feat: update lc problems

Author: yanglbme

solution/0000-0099/0050.Pow(x, n)/README_EN.md

@@ -36,6 +36,7 @@
 	<li><code>-100.0 &lt; x &lt; 100.0</code></li>
 	<li><code>-2<sup>31</sup> &lt;= n &lt;= 2<sup>31</sup>-1</code></li>
 	<li><code>n</code> is an integer.</li>
+	<li>Either <code>x</code> is not zero or <code>n &gt; 0</code>.</li>
 	<li><code>-10<sup>4</sup> &lt;= x<sup>n</sup> &lt;= 10<sup>4</sup></code></li>
 </ul>
 

solution/0800-0899/0802.Find Eventual Safe States/README.md

@@ -8,7 +8,7 @@
 
 <p>有一个有 <code>n</code> 个节点的有向图，节点按 <code>0</code> 到 <code>n - 1</code> 编号。图由一个 <strong>索引从 0 开始</strong> 的 2D 整数数组&nbsp;<code>graph</code>表示，&nbsp;<code>graph[i]</code>是与节点 <code>i</code> 相邻的节点的整数数组，这意味着从节点 <code>i</code> 到&nbsp;<code>graph[i]</code>中的每个节点都有一条边。</p>
 
-<p>如果一个节点没有连出的有向边，则它是 <strong>终端节点</strong> 。如果没有出边，则节点为终端节点。如果从该节点开始的所有可能路径都通向 <strong>终端节点</strong> ，则该节点为 <strong>安全节点</strong> 。</p>
+<p>如果一个节点没有连出的有向边，则该节点是 <strong>终端节点</strong> 。如果从该节点开始的所有可能路径都通向 <strong>终端节点</strong> ，则该节点为 <strong>安全节点</strong> 。</p>
 
 <p>返回一个由图中所有 <strong>安全节点</strong> 组成的数组作为答案。答案数组中的元素应当按 <strong>升序</strong> 排列。</p>
 

solution/1000-1099/1015.Smallest Integer Divisible by K/README.md

@@ -10,7 +10,7 @@
 
 <p>返回 <code>n</code>&nbsp;的长度。如果不存在这样的 <code>n</code>&nbsp;，就返回-1。</p>
 
-<p><strong>注意：</strong> <code>n</code>&nbsp;不符合 64 位带符号整数。</p>
+<p><strong>注意：</strong> <code>n</code> 可能不符合 64 位带符号整数。</p>
 
 <p>&nbsp;</p>
 

solution/1000-1099/1072.Flip Columns For Maximum Number of Equal Rows/README.md

@@ -10,7 +10,7 @@
 
 <p>你可以从中选出任意数量的列并翻转其上的&nbsp;<strong>每个&nbsp;</strong>单元格。（即翻转后，单元格的值从 <code>0</code> 变成 <code>1</code>，或者从 <code>1</code> 变为 <code>0</code> 。）</p>
 
-<p>返回 <em>经过一些翻转后，行与行之间所有值都相等的最大行数</em>&nbsp;。</p>
+<p>返回 <em>经过一些翻转后，行内所有值都相等的最大行数</em>&nbsp;。</p>
 
 <p>&nbsp;</p>
 

solution/1000-1099/1081.Smallest Subsequence of Distinct Characters/README.md

@@ -40,13 +40,13 @@
 
 **方法一：栈**
 
-我们用一个数组 `last` 记录每个字符最后一次出现的位置，用栈来保存结果字符串，用一个数组 `vis` 或者一个整型变量 `mask` 记录当前字符是否在栈中。
+我们用一个数组 $last$ 记录字符串 $s$ 每个字符最后一次出现的位置，用栈来保存结果字符串，用一个数组 $vis$ 或者一个整型变量 $mask$ 记录当前字符是否在栈中。
 
 遍历字符串 $s$，对于每个字符 $c$，如果 $c$ 不在栈中，我们就需要判断栈顶元素是否大于 $c$，如果大于 $c$，且栈顶元素在后面还会出现，我们就将栈顶元素弹出，将 $c$ 压入栈中。
 
 最后将栈中元素拼接成字符串作为结果返回。
 
-时间复杂度 $O(n)$。其中 $n$ 为字符串 $s$ 的长度。
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是字符串 $s$ 的长度。
 
 <!-- tabs:start -->
 
@@ -57,9 +57,7 @@
 ```python
 class Solution:
     def smallestSubsequence(self, s: str) -> str:
-        last = defaultdict(int)
-        for i, c in enumerate(s):
-            last[c] = i
+        last = {c: i for i, c in enumerate(s)}
         stk = []
         vis = set()
         for i, c in enumerate(s):
@@ -69,30 +67,7 @@ class Solution:
                 vis.remove(stk.pop())
             stk.append(c)
             vis.add(c)
-        return ''.join(stk)
-```
-
-```python
-class Solution:
-    def smallestSubsequence(self, s: str) -> str:
-        count, in_stack = [0] * 128, [False] * 128
-        stack = []
-        for c in s:
-            count[ord(c)] += 1
-
-        for c in s:
-            count[ord(c)] -= 1
-            if in_stack[ord(c)]:
-                continue
-            while len(stack) and stack[-1] > c:
-                peek = stack[-1]
-                if count[ord(peek)] < 1:
-                    break
-                in_stack[ord(peek)] = False
-                stack.pop()
-            stack.append(c)
-            in_stack[ord(c)] = True
-        return ''.join(stack)
+        return "".join(stk)
 ```
 
