feat: add solutions to lc problem: No.2243 (doocs#4148)

No.2243.Calculate Digit Sum of a String

Author: yanglbme

solution/2200-2299/2243.Calculate Digit Sum of a String/README.md

@@ -39,11 +39,11 @@ tags:
 <strong>输出：</strong>"135"
 <strong>解释：</strong>
 - 第一轮，将 s 分成："111"、"112"、"222" 和 "23" 。
-  接着，计算每一组的数字和：1 + 1 + 1 = 3、1 + 1 + 2 = 4、2 + 2 + 2 = 6 和 2 + 3 = 5 。 
+  接着，计算每一组的数字和：1 + 1 + 1 = 3、1 + 1 + 2 = 4、2 + 2 + 2 = 6 和 2 + 3 = 5 。
 &nbsp; 这样，s 在第一轮之后变成 "3" + "4" + "6" + "5" = "3465" 。
 - 第二轮，将 s 分成："346" 和 "5" 。
 &nbsp; 接着，计算每一组的数字和：3 + 4 + 6 = 13 、5 = 5 。
-&nbsp; 这样，s 在第二轮之后变成 "13" + "5" = "135" 。 
+&nbsp; 这样，s 在第二轮之后变成 "13" + "5" = "135" 。
 现在，s.length &lt;= k ，所以返回 "135" 作为答案。
 </pre>
 
@@ -53,7 +53,7 @@ tags:
 <strong>输出：</strong>"000"
 <strong>解释：</strong>
 将 "000", "000", and "00".
-接着，计算每一组的数字和：0 + 0 + 0 = 0 、0 + 0 + 0 = 0 和 0 + 0 = 0 。 
+接着，计算每一组的数字和：0 + 0 + 0 = 0 、0 + 0 + 0 = 0 和 0 + 0 = 0 。
 s 变为 "0" + "0" + "0" = "000" ，其长度等于 k ，所以返回 "000" 。
 </pre>
 
@@ -167,19 +167,65 @@ func digitSum(s string, k int) string {
 
 ```ts
 function digitSum(s: string, k: number): string {
-    let ans = [];
     while (s.length > k) {
+        const t: number[] = [];
         for (let i = 0; i < s.length; i += k) {
-            let cur = s.slice(i, i + k);
-            ans.push(cur.split('').reduce((a, c) => a + parseInt(c), 0));
+            const x = s
+                .slice(i, i + k)
+                .split('')
+                .reduce((a, b) => a + +b, 0);
+            t.push(x);
         }
-        s = ans.join('');
-        ans = [];
+        s = t.join('');
     }
     return s;
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn digit_sum(s: String, k: i32) -> String {
+        let mut s = s;
+        let k = k as usize;
+        while s.len() > k {
+            let mut t = Vec::new();
+            for chunk in s.as_bytes().chunks(k) {
+                let sum: i32 = chunk.iter().map(|&c| (c - b'0') as i32).sum();
+                t.push(sum.to_string());
+            }
+            s = t.join("");
+        }
+        s
+    }
+}
+```
+
+#### JavaScript
+
+```js
+/**
+ * @param {string} s
+ * @param {number} k
+ * @return {string}
+ */
+var digitSum = function (s, k) {
+    while (s.length > k) {
+        const t = [];
+        for (let i = 0; i < s.length; i += k) {
+            const x = s
+                .slice(i, i + k)
+                .split('')
+                .reduce((a, b) => a + +b, 0);
+            t.push(x);
+        }
+        s = t.join('');
+    }
+    return s;
+};
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2200-2299/2243.Calculate Digit Sum of a String/README_EN.md

@@ -37,13 +37,13 @@ tags:
 <pre>
 <strong>Input:</strong> s = &quot;11111222223&quot;, k = 3
 <strong>Output:</strong> &quot;135&quot;
-<strong>Explanation:</strong> 
+<strong>Explanation:</strong>
 - For the first round, we divide s into groups of size 3: &quot;111&quot;, &quot;112&quot;, &quot;222&quot;, and &quot;23&quot;.
-  ​​​​​Then we calculate the digit sum of each group: 1 + 1 + 1 = 3, 1 + 1 + 2 = 4, 2 + 2 + 2 = 6, and 2 + 3 = 5. 
+  ​​​​​Then we calculate the digit sum of each group: 1 + 1 + 1 = 3, 1 + 1 + 2 = 4, 2 + 2 + 2 = 6, and 2 + 3 = 5.
 &nbsp; So, s becomes &quot;3&quot; + &quot;4&quot; + &quot;6&quot; + &quot;5&quot; = &quot;3465&quot; after the first round.
 - For the second round, we divide s into &quot;346&quot; and &quot;5&quot;.
-&nbsp; Then we calculate the digit sum of each group: 3 + 4 + 6 = 13, 5 = 5. 
-&nbsp; So, s becomes &quot;13&quot; + &quot;5&quot; = &quot;135&quot; after second round. 
+&nbsp; Then we calculate the digit sum of each group: 3 + 4 + 6 = 13, 5 = 5.
+&nbsp; So, s becomes &quot;13&quot; + &quot;5&quot; = &quot;135&quot; after second round.
 Now, s.length &lt;= k, so we return &quot;135&quot; as the answer.
 </pre>
 
@@ -52,9 +52,9 @@ Now, s.length &lt;= k, so we return &quot;135&quot; as the answer.
 <pre>
 <strong>Input:</strong> s = &quot;00000000&quot;, k = 3
 <strong>Output:</strong> &quot;000&quot;
-<strong>Explanation:</strong> 
+<strong>Explanation:</strong>
 We divide s into &quot;000&quot;, &quot;000&quot;, and &quot;00&quot;.
-Then we calculate the digit sum of each group: 0 + 0 + 0 = 0, 0 + 0 + 0 = 0, and 0 + 0 = 0. 
+Then we calculate the digit sum of each group: 0 + 0 + 0 = 0, 0 + 0 + 0 = 0, and 0 + 0 = 0.
 s becomes &quot;0&quot; + &quot;0&quot; + &quot;0&quot; = &quot;000&quot;, whose length is equal to k, so we return &quot;000&quot;.
 </pre>
 
@@ -73,7 +73,11 @@ s becomes &quot;0&quot; + &quot;0&quot; + &quot;0&quot; = &quot;000&quot;, whose
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Simulation
+
+According to the problem statement, we can simulate the operations described in the problem until the length of the string is less than or equal to $k$. Finally, return the string.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the string $s$.
 
 <!-- tabs:start -->
 
@@ -163,19 +167,65 @@ func digitSum(s string, k int) string {
 
