[i]$ as the minimum number of operations required to obtain $target[0,..i]$, initially setting $f[0] = target[0]$.</p>
+<p>For $target[i]$, if $target[i] \leq target[i-1]$, then $f[i] = f[i-1]$; otherwise, $f[i] = f[i-1] + target[i] - target[i-1]$.</p>
+<p>The final answer is $f[n-1]$.</p>
+<p>We notice that $f[i]$ only depends on $f[i-1]$, so we can maintain the operation count using just one variable.</p>
+<p>The time complexity is $O(n)$, where $n$ is the length of the array $target$. The space complexity is $O(1)$.</p>
+<p>Similar problems:</p>
+<ul>
+<li><a href="https://github.com/doocs/leetcode/blob/main/solution/3200-3299/3229.Minimum%20Operations%20to%20Make%20Array%20Equal%20to%20Target/README_EN.md">3229. Minimum Operations to Make Array Equal to Target</a></li>
+</ul>
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