feat: add biweekly contest 133 and weekly contest 403

solution/3100-3199/3190.Find Minimum Operations to Make All Elements Divisible by Three/README.md

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+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3190.Find%20Minimum%20Operations%20to%20Make%20All%20Elements%20Divisible%20by%20Three/README.md
+---
+
+<!-- problem:start -->
+
+# [3190. 使所有元素都可以被 3 整除的最少操作数](https://leetcode.cn/problems/find-minimum-operations-to-make-all-elements-divisible-by-three)
+
+[English Version](/solution/3100-3199/3190.Find%20Minimum%20Operations%20to%20Make%20All%20Elements%20Divisible%20by%20Three/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你一个整数数组&nbsp;<code>nums</code>&nbsp;。一次操作中，你可以将&nbsp;<code>nums</code>&nbsp;中的&nbsp;<strong>任意</strong>&nbsp;一个元素增加或者减少 1 。</p>
+
+<p>请你返回将 <code>nums</code>&nbsp;中所有元素都可以被 3 整除的 <strong>最少</strong>&nbsp;操作次数。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>nums = [1,2,3,4]</span></p>
+
+<p><span class="example-io"><b>输出：</b>3</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>通过以下 3 个操作，数组中的所有元素都可以被 3 整除：</p>
+
+<ul>
+	<li>将 1 减少 1 。</li>
+	<li>将 2 增加 1 。</li>
+	<li>将 4 减少 1 。</li>
+</ul>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><span class="example-io"><b>输入：</b>nums = [3,6,9]</span></p>
+
+<p><span class="example-io"><b>输出：</b>0</span></p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 50</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 50</code></li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：数学
+
+我们直接遍历数组 $\text{nums}$，对于每个元素 $x$，我们计算 $x$ 除以 3 的余数 $x \bmod 3$，如果余数不为 0，我们需要将 $x$ 变为能被 3 整除且操作次数最少，那么我们可以选择将 $x$ 减少 $x \bmod 3$ 或者增加 $3 - x \bmod 3$，取两者的最小值累加到答案中。
+
+时间复杂度 $O(n)$，其中 $n$ 为数组 $\text{nums}$ 的长度。空间复杂度 $O(1)$。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def minimumOperations(self, nums: List[int]) -> int:
+        ans = 0
+        for x in nums:
+            if mod := x % 3:
+                ans += min(mod, 3 - mod)
+        return ans
+```
+
+#### Java
+
+```java
+class Solution {
+    public int minimumOperations(int[] nums) {
+        int ans = 0;
+        for (int x : nums) {
+            int mod = x % 3;
+            if (mod != 0) {
+                ans += Math.min(mod, 3 - mod);
+            }
+        }
+        return ans;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+    public int minimumOperations(int[] nums) {
+        int ans = 0;
+        for (int x : nums) {
+            int mod = x % 3;
+            if (mod != 0) {
+                ans += Math.min(mod, 3 - mod);
+            }
+        }
+        return ans;
+    }
+}
+```
+
+#### Go
+
+```go
+func minimumOperations(nums []int) (ans int) {
+	for _, x := range nums {
+		if mod := x % 3; mod > 0 {
+			ans += min(mod, 3-mod)
+		}
+	}
+	return
+}
+```
+
+#### TypeScript
+
+```ts
+function minimumOperations(nums: number[]): number {
+    let ans = 0;
+    for (const x of nums) {
+        const mod = x % 3;
+        if (mod) {
+            ans += Math.min(mod, 3 - mod);
+        }
+    }
+    return ans;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/3100-3199/3190.Find Minimum Operations to Make All Elements Divisible by Three/README_EN.md

