feat: add solutions to lc problem: No.0242

No.0242.Valid Anagram

Author: yanglbme

solution/0200-0299/0242.Valid Anagram/README.md

@@ -42,7 +42,13 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
-数组或哈希表累加 s 中每个字符出现的次数，再减去 t 中对应的每个字符出现的次数。遍历结束后，若字符中出现次数不为 0 的情况，返回 false，否则返回 true。
+**方法一：计数**
+
+我们先判断两个字符串的长度是否相等，如果不相等，说明两个字符串中的字符肯定不同，返回 `false`。
+
+否则，我们用哈希表或者一个长度为 $26$ 的数组来记录字符串 $s$ 中每个字符出现的次数，然后遍历另一个字符串 $t$，每遍历到一个字符，就将哈希表中对应的字符次数减一，如果减一后的次数小于 $0$，说明该字符在两个字符串中出现的次数不同，返回 `false`。如果遍历完两个字符串后，哈希表中的所有字符次数都为 $0$，说明两个字符串中的字符出现次数相同，返回 `true`。
+
+时间复杂度 $O(n)$，空间复杂度 $O(C)$，其中 $n$ 是字符串的长度；而 $C$ 是字符集的大小，本题中 $C=26$。
 
 <!-- tabs:start -->
 
@@ -55,11 +61,12 @@ class Solution:
     def isAnagram(self, s: str, t: str) -> bool:
         if len(s) != len(t):
             return False
-        chars = [0] * 26
-        for i in range(len(s)):
-            chars[ord(s[i]) - ord('a')] += 1
-            chars[ord(t[i]) - ord('a')] -= 1
-        return all(c == 0 for c in chars)
+        cnt = Counter(s)
+        for c in t:
+            cnt[c] -= 1
+            if cnt[c] < 0:
+                return False
+        return True
 ```
 
 ### **Java**
@@ -72,13 +79,13 @@ class Solution {
         if (s.length() != t.length()) {
             return false;
         }
-        int[] chars = new int[26];
+        int[] cnt = new int[26];
         for (int i = 0; i < s.length(); ++i) {
-            ++chars[s.charAt(i) - 'a'];
-            --chars[t.charAt(i) - 'a'];
+            ++cnt[s.charAt(i) - 'a'];
+            --cnt[t.charAt(i) - 'a'];
         }
-        for (int c : chars) {
-            if (c != 0) {
+        for (int i = 0; i < 26; ++i) {
+            if (cnt[i] != 0) {
                 return false;
             }
         }
@@ -93,18 +100,15 @@ class Solution {
 class Solution {
 public:
     bool isAnagram(string s, string t) {
-        if (s.size() != t.size())
+        if (s.size() != t.size()) {
             return false;
-        vector<int> chars(26, 0);
-        for (int i = 0, n = s.size(); i < n; ++i) {
-            ++chars[s[i] - 'a'];
-            --chars[t[i] - 'a'];
         }
-        for (int c : chars) {
-            if (c != 0)
-                return false;
+        vector<int> cnt(26);
+        for (int i = 0; i < s.size(); ++i) {
+            ++cnt[s[i] - 'a'];
+            --cnt[t[i] - 'a'];
         }
-        return true;
+        return all_of(cnt.begin(), cnt.end(), [](int x) { return x == 0; });
     }
 };
 ```
@@ -116,13 +120,13 @@ func isAnagram(s string, t string) bool {
 	if len(s) != len(t) {
 		return false
 	}
-	var chars [26]int
+	cnt := [26]int{}
 	for i := 0; i < len(s); i++ {
-		chars[s[i]-'a']++
-		chars[t[i]-'a']--
+		cnt[s[i]-'a']++
+		cnt[t[i]-'a']--
 	}
-	for _, c := range chars {
-		if c != 0 {
+	for _, v := range cnt {
+		if v != 0 {
 			return false
 		}
 	}
@@ -139,40 +143,49 @@ func isAnagram(s string, t string) bool {
  * @return {boolean}
  */
 var isAnagram = function (s, t) {
-    if (s.length != t.length) return false;
-    let record = new Array(26).fill(0);
-    let base = 'a'.charCodeAt(0);
+    if (s.length !== t.length) {
+        return false;
+    }
+    const cnt = new Array(26).fill(0);
     for (let i = 0; i < s.length; ++i) {
-        ++record[s.charCodeAt(i) - base];
-        --record[t.charCodeAt(i) - base];
+        ++cnt[s.charCodeAt(i) - 'a'.charCodeAt(0)];
+        --cnt[t.charCodeAt(i) - 'a'.charCodeAt(0)];
     }
-    return record.every(v => v == 0);
+    return cnt.every(x => x === 0);
 };
 ```
 
 ### **TypeScript**
 
