feat: add solutions to lc problem: No.3019 (doocs#2278)

No.3019.Number of Changing Keys

Author: yanglbme

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

@@ -46,28 +46,30 @@
 
 ## 解法
 
-### 方法一
+### 方法一：一次遍历
+
+我们可以遍历字符串，每次判断当前字符的小写形式是否与前一个字符的小写形式相同，如果不同则说明发生了按键变更，将答案加一即可。
+
+时间复杂度 $O(n)$，其中 $n$ 是字符串 $s$ 的长度。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
 ```python
 class Solution:
     def countKeyChanges(self, s: str) -> int:
-        return sum(s.lower()[i] != s.lower()[i-1] for i in range(1, len(s)))
+        return sum(a.lower() != b.lower() for a, b in pairwise(s))
 ```
 
 ```java
 class Solution {
     public int countKeyChanges(String s) {
-        int n = s.length();
-        int count = 0;
-        String lowerCaseString = s.toLowerCase();
-        for (int i = 0; i < n - 1; i++) {
-            if (lowerCaseString.charAt(i) != lowerCaseString.charAt(i + 1)) {
-                count++;
+        int ans = 0;
+        for (int i = 1; i < s.length(); ++i) {
+            if (Character.toLowerCase(s.charAt(i)) != Character.toLowerCase(s.charAt(i - 1))) {
+                ++ans;
             }
         }
-        return count;
+        return ans;
     }
 }
 ```
@@ -76,30 +78,38 @@ class Solution {
 class Solution {
 public:
     int countKeyChanges(string s) {
-        int n = s.size();
-        int c = 0;
         transform(s.begin(), s.end(), s.begin(), ::tolower);
-        for (int i = 0; i < n - 1; i++) {
-            if (s[i] != s[i + 1]) {
-                c++;
-            }
+        int ans = 0;
+        for (int i = 1; i < s.size(); ++i) {
+            ans += s[i] != s[i - 1];
         }
-        return c;
+        return ans;
     }
 };
 ```
 
 ```go
-func countKeyChanges(s string) int {
-	n := len(s)
-	count := 0
+func countKeyChanges(s string) (ans int) {
 	s = strings.ToLower(s)
-	for i := 0; i < n-1; i++ {
-		if s[i] != s[i+1] {
-			count++
+	for i, c := range s[1:] {
+		if byte(c) != s[i] {
+			ans++
 		}
 	}
-	return count
+	return
+}
+```
+
+```ts
+function countKeyChanges(s: string): number {
+    s = s.toLowerCase();
+    let ans = 0;
+    for (let i = 1; i < s.length; ++i) {
+        if (s[i] !== s[i - 1]) {
+            ++ans;
+        }
+    }
+    return ans;
 }
 ```
 

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

@@ -43,28 +43,30 @@ From s[4] = 'c' to s[5] = 'C', there is no change of key as caps
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Single Pass
+
+We can traverse the string, each time checking whether the lowercase form of the current character is the same as the lowercase form of the previous character. If they are different, it means that the key has been changed, and we can increment the answer accordingly.
+
+The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
 ```python
 class Solution:
     def countKeyChanges(self, s: str) -> int:
-        return sum(s.lower()[i] != s.lower()[i-1] for i in range(1, len(s)))
+        return sum(a.lower() != b.lower() for a, b in pairwise(s))
 ```
 
 ```java
 class Solution {
     public int countKeyChanges(String s) {
-        int n = s.length();
-        int count = 0;
-        String lowerCaseString = s.toLowerCase();
-        for (int i = 0; i < n - 1; i++) {
-            if (lowerCaseString.charAt(i) != lowerCaseString.charAt(i + 1)) {
-                count++;
+        int ans = 0;
+        for (int i = 1; i < s.length(); ++i) {
+            if (Character.toLowerCase(s.charAt(i)) != Character.toLowerCase(s.charAt(i - 1))) {
+                ++ans;
             }
         }
-        return count;
+        return ans;
     }
 }
 ```
@@ -73,30 +75,38 @@ class Solution {
 class Solution {
 public:
     int countKeyChanges(string s) {
-        int n = s.size();
-        int c = 0;
         transform(s.begin(), s.end(), s.begin(), ::tolower);
-        for (int i = 0; i < n - 1; i++) {
-            if (s[i] != s[i + 1]) {
-                c++;
-            }
+        int ans = 0;
+        for (int i = 1; i < s.size(); ++i) {
+            ans += s[i] != s[i - 1];
         }
-        return c;
+        return ans;
     }
 };
 ```
 
 ```go
-func countKeyChanges(s string) int {
-	n := len(s)
-	count := 0
+func countKeyChanges(s string) (ans int) {
 	s = strings.ToLower(s)
-	for i := 0; i < n-1; i++ {
-		if s[i] != s[i+1] {
-			count++
+	for i, c := range s[1:] {
+		if byte(c) != s[i] {
+			ans++
 		}
 	}
-	return count
+	return
+}
+```
+
+```ts
+function countKeyChanges(s: string): number {
+    s = s.toLowerCase();
+    let ans = 0;
+    for (let i = 1; i < s.length; ++i) {
+        if (s[i] !== s[i - 1]) {
+            ++ans;
+        }
+    }
+    return ans;
 }
 ```
 

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

@@ -1,14 +1,11 @@
 class Solution {
 public:
     int countKeyChanges(string s) {
-        int n = s.size();
-        int c = 0;
         transform(s.begin(), s.end(), s.begin(), ::tolower);
-        for (int i = 0; i < n - 1; i++) {
-            if (s[i] != s[i + 1]) {
-                c++;
-            }
+        int ans = 0;
+        for (int i = 1; i < s.size(); ++i) {
+            ans += s[i] != s[i - 1];
         }
-        return c;
+        return ans;
     }
-};
+};

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

@@ -1,11 +1,9 @@
-func countKeyChanges(s string) int {
-	n := len(s)
-	count := 0
+func countKeyChanges(s string) (ans int) {
 	s = strings.ToLower(s)
-	for i := 0; i < n-1; i++ {
-		if s[i] != s[i+1] {
-			count++
+	for i, c := range s[1:] {
+		if byte(c) != s[i] {
+			ans++
 		}
 	}
-	return count
+	return
 }

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

@@ -1,13 +1,11 @@
 class Solution {
     public int countKeyChanges(String s) {
-        int n = s.length();
-        int count = 0;
-        String lowerCaseString = s.toLowerCase();
-        for (int i = 0; i < n - 1; i++) {
-            if (lowerCaseString.charAt(i) != lowerCaseString.charAt(i + 1)) {
-                count++;
+        int ans = 0;
+        for (int i = 1; i < s.length(); ++i) {
+            if (Character.toLowerCase(s.charAt(i)) != Character.toLowerCase(s.charAt(i - 1))) {
+                ++ans;
             }
         }
-        return count;
+        return ans;
     }
-}
+}

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

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

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

@@ -0,0 +1,10 @@
+function countKeyChanges(s: string): number {
+    s = s.toLowerCase();
+    let ans = 0;
+    for (let i = 1; i < s.length; ++i) {
+        if (s[i] !== s[i - 1]) {
+            ++ans;
+        }
+    }
+    return ans;
+}