feat: add solutions to lcp problem: No.33

Author: yanglbme

lcp/LCP 33. 蓄水/README.md

@@ -43,22 +43,140 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：贪心 + 枚举**
+
+题目中涉及两个操作：升级水桶、蓄水。我们应该贪心地把升级水桶的操作放在前面，这样在蓄水时，每次能蓄水的量就会更多，操作次数就会更少。
+
+首先，如果最低蓄水量 $vat$ 中所有元素都为 $0$，说明不需要蓄水，直接返回 $0$ 即可。
+
+接下来，我们可以枚举蓄水的次数 $x$，其中 $x \in [1, \max(vat)]$，那么在开始蓄水前，每个水桶的容量至少应该为 $\lceil \frac{vat_i}{x} \rceil$，其中 $\lceil  \rceil$ 表示向上取整。因此，每个水桶的升级次数为 $\max(0, \lceil \frac{vat_i}{x} \rceil - bucket_i)$，我们将所有水桶的升级次数累加，记为 $y$，再加上蓄水的次数 $x$，就是总的操作次数。答案为所有 $x + y$ 中的最小值。
+
+时间复杂度 $O(n \times M)$，其中 $n$ 和 $M$ 分别为数组 $vat$ 的长度和数组 $vat$ 中的最大值。空间复杂度 $O(1)$。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def storeWater(self, bucket: List[int], vat: List[int]) -> int:
+        mx = max(vat)
+        if mx == 0:
+            return 0
+        ans = inf
+        for x in range(1, mx + 1):
+            y = sum(max(0, (v + x - 1) // x - b) for v, b in zip(vat, bucket))
+            ans = min(ans, x + y)
+        return ans
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public int storeWater(int[] bucket, int[] vat) {
+        int mx = Arrays.stream(vat).max().getAsInt();
+        if (mx == 0) {
+            return 0;
+        }
+        int n = vat.length;
+        int ans = 1 << 30;
+        for (int x = 1; x <= mx; ++x) {
+            int y = 0;
+            for (int i = 0; i < n; ++i) {
+                y += Math.max(0, (vat[i] + x - 1) / x - bucket[i]);
+            }
+            ans = Math.min(ans, x + y);
+        }
+        return ans;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int storeWater(vector<int>& bucket, vector<int>& vat) {
+        int mx = *max_element(vat.begin(), vat.end());
+        if (mx == 0) {
+            return 0;
+        }
+        int ans = 1 << 30;
+        int n = bucket.size();
+        for (int x = 1; x <= mx; ++x) {
+            int y = 0;
+            for (int i = 0; i < n; ++i) {
+                y += max(0, (vat[i] + x - 1) / x - bucket[i]);
+            }
+            ans = min(ans, x + y);
+        }
+        return ans;
+    }
+};
+```
+
+### **Go**
+
+```go
+func storeWater(bucket []int, vat []int) int {
+	mx := 0
+	for _, x := range vat {
+		mx = max(mx, x)
+	}
+	if mx == 0 {
+		return 0
+	}
+	ans := 1 << 30
+	for x := 1; x <= mx; x++ {
+		y := 0
+		for i, v := range vat {
+			y += max(0, (v+x-1)/x-bucket[i])
+		}
+		ans = min(ans, x+y)
+	}
+	return ans
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func min(a, b int) int {
+    if a < b {
+        return a
+    }
+    return b
+}
+```
 
