feat: add solutions to lc problem: No.2083

No.2083.Substrings That Begin and End With the Same Letter

Author: yanglbme

solution/2000-2099/2083.Substrings That Begin and End With the Same Letter/README.md

@@ -6,70 +6,131 @@
 
 <!-- 这里写题目描述 -->
 
-<p>You are given a <strong>0-indexed</strong> string <code>s</code> consisting of only lowercase English letters. Return <em>the number of <strong>substrings</strong> in </em><code>s</code> <em>that begin and end with the <strong>same</strong> character.</em></p>
+<p>给你一个仅由小写英文字母组成的，&nbsp; 下标从 <code>0</code> 开始的字符串 <code>s</code> 。返回 <code>s</code> 中以相同字符开头和结尾的子字符串总数。</p>
 
-<p>A <strong>substring</strong> is a contiguous non-empty sequence of characters within a string.</p>
+<p>子字符串是字符串中连续的非空字符序列。</p>
 
 <p>&nbsp;</p>
-<p><strong>Example 1:</strong></p>
+
+<p><strong>示例 1：</strong></p>
 
 <pre>
-<strong>Input:</strong> s = &quot;abcba&quot;
-<strong>Output:</strong> 7
-<strong>Explanation:</strong>
-The substrings of length 1 that start and end with the same letter are: &quot;a&quot;, &quot;b&quot;, &quot;c&quot;, &quot;b&quot;, and &quot;a&quot;.
-The substring of length 3 that starts and ends with the same letter is: &quot;bcb&quot;.
-The substring of length 5 that starts and ends with the same letter is: &quot;abcba&quot;.
+<strong>输入：</strong>s = "abcba"
+<strong>输出：</strong>7
+<strong>解释：</strong>
+以相同字母开头和结尾的长度为 1 的子串是："a"、"b"、"c"、"b" 和 "a" 。
+以相同字母开头和结尾的长度为 3 的子串是："bcb" 。
+以相同字母开头和结尾的长度为 5 的子串是："abcba" 。
 </pre>
 
-<p><strong>Example 2:</strong></p>
+<p><strong>示例 2：</strong></p>
 
 <pre>
-<strong>Input:</strong> s = &quot;abacad&quot;
-<strong>Output:</strong> 9
-<strong>Explanation:</strong>
-The substrings of length 1 that start and end with the same letter are: &quot;a&quot;, &quot;b&quot;, &quot;a&quot;, &quot;c&quot;, &quot;a&quot;, and &quot;d&quot;.
-The substrings of length 3 that start and end with the same letter are: &quot;aba&quot; and &quot;aca&quot;.
-The substring of length 5 that starts and ends with the same letter is: &quot;abaca&quot;.
+<strong>输入：</strong>s = "abacad"
+<strong>输出：</strong>9
+<strong>解释：</strong>
+以相同字母开头和结尾的长度为 1 的子串是："a"、"b"、"a"、"c"、"a" 和 "d" 。
+以相同字母开头和结尾的长度为 3 的子串是："aba" 和 "aca" 。
+以相同字母开头和结尾的长度为 5 的子串是："abaca" 。
 </pre>
 
-<p><strong>Example 3:</strong></p>
+<p><strong>示例 3：</strong></p>
 
 <pre>
-<strong>Input:</strong> s = &quot;a&quot;
-<strong>Output:</strong> 1
-<strong>Explanation:</strong>
-The substring of length 1 that starts and ends with the same letter is: &quot;a&quot;.
+<strong>输入：</strong>s = "a"
+<strong>输出：</strong>1
+<strong>解释：</strong>
+只有一个，以相同字母开头和结尾的长度为 1 的子串是："a"。
 </pre>
 
 <p>&nbsp;</p>
-<p><strong>Constraints:</strong></p>
+
+<p><strong>提示：</strong></p>
 
 <ul>
 	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
-	<li><code>s</code> consists only of lowercase English letters.</li>
+	<li><code>s</code> 仅包含小写英文字母。</li>
 </ul>
 