 ### **Java**
@@ -209,27 +184,33 @@ func smallestSubsequence(s string) string {
 }
 ```
 
-```go
-func smallestSubsequence(s string) string {
-	count, in_stack, stack := make([]int, 128), make([]bool, 128), make([]rune, 0)
-	for _, c := range s {
-		count[c] += 1
-	}
+### **TypeScript**
 
-	for _, c := range s {
-		count[c] -= 1
-		if in_stack[c] {
-			continue
-		}
-		for len(stack) > 0 && stack[len(stack)-1] > c && count[stack[len(stack)-1]] > 0 {
-			peek := stack[len(stack)-1]
-			stack = stack[0 : len(stack)-1]
-			in_stack[peek] = false
-		}
-		stack = append(stack, c)
-		in_stack[c] = true
-	}
-	return string(stack)
+```ts
+function smallestSubsequence(s: string): string {
+    const f = (c: string): number => c.charCodeAt(0) - 'a'.charCodeAt(0);
+    const last: number[] = new Array(26).fill(0);
+    for (const [i, c] of [...s].entries()) {
+        last[f(c)] = i;
+    }
+    const stk: string[] = [];
+    let mask = 0;
+    for (const [i, c] of [...s].entries()) {
+        const x = f(c);
+        if ((mask >> x) & 1) {
+            continue;
+        }
+        while (
+            stk.length &&
+            stk[stk.length - 1] > c &&
+            last[f(stk[stk.length - 1])] > i
+        ) {
+            mask ^= 1 << f(stk.pop()!);
+        }
+        stk.push(c);
+        mask |= 1 << x;
+    }
+    return stk.join('');
 }
 ```
 

solution/1000-1099/1081.Smallest Subsequence of Distinct Characters/README_EN.md

@@ -43,9 +43,7 @@
 ```python
 class Solution:
     def smallestSubsequence(self, s: str) -> str:
-        last = defaultdict(int)
-        for i, c in enumerate(s):
-            last[c] = i
+        last = {c: i for i, c in enumerate(s)}
         stk = []
         vis = set()
         for i, c in enumerate(s):
@@ -55,30 +53,7 @@ class Solution:
                 vis.remove(stk.pop())
             stk.append(c)
             vis.add(c)
-        return ''.join(stk)
-```
-
-```python
-class Solution:
-    def smallestSubsequence(self, s: str) -> str:
-        count, in_stack = [0] * 128, [False] * 128
-        stack = []
-        for c in s:
-            count[ord(c)] += 1
-
-        for c in s:
-            count[ord(c)] -= 1
-            if in_stack[ord(c)]:
-                continue
-            while len(stack) and stack[-1] > c:
-                peek = stack[-1]
-                if count[ord(peek)] < 1:
-                    break
-                in_stack[ord(peek)] = False
-                stack.pop()
-            stack.append(c)
-            in_stack[ord(c)] = True
-        return ''.join(stack)
+        return "".join(stk)
 ```
 
 ### **Java**
@@ -193,27 +168,33 @@ func smallestSubsequence(s string) string {
 }
 ```
 
-```go
-func smallestSubsequence(s string) string {
-	count, in_stack, stack := make([]int, 128), make([]bool, 128), make([]rune, 0)
-	for _, c := range s {
-		count[c] += 1
-	}
+### **TypeScript**
 
-	for _, c := range s {
-		count[c] -= 1
-		if in_stack[c] {
-			continue
-		}
-		for len(stack) > 0 && stack[len(stack)-1] > c && count[stack[len(stack)-1]] > 0 {
-			peek := stack[len(stack)-1]
-			stack = stack[0 : len(stack)-1]
-			in_stack[peek] = false
-		}
-		stack = append(stack, c)
-		in_stack[c] = true
-	}
-	return string(stack)
+```ts
+function smallestSubsequence(s: string): string {
+    const f = (c: string): number => c.charCodeAt(0) - 'a'.charCodeAt(0);
+    const last: number[] = new Array(26).fill(0);
+    for (const [i, c] of [...s].entries()) {
+        last[f(c)] = i;
+    }
+    const stk: string[] = [];
+    let mask = 0;
+    for (const [i, c] of [...s].entries()) {
+        const x = f(c);
+        if ((mask >> x) & 1) {
+            continue;
+        }
+        while (
+            stk.length &&
+            stk[stk.length - 1] > c &&
+            last[f(stk[stk.length - 1])] > i
+        ) {
+            mask ^= 1 << f(stk.pop()!);
+        }
+        stk.push(c);
+        mask |= 1 << x;
+    }
+    return stk.join('');
 }
 ```
 

