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feat: add solutions to lc problem: No.0553
No.0553.Optimal Division
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‎solution/0500-0599/0553.Optimal Division/README.md

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<!-- 这里可写通用的实现逻辑 -->
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贪心。
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要使得除法的结果最大，分子应该尽可能大，而分母应该尽可能小。
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分子最大应该是 nums[0]，而分母最大是 `nums[1] / nums[2] / ... / nums[n - 1]`，此时的除法结果最大。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def optimalDivision(self, nums: List[int]) -> str:
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n = len(nums)
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if n == 1:
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return str(nums[0])
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if n == 2:
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return f'{nums[0]}/{nums[1]}'
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return f'{nums[0]}/({"/".join(map(str, nums[1:]))})'
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public String optimalDivision(int[] nums) {
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int n = nums.length;
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if (n == 1) {
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return nums[0] + "";
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}
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if (n == 2) {
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return nums[0] + "/" + nums[1];
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}
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StringBuilder ans = new StringBuilder(nums[0] + "/(");
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for (int i = 1; i < n - 1; ++i) {
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ans.append(nums[i] + "/");
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}
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ans.append(nums[n - 1] + ")");
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return ans.toString();
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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string optimalDivision(vector<int>& nums) {
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int n = nums.size();
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if (n == 1) return to_string(nums[0]);
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if (n == 2) return to_string(nums[0]) + "/" + to_string(nums[1]);
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string ans = to_string(nums[0]) + "/(";
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for (int i = 1; i < n - 1; i++) ans.append(to_string(nums[i]) + "/");
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ans.append(to_string(nums[n - 1]) + ")");
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return ans;
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}
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};
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```
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### **Go**
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```go
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func optimalDivision(nums []int) string {
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n := len(nums)
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if n == 1 {
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return strconv.Itoa(nums[0])
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}
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if n == 2 {
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return fmt.Sprintf("%d/%d", nums[0], nums[1])
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}
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ans := &strings.Builder{}
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ans.WriteString(fmt.Sprintf("%d/(", nums[0]))
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for _, num := range nums[1 : n-1] {
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ans.WriteString(strconv.Itoa(num))
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ans.WriteByte('/')
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}
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ans.WriteString(fmt.Sprintf("%d)", nums[n-1]))
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return ans.String()
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}
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```
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### **...**