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
 
+前缀和 + 计数器。
+
+用 counter 累计以每个字符结尾的字符串的个数。
+
+遍历字符串 s 中每个字符 c，累加以 c 字符串结尾的字符串个数。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def numberOfSubstrings(self, s: str) -> int:
+        counter = [0] * 26
+        ans = 0
+        for c in s:
+            i = ord(c) - ord('a')
+            counter[i] += 1
+            ans += counter[i]
+        return ans
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public long numberOfSubstrings(String s) {
+        int[] counter = new int[26];
+        long ans = 0;
+        for (char c : s.toCharArray()) {
+            int i = c - 'a';
+            ++counter[i];
+            ans += counter[i];
+        }
+        return ans;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    long long numberOfSubstrings(string s) {
+        vector<int> counter(26);
+        long long ans = 0;
+        for (char c : s)
+        {
+            int i = c - 'a';
+            ++counter[i];
+            ans += counter[i];
+        }
+        return ans;
+    }
+};
+```
 
+### **Go**
+
+```go
+func numberOfSubstrings(s string) int64 {
+	var ans int64
+	counter := make([]int64, 26)
+	for _, c := range s {
+		i := c - 'a'
+		counter[i]++
+		ans += counter[i]
+	}
+	return ans
+}
 ```
 
 ### **...**

solution/2000-2099/2083.Substrings That Begin and End With the Same Letter/README_EN.md

@@ -55,13 +55,66 @@ The substring of length 1 that starts and ends with the same letter is: "a&
 ### **Python3**
 
 ```python
-
+class Solution:
+    def numberOfSubstrings(self, s: str) -> int:
+        counter = [0] * 26
+        ans = 0
+        for c in s:
+            i = ord(c) - ord('a')
+            counter[i] += 1
+            ans += counter[i]
+        return ans
 ```
 
 ### **Java**
 
 ```java
+class Solution {
+    public long numberOfSubstrings(String s) {
+        int[] counter = new int[26];
+        long ans = 0;
+        for (char c : s.toCharArray()) {
+            int i = c - 'a';
+            ++counter[i];
+            ans += counter[i];
+        }
+        return ans;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    long long numberOfSubstrings(string s) {
+        vector<int> counter(26);
+        long long ans = 0;
+        for (char c : s)
+        {
+            int i = c - 'a';
+            ++counter[i];
+            ans += counter[i];
+        }
+        return ans;
+    }
+};
+```
 
+### **Go**
+
+```go
+func numberOfSubstrings(s string) int64 {
+	var ans int64
+	counter := make([]int64, 26)
+	for _, c := range s {
+		i := c - 'a'
+		counter[i]++
+		ans += counter[i]
+	}
+	return ans
+}
 ```
 
 ### **...**

solution/2000-2099/2083.Substrings That Begin and End With the Same Letter/Solution.cpp

@@ -0,0 +1,14 @@
+class Solution {
+public:
+    long long numberOfSubstrings(string s) {
+        vector<int> counter(26);
+        long long ans = 0;
+        for (char c : s)
+        {
+            int i = c - 'a';
+            ++counter[i];
+            ans += counter[i];
+        }
+        return ans;
+    }
+};

solution/2000-2099/2083.Substrings That Begin and End With the Same Letter/Solution.go

@@ -0,0 +1,10 @@
+func numberOfSubstrings(s string) int64 {
+	var ans int64
+	counter := make([]int64, 26)
+	for _, c := range s {
+		i := c - 'a'
+		counter[i]++
+		ans += counter[i]
+	}
+	return ans
+}

solution/2000-2099/2083.Substrings That Begin and End With the Same Letter/Solution.java

@@ -0,0 +1,12 @@
+class Solution {
+    public long numberOfSubstrings(String s) {
+        int[] counter = new int[26];
+        long ans = 0;
+        for (char c : s.toCharArray()) {
+            int i = c - 'a';
+            ++counter[i];
+            ans += counter[i];
+        }
+        return ans;
+    }
+}

solution/2000-2099/2083.Substrings That Begin and End With the Same Letter/Solution.py

@@ -0,0 +1,9 @@
+class Solution:
+    def numberOfSubstrings(self, s: str) -> int:
+        counter = [0] * 26
+        ans = 0
+        for c in s:
+            i = ord(c) - ord('a')
+            counter[i] += 1
+            ans += counter[i]
+        return ans