 ```ts
 function isAnagram(s: string, t: string): boolean {
-    const n = s.length;
-    const m = t.length;
-    return n === m && [...s].sort().join('') === [...t].sort().join('');
+    if (s.length !== t.length) {
+        return false;
+    }
+    const cnt = new Array(26).fill(0);
+    for (let i = 0; i < s.length; ++i) {
+        ++cnt[s.charCodeAt(i) - 'a'.charCodeAt(0)];
+        --cnt[t.charCodeAt(i) - 'a'.charCodeAt(0)];
+    }
+    return cnt.every(x => x === 0);
 }
 ```
 
-```ts
-function isAnagram(s: string, t: string): boolean {
-    const n = s.length;
-    const m = t.length;
-    if (n !== m) {
-        return false;
-    }
-    const count = new Array(26).fill(0);
-    for (let i = 0; i < n; i++) {
-        count[s.charCodeAt(i) - 'a'.charCodeAt(0)]++;
-        count[t.charCodeAt(i) - 'a'.charCodeAt(0)]--;
+### **C#**
+
+```cs
+public class Solution {
+    public bool IsAnagram(string s, string t) {
+        if (s.Length != t.Length) {
+            return false;
+        }
+        int[] cnt = new int[26];
+        for (int i = 0; i < s.Length; ++i) {
+            ++cnt[s[i] - 'a'];
+            --cnt[t[i] - 'a'];
+        }
+        return cnt.All(x => x == 0);
     }
-    return count.every(v => v === 0);
 }
 ```
 

solution/0200-0299/0242.Valid Anagram/README_EN.md

@@ -29,6 +29,14 @@
 
 ## Solutions
 
+**Approach 1: Counting**
+
+We first determine whether the length of the two strings is equal. If they are not equal, the characters in the two strings must be different, so return `false`.
+
+Otherwise, we use a hash table or an array of length $26$ to record the number of times each character appears in the string $s$, and then traverse the other string $t$. Each time we traverse a character, we subtract the number of times the corresponding character appears in the hash table by one. If the number of times after subtraction is less than $0$, the number of times the character appears in the two strings is different, return `false`. If after traversing the two strings, all the character counts in the hash table are $0$, it means that the characters in the two strings appear the same number of times, return `true`.
+
+The time complexity is $O(n)$, the space complexity is $O(C)$, where $n$ is the length of the string; and $C$ is the size of the character set, which is $C=26$ in this problem.
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -38,11 +46,12 @@ class Solution:
     def isAnagram(self, s: str, t: str) -> bool:
         if len(s) != len(t):
             return False
-        chars = [0] * 26
-        for i in range(len(s)):
-            chars[ord(s[i]) - ord('a')] += 1
-            chars[ord(t[i]) - ord('a')] -= 1
-        return all(c == 0 for c in chars)
+        cnt = Counter(s)
+        for c in t:
+            cnt[c] -= 1
+            if cnt[c] < 0:
+                return False
+        return True
 ```
 
 ### **Java**
@@ -53,13 +62,13 @@ class Solution {
         if (s.length() != t.length()) {
             return false;
         }
-        int[] chars = new int[26];
+        int[] cnt = new int[26];
         for (int i = 0; i < s.length(); ++i) {
-            ++chars[s.charAt(i) - 'a'];
-            --chars[t.charAt(i) - 'a'];
+            ++cnt[s.charAt(i) - 'a'];
+            --cnt[t.charAt(i) - 'a'];
         }
-        for (int c : chars) {
-            if (c != 0) {
+        for (int i = 0; i < 26; ++i) {
+            if (cnt[i] != 0) {
                 return false;
             }
         }
@@ -74,18 +83,15 @@ class Solution {
 class Solution {
 public:
     bool isAnagram(string s, string t) {
-        if (s.size() != t.size())
+        if (s.size() != t.size()) {
             return false;
-        vector<int> chars(26, 0);
-        for (int i = 0, n = s.size(); i < n; ++i) {
-            ++chars[s[i] - 'a'];
-            --chars[t[i] - 'a'];
         }
-        for (int c : chars) {
-            if (c != 0)
-                return false;
+        vector<int> cnt(26);
+        for (int i = 0; i < s.size(); ++i) {
+            ++cnt[s[i] - 'a'];
+            --cnt[t[i] - 'a'];
         }
-        return true;
+        return all_of(cnt.begin(), cnt.end(), [](int x) { return x == 0; });
     }
 };
 ```
@@ -97,13 +103,13 @@ func isAnagram(s string, t string) bool {
 	if len(s) != len(t) {
 		return false
 	}
-	var chars [26]int
+	cnt := [26]int{}
 	for i := 0; i < len(s); i++ {
-		chars[s[i]-'a']++
-		chars[t[i]-'a']--
+		cnt[s[i]-'a']++
+		cnt[t[i]-'a']--
 	}
-	for _, c := range chars {
-		if c != 0 {
+	for _, v := range cnt {
+		if v != 0 {
 			return false
 		}
 	}
@@ -120,40 +126,49 @@ func isAnagram(s string, t string) bool {
  * @return {boolean}
  */
 var isAnagram = function (s, t) {
-    if (s.length != t.length) return false;
-    let record = new Array(26).fill(0);
-    let base = 'a'.charCodeAt(0);
+    if (s.length !== t.length) {
+        return false;
+    }
+    const cnt = new Array(26).fill(0);
     for (let i = 0; i < s.length; ++i) {
-        ++record[s.charCodeAt(i) - base];
-        --record[t.charCodeAt(i) - base];
+        ++cnt[s.charCodeAt(i) - 'a'.charCodeAt(0)];
+        --cnt[t.charCodeAt(i) - 'a'.charCodeAt(0)];
     }
-    return record.every(v => v == 0);
+    return cnt.every(x => x === 0);
 };
 ```
 
 ### **TypeScript**
 
