perf: update python solution to lc problem: No.3019

solution/3000-3099/3019.Number of Changing Keys/README.md

@@ -31,7 +31,7 @@ tags:
 <pre>
 <strong>输入：</strong>s = "aAbBcC"
 <strong>输出：</strong>2
-<strong>解释：</strong> 
+<strong>解释：</strong>
 从 s[0] = 'a' 到 s[1] = 'A'，不存在按键变更，因为不计入 caps lock 或 shift 。
 从 s[1] = 'A' 到 s[2] = 'b'，按键变更。
 从 s[2] = 'b' 到 s[3] = 'B'，不存在按键变更，因为不计入 caps lock 或 shift 。
@@ -75,7 +75,7 @@ tags:
 ```python
 class Solution:
     def countKeyChanges(self, s: str) -> int:
-        return sum(a.lower() != b.lower() for a, b in pairwise(s))
+        return sum(a != b for a, b in pairwise(s.lower()))
 ```
 
 #### Java

solution/3000-3099/3019.Number of Changing Keys/README_EN.md

@@ -30,7 +30,7 @@ tags:
 <pre>
 <strong>Input:</strong> s = &quot;aAbBcC&quot;
 <strong>Output:</strong> 2
-<strong>Explanation:</strong> 
+<strong>Explanation:</strong>
 From s[0] = &#39;a&#39; to s[1] = &#39;A&#39;, there is no change of key as caps lock or shift is not counted.
 From s[1] = &#39;A&#39; to s[2] = &#39;b&#39;, there is a change of key.
 From s[2] = &#39;b&#39; to s[3] = &#39;B&#39;, there is no change of key as caps lock or shift is not counted.
@@ -74,7 +74,7 @@ The time complexity is $O(n)$, where $n$ is the length of the string $s$. The sp
 ```python
 class Solution:
     def countKeyChanges(self, s: str) -> int:
-        return sum(a.lower() != b.lower() for a, b in pairwise(s))
+        return sum(a != b for a, b in pairwise(s.lower()))
 ```
 
 #### Java

solution/3000-3099/3019.Number of Changing Keys/Solution.py

@@ -1,3 +1,3 @@
 class Solution:
     def countKeyChanges(self, s: str) -> int:
-        return sum(a.lower() != b.lower() for a, b in pairwise(s))
+        return sum(a != b for a, b in pairwise(s.lower()))