feat: add weekly contest 402 (doocs#3113)

Author: yanglbme

solution/3100-3199/3184.Count Pairs That Form a Complete Day I/README.md

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+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3184.Count%20Pairs%20That%20Form%20a%20Complete%20Day%20I/README.md
+---
+
+<!-- problem:start -->
+
+# [3184. 构成整天的下标对数目 I](https://leetcode.cn/problems/count-pairs-that-form-a-complete-day-i)
+
+[English Version](/solution/3100-3199/3184.Count%20Pairs%20That%20Form%20a%20Complete%20Day%20I/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你一个整数数组 <code>hours</code>，表示以 <strong>小时 </strong>为单位的时间，返回一个整数，表示满足 <code>i &lt; j</code> 且 <code>hours[i] + hours[j]</code> 构成 <strong>整天 </strong>的下标对&nbsp;<code>i</code>, <code>j</code> 的数目。</p>
+
+<p><strong>整天 </strong>定义为时间持续时间是 24 小时的 <strong>整数倍 </strong>。</p>
+
+<p>例如，1 天是 24 小时，2 天是 48 小时，3 天是 72 小时，以此类推。</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">hours = [12,12,30,24,24]</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">2</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>构成整天的下标对分别是 <code>(0, 1)</code> 和 <code>(3, 4)</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">hours = [72,48,24,3]</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">3</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>构成整天的下标对分别是 <code>(0, 1)</code>、<code>(0, 2)</code> 和 <code>(1, 2)</code>。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= hours.length &lt;= 100</code></li>
+	<li><code>1 &lt;= hours[i] &lt;= 10<sup>9</sup></code></li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：计数
+
+我们可以用一个哈希表或者一个长度为 $24$ 的数组 $\text{cnt}$ 来记录每个小时数模 $24$ 的出现次数。
+
+遍历数组 $\text{hours}$，对于每个小时数 $x$，我们可以得出与 $x$ 相加为 $24$ 的倍数，且模 $24$ 之后的数为 $(24 - x \bmod 24) \bmod 24$。累加这个数在哈希表或者数组中的出现次数即可。然后我们将 $x$ 的模 $24$ 的出现次数加一。
+
+遍历完数组 $\text{hours}$ 后，我们就可以得到满足题意的下标对数目。
+
+时间复杂度 $O(n)$，其中 $n$ 为数组 $\text{hours}$ 的长度。空间复杂度 $O(C)$，其中 $C=24$。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def countCompleteDayPairs(self, hours: List[int]) -> int:
+        cnt = Counter()
+        ans = 0
+        for x in hours:
+            ans += cnt[(24 - (x % 24)) % 24]
+            cnt[x % 24] += 1
+        return ans
+```
+
+#### Java
+
+```java
+class Solution {
+    public int countCompleteDayPairs(int[] hours) {
+        int[] cnt = new int[24];
+        int ans = 0;
+        for (int x : hours) {
+            ans += cnt[(24 - x % 24) % 24];
+            ++cnt[x % 24];
+        }
+        return ans;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int countCompleteDayPairs(vector<int>& hours) {
+        int cnt[24]{};
+        int ans = 0;
+        for (int x : hours) {
+            ans += cnt[(24 - x % 24) % 24];
+            ++cnt[x % 24];
+        }
+        return ans;
+    }
+};
+```
+
+#### Go
+
+```go
+func countCompleteDayPairs(hours []int) (ans int) {
+	cnt := [24]int{}
+	for _, x := range hours {
+		ans += cnt[(24-x%24)%24]
+		cnt[x%24]++
+	}
+	return
+}
+```
+
+#### TypeScript
+
+```ts
+function countCompleteDayPairs(hours: number[]): number {
+    const cnt: number[] = Array(24).fill(0);
+    let ans: number = 0;
+    for (const x of hours) {
+        ans += cnt[(24 - (x % 24)) % 24];
+        ++cnt[x % 24];
+    }
+    return ans;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/3100-3199/3184.Count Pairs That Form a Complete Day I/README_EN.md