 ```ts
 function isAnagram(s: string, t: string): boolean {
-    const n = s.length;
-    const m = t.length;
-    return n === m && [...s].sort().join('') === [...t].sort().join('');
+    if (s.length !== t.length) {
+        return false;
+    }
+    const cnt = new Array(26).fill(0);
+    for (let i = 0; i < s.length; ++i) {
+        ++cnt[s.charCodeAt(i) - 'a'.charCodeAt(0)];
+        --cnt[t.charCodeAt(i) - 'a'.charCodeAt(0)];
+    }
+    return cnt.every(x => x === 0);
 }
 ```
 
-```ts
-function isAnagram(s: string, t: string): boolean {
-    const n = s.length;
-    const m = t.length;
-    if (n !== m) {
-        return false;
-    }
-    const count = new Array(26).fill(0);
-    for (let i = 0; i < n; i++) {
-        count[s.charCodeAt(i) - 'a'.charCodeAt(0)]++;
-        count[t.charCodeAt(i) - 'a'.charCodeAt(0)]--;
+### **C#**
+
+```cs
+public class Solution {
+    public bool IsAnagram(string s, string t) {
+        if (s.Length != t.Length) {
+            return false;
+        }
+        int[] cnt = new int[26];
+        for (int i = 0; i < s.Length; ++i) {
+            ++cnt[s[i] - 'a'];
+            --cnt[t[i] - 'a'];
+        }
+        return cnt.All(x => x == 0);
     }
-    return count.every(v => v === 0);
 }
 ```
 

solution/0200-0299/0242.Valid Anagram/Solution.cpp

@@ -1,17 +1,14 @@
 class Solution {
 public:
     bool isAnagram(string s, string t) {
-        if (s.size() != t.size())
+        if (s.size() != t.size()) {
             return false;
-        vector<int> chars(26, 0);
-        for (int i = 0, n = s.size(); i < n; ++i) {
-            ++chars[s[i] - 'a'];
-            --chars[t[i] - 'a'];
         }
-        for (int c : chars) {
-            if (c != 0)
-                return false;
+        vector<int> cnt(26);
+        for (int i = 0; i < s.size(); ++i) {
+            ++cnt[s[i] - 'a'];
+            --cnt[t[i] - 'a'];
         }
-        return true;
+        return all_of(cnt.begin(), cnt.end(), [](int x) { return x == 0; });
     }
 };

solution/0200-0299/0242.Valid Anagram/Solution.cs

@@ -0,0 +1,13 @@
+public class Solution {
+    public bool IsAnagram(string s, string t) {
+        if (s.Length != t.Length) {
+            return false;
+        }
+        int[] cnt = new int[26];
+        for (int i = 0; i < s.Length; ++i) {
+            ++cnt[s[i] - 'a'];
+            --cnt[t[i] - 'a'];
+        }
+        return cnt.All(x => x == 0);
+    }
+}

solution/0200-0299/0242.Valid Anagram/Solution.go

@@ -2,13 +2,13 @@ func isAnagram(s string, t string) bool {
 	if len(s) != len(t) {
 		return false
 	}
-	var chars [26]int
+	cnt := [26]int{}
 	for i := 0; i < len(s); i++ {
-		chars[s[i]-'a']++
-		chars[t[i]-'a']--
+		cnt[s[i]-'a']++
+		cnt[t[i]-'a']--
 	}
-	for _, c := range chars {
-		if c != 0 {
+	for _, v := range cnt {
+		if v != 0 {
 			return false
 		}
 	}