feat: add weekly contest 489 and biweekly contest 176

README.md

@@ -52,7 +52,7 @@
 - [乘积小于 K 的子数组](/solution/0700-0799/0713.Subarray%20Product%20Less%20Than%20K/README.md) - `双指针`
 - [位 1 的个数](/solution/0100-0199/0191.Number%20of%201%20Bits/README.md) - `位运算`、`lowbit`
 - [合并区间](/solution/0000-0099/0056.Merge%20Intervals/README.md) - `区间合并`
-  <!-- 排序算法、待补充 -->
+      <!-- 排序算法、待补充 -->
 
 ### 2. 数据结构
 

solution/0700-0799/0799.Champagne Tower/Solution.ts

@@ -12,4 +12,4 @@ function champagneTower(poured: number, query_row: number, query_glass: number):
         }
     }
     return f[query_row][query_glass];
-}
+}

solution/0700-0799/0799.Champagne Tower/Solution2.ts

@@ -12,4 +12,4 @@ function champagneTower(poured: number, query_row: number, query_glass: number):
         f = g;
     }
     return Math.min(1, f[query_glass]);
-}
+}

solution/1900-1999/1959.Minimum Total Space Wasted With K Resizing Operations/README_EN.md

@@ -44,7 +44,7 @@ The total wasted space is (20 - 10) + (20 - 20) = 10.
 <strong>Input:</strong> nums = [10,20,30], k = 1
 <strong>Output:</strong> 10
 <strong>Explanation:</strong> size = [20,20,30].
-We can set the initial size to be 20 and resize to 30 at time 2.
+We can set the initial size to be 20 and resize to 30 at time 2. 
 The total wasted space is (20 - 10) + (20 - 20) + (30 - 30) = 10.
 </pre>
 

solution/2500-2599/2535.Difference Between Element Sum and Digit Sum of an Array/README_EN.md

@@ -36,7 +36,7 @@ tags:
 <pre>
 <strong>Input:</strong> nums = [1,15,6,3]
 <strong>Output:</strong> 9
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 The element sum of nums is 1 + 15 + 6 + 3 = 25.
 The digit sum of nums is 1 + 1 + 5 + 6 + 3 = 16.
 The absolute difference between the element sum and digit sum is |25 - 16| = 9.

solution/3800-3899/3833.Count Dominant Indices/README.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: 简单
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3800-3899/3833.Count%20Dominant%20Indices/README.md
+rating: 1171
+source: 第 488 场周赛 Q1
 ---
 
 <!-- problem:start -->

solution/3800-3899/3833.Count Dominant Indices/README_EN.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: Easy
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3800-3899/3833.Count%20Dominant%20Indices/README_EN.md
+rating: 1171
+source: Weekly Contest 488 Q1
 ---
 
 <!-- problem:start -->

solution/3800-3899/3834.Merge Adjacent Equal Elements/README.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: 中等
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3800-3899/3834.Merge%20Adjacent%20Equal%20Elements/README.md
+rating: 1428
+source: 第 488 场周赛 Q2
 ---
 
 <!-- problem:start -->

solution/3800-3899/3834.Merge Adjacent Equal Elements/README_EN.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: Medium
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3800-3899/3834.Merge%20Adjacent%20Equal%20Elements/README_EN.md
+rating: 1428
+source: Weekly Contest 488 Q2
 ---
 
 <!-- problem:start -->

solution/3800-3899/3835.Count Subarrays With Cost Less Than or Equal to K/README.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: 中等
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3800-3899/3835.Count%20Subarrays%20With%20Cost%20Less%20Than%20or%20Equal%20to%20K/README.md
+rating: 1759
+source: 第 488 场周赛 Q3
 ---
 
 <!-- problem:start -->

solution/3800-3899/3835.Count Subarrays With Cost Less Than or Equal to K/README_EN.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: Medium
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3800-3899/3835.Count%20Subarrays%20With%20Cost%20Less%20Than%20or%20Equal%20to%20K/README_EN.md
+rating: 1759
+source: Weekly Contest 488 Q3
 ---
 
 <!-- problem:start -->

solution/3800-3899/3836.Maximum Score Using Exactly K Pairs/README.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: 困难
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3800-3899/3836.Maximum%20Score%20Using%20Exactly%20K%20Pairs/README.md
+rating: 1987
+source: 第 488 场周赛 Q4
 ---
 
 <!-- problem:start -->

solution/3800-3899/3836.Maximum Score Using Exactly K Pairs/README_EN.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: Hard
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3800-3899/3836.Maximum%20Score%20Using%20Exactly%20K%20Pairs/README_EN.md
+rating: 1987
+source: Weekly Contest 488 Q4
 ---
 