solution/1000-1099/1081.Smallest Subsequence of Distinct Characters/Solution.py

@@ -1,8 +1,6 @@
 class Solution:
     def smallestSubsequence(self, s: str) -> str:
-        last = defaultdict(int)
-        for i, c in enumerate(s):
-            last[c] = i
+        last = {c: i for i, c in enumerate(s)}
         stk = []
         vis = set()
         for i, c in enumerate(s):
@@ -12,4 +10,4 @@ def smallestSubsequence(self, s: str) -> str:
                 vis.remove(stk.pop())
             stk.append(c)
             vis.add(c)
-        return ''.join(stk)
+        return "".join(stk)

solution/1000-1099/1081.Smallest Subsequence of Distinct Characters/Solution.ts

@@ -0,0 +1,25 @@
+function smallestSubsequence(s: string): string {
+    const f = (c: string): number => c.charCodeAt(0) - 'a'.charCodeAt(0);
+    const last: number[] = new Array(26).fill(0);
+    for (const [i, c] of [...s].entries()) {
+        last[f(c)] = i;
+    }
+    const stk: string[] = [];
+    let mask = 0;
+    for (const [i, c] of [...s].entries()) {
+        const x = f(c);
+        if ((mask >> x) & 1) {
+            continue;
+        }
+        while (
+            stk.length &&
+            stk[stk.length - 1] > c &&
+            last[f(stk[stk.length - 1])] > i
+        ) {
+            mask ^= 1 << f(stk.pop()!);
+        }
+        stk.push(c);
+        mask |= 1 << x;
+    }
+    return stk.join('');
+}

solution/1000-1099/1082.Sales Analysis I/README.md

@@ -87,7 +87,14 @@ Product 表：
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT  seller_id
+FROM Sales
+GROUP BY  seller_id
+HAVING SUM(price) >= ALL (
+SELECT  SUM(price)
+FROM Sales
+GROUP BY  seller_id )
 ```
 
 <!-- tabs:end -->

solution/1000-1099/1082.Sales Analysis I/README_EN.md

@@ -83,7 +83,14 @@ Sales table:
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT  seller_id
+FROM Sales
+GROUP BY  seller_id
+HAVING SUM(price) >= ALL (
+SELECT  SUM(price)
+FROM Sales
+GROUP BY  seller_id )
 ```
 
 <!-- tabs:end -->

solution/1000-1099/1082.Sales Analysis I/Solution.sql

@@ -0,0 +1,8 @@
+# Write your MySQL query statement below
+SELECT  seller_id
+FROM Sales
+GROUP BY  seller_id
+HAVING SUM(price) >= ALL (
+SELECT  SUM(price)
+FROM Sales
+GROUP BY  seller_id )

solution/1000-1099/1083.Sales Analysis II/README.md

@@ -76,7 +76,19 @@ id 为 1 的买家购买了一部 S8，但是却没有购买 iPhone，而 id 为
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT  buyer_id
+FROM
+(
+	SELECT  buyer_id
+	       ,CASE WHEN p.product_name = 'S8' THEN 1  ELSE 0 END     AS s8
+	       ,CASE WHEN p.product_name = 'iPhone' THEN 1  ELSE 0 END AS iPhone
+	FROM Product p
+	JOIN Sales s
+	ON p.product_id = s.product_id
+) t
+GROUP BY  buyer_id
+HAVING SUM(S8) > 0 AND SUM(iPhone) = 0;
 ```
 
 <!-- tabs:end -->

solution/1000-1099/1083.Sales Analysis II/README_EN.md

@@ -84,7 +84,19 @@ Sales table:
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT  buyer_id
+FROM
+(
+	SELECT  buyer_id
+	       ,CASE WHEN p.product_name = 'S8' THEN 1  ELSE 0 END     AS s8
+	       ,CASE WHEN p.product_name = 'iPhone' THEN 1  ELSE 0 END AS iPhone
+	FROM Product p
+	JOIN Sales s
+	ON p.product_id = s.product_id
+) t
+GROUP BY  buyer_id
+HAVING SUM(S8) > 0 AND SUM(iPhone) = 0;
 ```
 
 <!-- tabs:end -->

solution/1000-1099/1083.Sales Analysis II/Solution.sql

@@ -0,0 +1,13 @@
+# Write your MySQL query statement below
+SELECT  buyer_id
+FROM
+(
+	SELECT  buyer_id
+	       ,CASE WHEN p.product_name = 'S8' THEN 1  ELSE 0 END     AS s8
+	       ,CASE WHEN p.product_name = 'iPhone' THEN 1  ELSE 0 END AS iPhone
+	FROM Product p
+	JOIN Sales s
+	ON p.product_id = s.product_id
+) t
+GROUP BY  buyer_id
+HAVING SUM(S8) > 0 AND SUM(iPhone) = 0;