Merge branch 'main' into main

Author: samarthswami1016

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;
 }

solution/3400-3499/3477.Fruits Into Baskets II/README.md

@@ -80,32 +80,118 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3477.Fr
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：模拟
+
+我们用一个长度为 $n$ 的布尔数组 $\textit{vis}$ 记录已经被使用的篮子，用一个答案变量 $\textit{ans}$ 记录所有未被放置的水果，初始时 $\textit{ans} = n$。
+
+接下来，我们遍历每一种水果 $x$，对于当前水果，我们遍历所有的篮子，找出第一个未被使用，且容量大于等于 $x$ 的篮子 $i$。如果找到了，那么答案 $\textit{ans}$ 减 $1$。
+
+遍历结束后，返回答案即可。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是数组 $\textit{fruits}$ 的长度。
 
 <!-- tabs:start -->
 
 #### Python3
 
 ```python
-
+class Solution:
+    def numOfUnplacedFruits(self, fruits: List[int], baskets: List[int]) -> int:
+        n = len(fruits)
+        vis = [False] * n
+        ans = n
+        for x in fruits:
+            for i, y in enumerate(baskets):
+                if y >= x and not vis[i]:
+                    vis[i] = True
+                    ans -= 1
+                    break
+        return ans
 ```
 
 #### Java
 
 ```java
-
+class Solution {
+    public int numOfUnplacedFruits(int[] fruits, int[] baskets) {
+        int n = fruits.length;
+        boolean[] vis = new boolean[n];
+        int ans = n;
+        for (int x : fruits) {
+            for (int i = 0; i < n; ++i) {
+                if (baskets[i] >= x && !vis[i]) {
+                    vis[i] = true;
+                    --ans;
+                    break;
+                }
+            }
+        }
+        return ans;
+    }
+}
 ```
 
 #### C++
 
 ```cpp
-
+class Solution {
+public:
+    int numOfUnplacedFruits(vector<int>& fruits, vector<int>& baskets) {
+        int n = fruits.size();
+        vector<bool> vis(n);
+        int ans = n;
+        for (int x : fruits) {
+            for (int i = 0; i < n; ++i) {
+                if (baskets[i] >= x && !vis[i]) {
+                    vis[i] = true;
+                    --ans;
+                    break;
+                }
+            }
+        }
+        return ans;
+    }
+};
 ```
 
 #### Go
 
 ```go
+func numOfUnplacedFruits(fruits []int, baskets []int) int {
+	n := len(fruits)
+	ans := n
+	vis := make([]bool, n)
+	for _, x := range fruits {
+		for i, y := range baskets {
+			if y >= x && !vis[i] {
+				vis[i] = true
+				ans--
+				break
+			}
+		}
+	}
+	return ans
+}
+```
 
+#### TypeScript
+
+```ts
+function numOfUnplacedFruits(fruits: number[], baskets: number[]): number {
+    const n = fruits.length;
+    const vis: boolean[] = Array(n).fill(false);
+    let ans = n;
+    for (const x of fruits) {
+        for (let i = 0; i < n; ++i) {
+            if (baskets[i] >= x && !vis[i]) {
+                vis[i] = true;
+                --ans;
+                break;
+            }
+        }
+    }
+    return ans;
+}
 ```
 
 <!-- tabs:end -->