deploy: da2ffb8

Author: yanglbme

en/lc/3325/index.html

@@ -79778,9 +79778,9 @@ <h2 id="solutions">Solutions</h2>
 <!-- solution:start -->
 
 <h3 id="solution-1-mathematics">Solution 1: Mathematics</h3>
-<p>According to the problem description, we can add $k$ to the maximum value in the array and subtract $k$ from the minimum value. This can reduce the difference between the maximum and minimum values in the array.</p>
-<p>Therefore, the final answer is $\max(nums) - \min(nums) - 2 \times k$.</p>
-<p>The time complexity is $O(n)$, where $n$ is the length of the array <code>nums</code>. The space complexity is $O(1)$.</p>
+<p>According to the problem description, we can subtract $k$ from the maximum value in the array and add $k$ to the minimum value in the array, which can reduce the difference between the maximum and minimum values in the array.</p>
+<p>Therefore, the final answer is the larger value between $\max(\textit{nums}) - \min(\textit{nums}) - 2 \times k$ and $0$.</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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