+### **TypeScript**
+
+```ts
+function storeWater(bucket: number[], vat: number[]): number {
+    const mx = Math.max(...vat);
+    if (mx === 0) {
+        return 0;
+    }
+    const n = vat.length;
+    let ans = 1 << 30;
+    for (let x = 1; x <= mx; ++x) {
+        let y = 0;
+        for (let i = 0; i < n; ++i) {
+            y += Math.max(0, Math.ceil(vat[i] / x) - bucket[i]);
+        }
+        ans = Math.min(ans, x + y);
+    }
+    return ans;
+}
 ```
 
 ### **...**

lcp/LCP 33. 蓄水/Solution.cpp

@@ -0,0 +1,19 @@
+class Solution {
+public:
+    int storeWater(vector<int>& bucket, vector<int>& vat) {
+        int mx = *max_element(vat.begin(), vat.end());
+        if (mx == 0) {
+            return 0;
+        }
+        int ans = 1 << 30;
+        int n = bucket.size();
+        for (int x = 1; x <= mx; ++x) {
+            int y = 0;
+            for (int i = 0; i < n; ++i) {
+                y += max(0, (vat[i] + x - 1) / x - bucket[i]);
+            }
+            ans = min(ans, x + y);
+        }
+        return ans;
+    }
+};

lcp/LCP 33. 蓄水/Solution.go

@@ -0,0 +1,32 @@
+func storeWater(bucket []int, vat []int) int {
+	mx := 0
+	for _, x := range vat {
+		mx = max(mx, x)
+	}
+	if mx == 0 {
+		return 0
+	}
+	ans := 1 << 30
+	for x := 1; x <= mx; x++ {
+		y := 0
+		for i, v := range vat {
+			y += max(0, (v+x-1)/x-bucket[i])
+		}
+		ans = min(ans, x+y)
+	}
+	return ans
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func min(a, b int) int {
+    if a < b {
+        return a
+    }
+    return b
+}

lcp/LCP 33. 蓄水/Solution.java

@@ -0,0 +1,18 @@
+class Solution {
+    public int storeWater(int[] bucket, int[] vat) {
+        int mx = Arrays.stream(vat).max().getAsInt();
+        if (mx == 0) {
+            return 0;
+        }
+        int n = vat.length;
+        int ans = 1 << 30;
+        for (int x = 1; x <= mx; ++x) {
+            int y = 0;
+            for (int i = 0; i < n; ++i) {
+                y += Math.max(0, (vat[i] + x - 1) / x - bucket[i]);
+            }
+            ans = Math.min(ans, x + y);
+        }
+        return ans;
+    }
+}

lcp/LCP 33. 蓄水/Solution.py

@@ -0,0 +1,10 @@
+class Solution:
+    def storeWater(self, bucket: List[int], vat: List[int]) -> int:
+        mx = max(vat)
+        if mx == 0:
+            return 0
+        ans = inf
+        for x in range(1, mx + 1):
+            y = sum(max(0, (v + x - 1) // x - b) for v, b in zip(vat, bucket))
+            ans = min(ans, x + y)
+        return ans

lcp/LCP 33. 蓄水/Solutoin.ts

@@ -0,0 +1,16 @@
+function storeWater(bucket: number[], vat: number[]): number {
+    const mx = Math.max(...vat);
+    if (mx === 0) {
+        return 0;
+    }
+    const n = vat.length;
+    let ans = 1 << 30;
+    for (let x = 1; x <= mx; ++x) {
+        let y = 0;
+        for (let i = 0; i < n; ++i) {
+            y += Math.max(0, Math.ceil(vat[i] / x) - bucket[i]);
+        }
+        ans = Math.min(ans, x + y);
+    }
+    return ans;
+}

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

@@ -10,21 +10,21 @@
 
 <p>&nbsp;</p>
 
-<p><strong>示例 1：</strong></p>
+<p><strong class="example">示例 1：</strong></p>
 
 <pre>
 <strong>输入：</strong>x = 2.00000, n = 10
 <strong>输出：</strong>1024.00000
 </pre>
 
-<p><strong>示例 2：</strong></p>
+<p><strong class="example">示例 2：</strong></p>
 
 <pre>
 <strong>输入：</strong>x = 2.10000, n = 3
 <strong>输出：</strong>9.26100
 </pre>
 
-<p><strong>示例 3：</strong></p>
+<p><strong class="example">示例 3：</strong></p>
 
 <pre>
 <strong>输入：</strong>x = 2.00000, n = -2
@@ -40,6 +40,7 @@
 	<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>&nbsp;是一个整数</li>
+	<li>要么 <code>x</code> 不为零，要么 <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/0500-0599/0532.K-diff Pairs in an Array/README_EN.md

@@ -11,7 +11,7 @@
 <ul>
 	<li><code>0 &lt;= i, j &lt; nums.length</code></li>
 	<li><code>i != j</code></li>
-	<li><code>nums[i] - nums[j] == k</code></li>
+	<li><code>|nums[i] - nums[j]| == k</code></li>
 </ul>
 
 <p><strong>Notice</strong> that <code>|val|</code> denotes the absolute value of <code>val</code>.</p>

solution/1300-1399/1336.Number of Transactions per Visit/README_EN.md

@@ -97,7 +97,7 @@ Transactions table:
 +--------------------+--------------+
 <strong>Explanation:</strong> The chart drawn for this example is shown above.
 * For transactions_count = 0, The visits (1, &quot;2020-01-01&quot;), (2, &quot;2020-01-02&quot;), (12, &quot;2020-01-01&quot;) and (19, &quot;2020-01-03&quot;) did no transactions so visits_count = 4.
-* For transactions_count = 1, The visits (2, &quot;2020-01-03&quot;), (7, &amp;quo;2020-01-11&quot;), (8, &quot;2020-01-28&quot;), (1, &quot;2020-01-02&quot;) and (1, &quot;2020-01-04&quot;) did one transaction so visits_count = 5.
+* For transactions_count = 1, The visits (2, &quot;2020-01-03&quot;), (7, &quot;2020-01-11&quot;), (8, &quot;2020-01-28&quot;), (1, &quot;2020-01-02&quot;) and (1, &quot;2020-01-04&quot;) did one transaction so visits_count = 5.
 * For transactions_count = 2, No customers visited the bank and did two transactions so visits_count = 0.
 * For transactions_count = 3, The visit (9, &quot;2020-01-25&quot;) did three transactions so visits_count = 1.
 * For transactions_count &gt;= 4, No customers visited the bank and did more than three transactions so we will stop at transactions_count = 3

solution/1500-1599/1528.Shuffle String/README.md

@@ -121,7 +121,7 @@ var restoreString = function (s, indices) {
     for (let i = 0; i < s.length; i++) {
         rs[indices[i]] = s[i];
     }
-    return rs.join("");
+    return rs.join('');
 };
 ```
 