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+---
+comments: true
+difficulty: Easy
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3190.Find%20Minimum%20Operations%20to%20Make%20All%20Elements%20Divisible%20by%20Three/README_EN.md
+---
+
+<!-- problem:start -->
+
+# [3190. Find Minimum Operations to Make All Elements Divisible by Three](https://leetcode.com/problems/find-minimum-operations-to-make-all-elements-divisible-by-three)
+
+[中文文档](/solution/3100-3199/3190.Find%20Minimum%20Operations%20to%20Make%20All%20Elements%20Divisible%20by%20Three/README.md)
+
+## Description
+
+<!-- description:start -->
+
+<p>You are given an integer array <code>nums</code>. In one operation, you can add or subtract 1 from <strong>any</strong> element of <code>nums</code>.</p>
+
+<p>Return the <strong>minimum</strong> number of operations to make all elements of <code>nums</code> divisible by 3.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [1,2,3,4]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">3</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>All array elements can be made divisible by 3 using 3 operations:</p>
+
+<ul>
+	<li>Subtract 1 from 1.</li>
+	<li>Add 1 to 2.</li>
+	<li>Subtract 1 from 4.</li>
+</ul>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">nums = [3,6,9]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">0</span></p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 50</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 50</code></li>
+</ul>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### Solution 1: Mathematics
+
+We directly iterate through the array $\text{nums}$. For each element $x$, we calculate the remainder of $x$ divided by 3, $x \bmod 3$. If the remainder is not 0, we need to make $x$ divisible by 3 with the minimum number of operations. Therefore, we can choose to either decrease $x$ by $x \bmod 3$ or increase $x$ by $3 - x \bmod 3$, and we accumulate the minimum of these two values to the answer.
+
+The time complexity is $O(n)$, where $n$ is the length of the array $\text{nums}$. The space complexity is $O(1)$.
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def minimumOperations(self, nums: List[int]) -> int:
+        ans = 0
+        for x in nums:
+            if mod := x % 3:
+                ans += min(mod, 3 - mod)
+        return ans
+```
+
+#### Java
+
+```java
+class Solution {
+    public int minimumOperations(int[] nums) {
+        int ans = 0;
+        for (int x : nums) {
+            int mod = x % 3;
+            if (mod != 0) {
+                ans += Math.min(mod, 3 - mod);
+            }
+        }
+        return ans;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+    public int minimumOperations(int[] nums) {
+        int ans = 0;
+        for (int x : nums) {
+            int mod = x % 3;
+            if (mod != 0) {
+                ans += Math.min(mod, 3 - mod);
+            }
+        }
+        return ans;
+    }
+}
+```
+
+#### Go
+
+```go
+func minimumOperations(nums []int) (ans int) {
+	for _, x := range nums {
+		if mod := x % 3; mod > 0 {
+			ans += min(mod, 3-mod)
+		}
+	}
+	return
+}
+```
+
+#### TypeScript
+
+```ts
+function minimumOperations(nums: number[]): number {
+    let ans = 0;
+    for (const x of nums) {
+        const mod = x % 3;
+        if (mod) {
+            ans += Math.min(mod, 3 - mod);
+        }
+    }
+    return ans;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->

solution/3100-3199/3190.Find Minimum Operations to Make All Elements Divisible by Three/Solution.cpp

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+class Solution {
+    public int minimumOperations(int[] nums) {
+        int ans = 0;
+        for (int x : nums) {
+            int mod = x % 3;
+            if (mod != 0) {
+                ans += Math.min(mod, 3 - mod);
+            }
+        }
+        return ans;
+    }
+}

solution/3100-3199/3190.Find Minimum Operations to Make All Elements Divisible by Three/Solution.go

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+func minimumOperations(nums []int) (ans int) {
+	for _, x := range nums {
+		if mod := x % 3; mod > 0 {
+			ans += min(mod, 3-mod)
+		}
+	}
+	return
+}

solution/3100-3199/3190.Find Minimum Operations to Make All Elements Divisible by Three/Solution.java

@@ -0,0 +1,12 @@
+class Solution {
+    public int minimumOperations(int[] nums) {
+        int ans = 0;
+        for (int x : nums) {
+            int mod = x % 3;
+            if (mod != 0) {
+                ans += Math.min(mod, 3 - mod);
+            }
+        }
+        return ans;
+    }
+}

solution/3100-3199/3190.Find Minimum Operations to Make All Elements Divisible by Three/Solution.py

@@ -0,0 +1,7 @@
+class Solution:
+    def minimumOperations(self, nums: List[int]) -> int:
+        ans = 0
+        for x in nums:
+            if mod := x % 3:
+                ans += min(mod, 3 - mod)
+        return ans

solution/3100-3199/3190.Find Minimum Operations to Make All Elements Divisible by Three/Solution.ts

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+function minimumOperations(nums: number[]): number {
+    let ans = 0;
+    for (const x of nums) {
+        const mod = x % 3;
+        if (mod) {
+            ans += Math.min(mod, 3 - mod);
+        }
+    }
+    return ans;
+}