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+---
+comments: true
+difficulty: Easy
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3184.Count%20Pairs%20That%20Form%20a%20Complete%20Day%20I/README_EN.md
+---
+
+<!-- problem:start -->
+
+# [3184. Count Pairs That Form a Complete Day I](https://leetcode.com/problems/count-pairs-that-form-a-complete-day-i)
+
+[中文文档](/solution/3100-3199/3184.Count%20Pairs%20That%20Form%20a%20Complete%20Day%20I/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>Given an integer array <code>hours</code> representing times in <strong>hours</strong>, return an integer denoting the number of pairs <code>i</code>, <code>j</code> where <code>i &lt; j</code> and <code>hours[i] + hours[j]</code> forms a <strong>complete day</strong>.</p>
+
+<p>A <strong>complete day</strong> is defined as a time duration that is an <strong>exact</strong> <strong>multiple</strong> of 24 hours.</p>
+
+<p>For example, 1 day is 24 hours, 2 days is 48 hours, 3 days is 72 hours, and so on.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">hours = [12,12,30,24,24]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">2</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>The pairs of indices that form a complete day are <code>(0, 1)</code> and <code>(3, 4)</code>.</p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">hours = [72,48,24,3]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">3</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>The pairs of indices that form a complete day are <code>(0, 1)</code>, <code>(0, 2)</code>, and <code>(1, 2)</code>.</p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= hours.length &lt;= 100</code></li>
+	<li><code>1 &lt;= hours[i] &lt;= 10<sup>9</sup></code></li>
+</ul>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### Solution 1: Counting
+
+We can use a hash table or an array $\text{cnt}$ of length $24$ to record the occurrence count of each hour modulo $24$.
+
+Iterate through the array $\text{hours}$. For each hour $x$, we can find the number that, when added to $x$, results in a multiple of $24$, and after modulo $24$, this number is $(24 - x \bmod 24) \bmod 24$. We then accumulate the occurrence count of this number from the hash table or array. After that, we increment the occurrence count of $x$ modulo $24$ by one.
+
+After iterating through the array $\text{hours}$, we can obtain the number of index pairs that meet the problem requirements.
+
+The time complexity is $O(n)$, where $n$ is the length of the array $\text{hours}$. The space complexity is $O(C)$, where $C=24$.
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def countCompleteDayPairs(self, hours: List[int]) -> int:
+        cnt = Counter()
+        ans = 0
+        for x in hours:
+            ans += cnt[(24 - (x % 24)) % 24]
+            cnt[x % 24] += 1
+        return ans
+```
+
+#### Java
+
+```java
+class Solution {
+    public int countCompleteDayPairs(int[] hours) {
+        int[] cnt = new int[24];
+        int ans = 0;
+        for (int x : hours) {
+            ans += cnt[(24 - x % 24) % 24];
+            ++cnt[x % 24];
+        }
+        return ans;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int countCompleteDayPairs(vector<int>& hours) {
+        int cnt[24]{};
+        int ans = 0;
+        for (int x : hours) {
+            ans += cnt[(24 - x % 24) % 24];
+            ++cnt[x % 24];
+        }
+        return ans;
+    }
+};
+```
+
+#### Go
+
+```go
+func countCompleteDayPairs(hours []int) (ans int) {
+	cnt := [24]int{}
+	for _, x := range hours {
+		ans += cnt[(24-x%24)%24]
+		cnt[x%24]++
+	}
+	return
+}
+```
+
+#### TypeScript
+
+```ts
+function countCompleteDayPairs(hours: number[]): number {
+    const cnt: number[] = Array(24).fill(0);
+    let ans: number = 0;
+    for (const x of hours) {
+        ans += cnt[(24 - (x % 24)) % 24];
+        ++cnt[x % 24];
+    }
+    return ans;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/3100-3199/3184.Count Pairs That Form a Complete Day I/Solution.cpp

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+class Solution {
+public:
+    int countCompleteDayPairs(vector<int>& hours) {
+        int cnt[24]{};
+        int ans = 0;
+        for (int x : hours) {
+            ans += cnt[(24 - x % 24) % 24];
+            ++cnt[x % 24];
+        }
+        return ans;
+    }
+};

solution/3100-3199/3184.Count Pairs That Form a Complete Day I/Solution.go

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+func countCompleteDayPairs(hours []int) (ans int) {
+	cnt := [24]int{}
+	for _, x := range hours {
+		ans += cnt[(24-x%24)%24]
+		cnt[x%24]++
+	}
+	return
+}

solution/3100-3199/3184.Count Pairs That Form a Complete Day I/Solution.java

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+class Solution {
+    public int countCompleteDayPairs(int[] hours) {
+        int[] cnt = new int[24];
+        int ans = 0;
+        for (int x : hours) {
+            ans += cnt[(24 - x % 24) % 24];
+            ++cnt[x % 24];
+        }
+        return ans;
+    }
+}

solution/3100-3199/3184.Count Pairs That Form a Complete Day I/Solution.py

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+class Solution:
+    def countCompleteDayPairs(self, hours: List[int]) -> int:
+        cnt = Counter()
+        ans = 0
+        for x in hours:
+            ans += cnt[(24 - (x % 24)) % 24]
+            cnt[x % 24] += 1
+        return ans

solution/3100-3199/3184.Count Pairs That Form a Complete Day I/Solution.ts

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+function countCompleteDayPairs(hours: number[]): number {
+    const cnt: number[] = Array(24).fill(0);
+    let ans: number = 0;
+    for (const x of hours) {
+        ans += cnt[(24 - (x % 24)) % 24];
+        ++cnt[x % 24];
+    }
+    return ans;
+}