 ```ts
 function digitSum(s: string, k: number): string {
-    let ans = [];
     while (s.length > k) {
+        const t: number[] = [];
         for (let i = 0; i < s.length; i += k) {
-            let cur = s.slice(i, i + k);
-            ans.push(cur.split('').reduce((a, c) => a + parseInt(c), 0));
+            const x = s
+                .slice(i, i + k)
+                .split('')
+                .reduce((a, b) => a + +b, 0);
+            t.push(x);
         }
-        s = ans.join('');
-        ans = [];
+        s = t.join('');
     }
     return s;
 }
 ```
 
+#### Rust
+
+```rust
+impl Solution {
+    pub fn digit_sum(s: String, k: i32) -> String {
+        let mut s = s;
+        let k = k as usize;
+        while s.len() > k {
+            let mut t = Vec::new();
+            for chunk in s.as_bytes().chunks(k) {
+                let sum: i32 = chunk.iter().map(|&c| (c - b'0') as i32).sum();
+                t.push(sum.to_string());
+            }
+            s = t.join("");
+        }
+        s
+    }
+}
+```
+
+#### JavaScript
+
+```js
+/**
+ * @param {string} s
+ * @param {number} k
+ * @return {string}
+ */
+var digitSum = function (s, k) {
+    while (s.length > k) {
+        const t = [];
+        for (let i = 0; i < s.length; i += k) {
+            const x = s
+                .slice(i, i + k)
+                .split('')
+                .reduce((a, b) => a + +b, 0);
+            t.push(x);
+        }
+        s = t.join('');
+    }
+    return s;
+};
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2200-2299/2243.Calculate Digit Sum of a String/Solution.js

@@ -0,0 +1,19 @@
+/**
+ * @param {string} s
+ * @param {number} k
+ * @return {string}
+ */
+var digitSum = function (s, k) {
+    while (s.length > k) {
+        const t = [];
+        for (let i = 0; i < s.length; i += k) {
+            const x = s
+                .slice(i, i + k)
+                .split('')
+                .reduce((a, b) => a + +b, 0);
+            t.push(x);
+        }
+        s = t.join('');
+    }
+    return s;
+};

solution/2200-2299/2243.Calculate Digit Sum of a String/Solution.rs

@@ -0,0 +1,15 @@
+impl Solution {
+    pub fn digit_sum(s: String, k: i32) -> String {
+        let mut s = s;
+        let k = k as usize;
+        while s.len() > k {
+            let mut t = Vec::new();
+            for chunk in s.as_bytes().chunks(k) {
+                let sum: i32 = chunk.iter().map(|&c| (c - b'0') as i32).sum();
+                t.push(sum.to_string());
+            }
+            s = t.join("");
+        }
+        s
+    }
+}

solution/2200-2299/2243.Calculate Digit Sum of a String/Solution.ts

@@ -1,12 +1,14 @@
 function digitSum(s: string, k: number): string {
-    let ans = [];
     while (s.length > k) {
+        const t: number[] = [];
         for (let i = 0; i < s.length; i += k) {
-            let cur = s.slice(i, i + k);
-            ans.push(cur.split('').reduce((a, c) => a + parseInt(c), 0));
+            const x = s
+                .slice(i, i + k)
+                .split('')
+                .reduce((a, b) => a + +b, 0);
+            t.push(x);
         }
-        s = ans.join('');
-        ans = [];
+        s = t.join('');
     }
     return s;
 }