 <!-- problem:start -->

solution/3800-3899/3838.Weighted Word Mapping/README.md

@@ -0,0 +1,186 @@
+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3800-3899/3838.Weighted%20Word%20Mapping/README.md
+---
+
+<!-- problem:start -->
+
+# [3838. 带权单词映射](https://leetcode.cn/problems/weighted-word-mapping)
+
+[English Version](/solution/3800-3899/3838.Weighted%20Word%20Mapping/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你一个字符串数组 <code>words</code>，其中每个字符串表示一个由小写英文字母组成的单词。</p>
+
+<p>同时给你一个长度为 26 的整数数组 <code>weights</code>，其中 <code>weights[i]</code> 表示第 <code>i</code> 个小写英文字母的权重。</p>
+
+<p>单词的 <strong>权重</strong> 定义为其所有字符权重的 <strong>总和</strong>。</p>
+
+<p>对于每个单词，将其权重对 26 取模，并将结果按字母倒序映射到一个小写英文字母（<code>0 -&gt; 'z', 1 -&gt; 'y', ..., 25 -&gt; 'a'</code>）。</p>
+
+<p>返回一个由所有单词映射后的字符按顺序连接而成的字符串。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong> <span class="example-io">words = ["abcd","def","xyz"], weights = [5,3,12,14,1,2,3,2,10,6,6,9,7,8,7,10,8,9,6,9,9,8,3,7,7,2]</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">"rij"</span></p>
+
+<p><strong>解释：</strong></p>
+
+<ul>
+	<li><code>"abcd"</code> 的权重是 <code>5 + 3 + 12 + 14 = 34</code>。对 26 取模的结果是 <code>34 % 26 = 8</code>，映射为 <code>'r'</code>。</li>
+	<li><code>"def"</code> 的权重是 <code>14 + 1 + 2 = 17</code>。对 26 取模的结果是 <code>17 % 26 = 17</code>，映射为 <code>'i'</code>。</li>
+	<li><code>"xyz"</code> 的权重是 <code>7 + 7 + 2 = 16</code>。对&nbsp;26 取模的结果是 <code>16 % 26 = 16</code>，映射为 <code>'j'</code>。</li>
+</ul>
+
+<p>因此，连接映射字符后形成的字符串是 <code>"rij"</code>。</p>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong> <span class="example-io">words = ["a","b","c"], weights = [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">"yyy"</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>每个单词的权重均为 1。对 26 取模的结果是 <code>1 % 26 = 1</code>，映射为 <code>'y'</code>。</p>
+
+<p>因此，连接映射字符后形成的字符串是 <code>"yyy"</code>。</p>
+</div>
+
+<p><strong class="example">示例 3：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong> <span class="example-io">words = ["abcd"], weights = [7,5,3,4,3,5,4,9,4,2,2,7,10,2,5,10,6,1,2,2,4,1,3,4,4,5]</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">"g"</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p><code>"abcd"</code> 的权重是 <code>7 + 5 + 3 + 4 = 19</code>。对 26 取模的结果是 <code>19 % 26 = 19</code>，映射为 <code>'g'</code>。</p>
+
+<p>因此，连接映射字符后形成的字符串是 <code>"g"</code>。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= words.length &lt;= 100</code></li>
+	<li><code>1 &lt;= words[i].length &lt;= 10</code></li>
+	<li><code>weights.length == 26</code></li>
+	<li><code>1 &lt;= weights[i] &lt;= 100</code></li>
+	<li><code>words[i]</code> 仅由小写英文字母组成。</li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：模拟
+
+我们遍历 $\textit{words}$ 中的每个单词 $w$，计算其权重 $s$，即单词中每个字符的权重之和。然后计算 $s$ 对 26 取模的结果，并将结果映射到一个小写英文字母上，最后连接所有映射后的字符并返回。
+
+时间复杂度 $O(L)$，其中 $L$ 是 $\textit{words}$ 中所有单词的长度之和。空间复杂度 $O(W)$，其中 $W$ 是 $\textit{words}$ 的长度。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def mapWordWeights(self, words: List[str], weights: List[int]) -> str:
+        ans = []
+        for w in words:
+            s = sum(weights[ord(c) - ord('a')] for c in w)
+            ans.append(ascii_lowercase[25 - s % 26])
+        return ''.join(ans)
+```
+
+#### Java
+
+```java
+class Solution {
+    public String mapWordWeights(String[] words, int[] weights) {
+        var ans = new StringBuilder();
+        for (var w : words) {
+            int s = 0;
+            for (char c : w.toCharArray()) {
+                s = (s + weights[c - 'a']) % 26;
+            }
+            ans.append((char) ('a' + (25 - s)));
+        }
+        return ans.toString();
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    string mapWordWeights(vector<string>& words, vector<int>& weights) {
+        string ans;
+        for (const string& w : words) {
+            int s = 0;
+            for (char c : w) {
+                s = (s + weights[c - 'a']) % 26;
+            }
+            ans.push_back(char('a' + (25 - s)));
+        }
+        return ans;
+    }
+};
+```
+
+#### Go
+
+```go
+func mapWordWeights(words []string, weights []int) string {
+	ans := make([]byte, 0, len(words))
+	for _, w := range words {
+		s := 0
+		for i := 0; i < len(w); i++ {
+			s = (s + weights[int(w[i]-'a')]) % 26
+		}
+		ans = append(ans, byte('a'+(25-s)))
+	}
+	return string(ans)
+}
+```
+
+#### TypeScript
+
+```ts
+function mapWordWeights(words: string[], weights: number[]): string {
+    const ans: string[] = [];
+    for (const w of words) {
+        let s = 0;
+        for (const c of w) {
+            s = (s + weights[c.charCodeAt(0) - 97]) % 26;
+        }
+        ans.push(String.fromCharCode(97 + (25 - s)));
+    }
+    return ans.join('');
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->