solution/1500-1599/1528.Shuffle String/README_EN.md

@@ -107,7 +107,7 @@ var restoreString = function (s, indices) {
     for (let i = 0; i < s.length; i++) {
         rs[indices[i]] = s[i];
     }
-    return rs.join("");
+    return rs.join('');
 };
 ```
 

solution/2600-2699/2689.Extract Kth Character From The Rope Tree/README_EN.md

@@ -170,7 +170,7 @@ public:
             string left = dfs(root->left);
             string right = dfs(root->right);
             return left + right;
-        }; 
+        };
         return dfs(root)[k - 1];
     }
 };

solution/2600-2699/2690.Infinite Method Object/README.md

@@ -1,37 +1,39 @@
-# [2690. Infinite Method Object](https://leetcode.cn/problems/infinite-method-object)
+# [2690. 无穷方法对象](https://leetcode.cn/problems/infinite-method-object)
 
 [English Version](/solution/2600-2699/2690.Infinite%20Method%20Object/README_EN.md)
 
 ## 题目描述
 
 <!-- 这里写题目描述 -->
 
-<p>Write a function that&nbsp;returns an&nbsp;<strong>infinite-method</strong><strong>&nbsp;object</strong>.</p>
+<p>请你编写一个函数，返回一个 <strong>无穷方法对象</strong> 。</p>
 
-<p>An&nbsp;<strong>infinite-method</strong><strong>&nbsp;object</strong>&nbsp;is defined as an object that allows you to call any method and it will always return the name of the method.</p>
+<p><strong>无穷方法对象</strong> 被定义为一个对象，它允许您调用任何方法，并始终返回方法的名称。</p>
 
-<p>For example, if you execute&nbsp;<code>obj.abc123()</code>, it will return&nbsp;<code>&quot;abc123&quot;</code>.</p>
+<p>例如，如果执行 <code>obj.abc123()</code> ，它将返回 <code>"abc123"</code> 。</p>
 
 <p>&nbsp;</p>
-<p><strong class="example">Example 1:</strong></p>
+
+<p><strong class="example">示例 1：</strong></p>
 
 <pre>
-<strong>Input:</strong> method = &quot;abc123&quot;
-<strong>Output:</strong> &quot;abc123&quot;
-<strong>Explanation:</strong>
+<b>输入：</b>method = "abc123"
+<b>输出：</b>"abc123"
+<strong>解释：</strong>
 const obj = createInfiniteObject();
-obj[&#39;abc123&#39;](); // &quot;abc123&quot;
-The returned string should always match the method name.</pre>
+obj['abc123'](); // "abc123"
+返回的字符串应始终与方法名称匹配。</pre>
 
-<p><strong class="example">Example 2:</strong></p>
+<p><strong class="example">示例 2：</strong></p>
 
 <pre>
-<strong>Input:</strong> method = &quot;.-qw73n|^2It&quot;
-<strong>Output:</strong> &quot;.-qw73n|^2It&quot;
-<strong>Explanation:</strong> The returned string should always match the method name.</pre>
+<b>输入：</b>method = ".-qw73n|^2It"
+<strong>输出：</strong>".-qw73n|^2It"
+<b>解释：</b>返回的字符串应始终与方法名称匹配。</pre>
 
 <p>&nbsp;</p>
-<p><strong>Constraints:</strong></p>
+
+<p><strong>提示：</strong></p>
 
 <ul>
 	<li><code>0 &lt;= method.length &lt;= 1000</code></li>