‎solution/0500-0599/0553.Optimal Division/README_EN.md

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## Description
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<p>Given a list of <b>positive integers</b>, the adjacent integers will perform the float division. For example, [2,3,4] -> 2 / 3 / 4.</p>
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<p>You are given an integer array <code>nums</code>. The adjacent integers in <code>nums</code> will perform the float division.</p>
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<p>However, you can add any number of parenthesis at any position to change the priority of operations. You should find out how to add parenthesis to get the <b>maximum</b> result, and return the corresponding expression in string format. <b>Your expression should NOT contain redundant parenthesis.</b></p>
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<ul>
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<li>For example, for <code>nums = [2,3,4]</code>, we will evaluate the expression <code>&quot;2/3/4&quot;</code>.</li>
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</ul>
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<p><b>Example:</b><br />
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<p>However, you can add any number of parenthesis at any position to change the priority of operations. You want to add these parentheses such the value of the expression after the evaluation is maximum.</p>
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<pre>
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<b>Input:</b> [1000,100,10,2]
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<p>Return <em>the corresponding expression that has the maximum value in string format</em>.</p>
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<b>Output:</b> "1000/(100/10/2)"
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<p><strong>Note:</strong> your expression should not contain redundant parenthesis.</p>
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<b>Explanation:</b>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [1000,100,10,2]
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<strong>Output:</strong> &quot;1000/(100/10/2)&quot;
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<strong>Explanation:</strong>
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1000/(100/10/2) = 1000/((100/10)/2) = 200
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However, the bold parenthesis in "1000/(<b>(</b>100/10<b>)</b>/2)" are redundant, <br/>since they don't influence the operation priority. So you should return "1000/(100/10/2)".
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However, the bold parenthesis in &quot;1000/((100/10)/2)&quot; are redundant, since they don&#39;t influence the operation priority. So you should return &quot;1000/(100/10/2)&quot;.
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Other cases:
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1000/(100/10)/2 = 50
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1000/(100/(10/2)) = 50
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1000/100/10/2 = 0.5
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1000/100/(10/2) = 2
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</pre>
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</p>
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<p><b>Note:</b>
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<ol>
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<p><strong>Example 2:</strong></p>
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<li>The length of the input array is [1, 10].</li>
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<pre>
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<strong>Input:</strong> nums = [2,3,4]
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<strong>Output:</strong> &quot;2/(3/4)&quot;
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</pre>
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<li>Elements in the given array will be in range [2, 1000].</li>
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<p><strong>Example 3:</strong></p>
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<li>There is only one optimal division for each test case.</li>
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<pre>
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<strong>Input:</strong> nums = [2]
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<strong>Output:</strong> &quot;2&quot;
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</pre>
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</ol>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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</p>
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<ul>
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<li><code>1 &lt;= nums.length &lt;= 10</code></li>
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<li><code>2 &lt;= nums[i] &lt;= 1000</code></li>
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<li>There is only one optimal division for the given iput.</li>
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</ul>
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## Solutions
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### **Python3**
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```python
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class Solution:
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def optimalDivision(self, nums: List[int]) -> str:
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n = len(nums)
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if n == 1:
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return str(nums[0])
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if n == 2:
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return f'{nums[0]}/{nums[1]}'
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return f'{nums[0]}/({"/".join(map(str, nums[1:]))})'
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```
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### **Java**
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```java
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class Solution {
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public String optimalDivision(int[] nums) {
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int n = nums.length;
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if (n == 1) {
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return nums[0] + "";
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}
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if (n == 2) {
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return nums[0] + "/" + nums[1];
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}
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StringBuilder ans = new StringBuilder(nums[0] + "/(");
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for (int i = 1; i < n - 1; ++i) {
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ans.append(nums[i] + "/");
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}
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ans.append(nums[n - 1] + ")");
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return ans.toString();
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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string optimalDivision(vector<int>& nums) {
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int n = nums.size();
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if (n == 1) return to_string(nums[0]);
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if (n == 2) return to_string(nums[0]) + "/" + to_string(nums[1]);
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string ans = to_string(nums[0]) + "/(";
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for (int i = 1; i < n - 1; i++) ans.append(to_string(nums[i]) + "/");
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ans.append(to_string(nums[n - 1]) + ")");
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return ans;
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}
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};
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```
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### **Go**
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```go
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func optimalDivision(nums []int) string {
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n := len(nums)
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if n == 1 {
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return strconv.Itoa(nums[0])
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}
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if n == 2 {
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return fmt.Sprintf("%d/%d", nums[0], nums[1])
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}
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ans := &strings.Builder{}
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ans.WriteString(fmt.Sprintf("%d/(", nums[0]))
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for _, num := range nums[1 : n-1] {
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ans.WriteString(strconv.Itoa(num))
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ans.WriteByte('/')
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}
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ans.WriteString(fmt.Sprintf("%d)", nums[n-1]))
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return ans.String()
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}
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```
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### **...**

‎solution/0500-0599/0553.Optimal Division/Solution.cpp

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class Solution {
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public:
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string optimalDivision(vector<int>& nums) {
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int n = nums.size();
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if (n == 1) return to_string(nums[0]);
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if (n == 2) return to_string(nums[0]) + "/" + to_string(nums[1]);
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string ans = to_string(nums[0]) + "/(";
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for (int i = 1; i < n - 1; i++) ans.append(to_string(nums[i]) + "/");
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ans.append(to_string(nums[n - 1]) + ")");
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return ans;
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}
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};

‎solution/0500-0599/0553.Optimal Division/Solution.go

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func optimalDivision(nums []int) string {
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n := len(nums)
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if n == 1 {
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return strconv.Itoa(nums[0])
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}
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if n == 2 {
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return fmt.Sprintf("%d/%d", nums[0], nums[1])
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}
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ans := &strings.Builder{}
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ans.WriteString(fmt.Sprintf("%d/(", nums[0]))
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for _, num := range nums[1 : n-1] {
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ans.WriteString(strconv.Itoa(num))
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ans.WriteByte('/')
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}
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ans.WriteString(fmt.Sprintf("%d)", nums[n-1]))
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return ans.String()
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}

‎solution/0500-0599/0553.Optimal Division/Solution.java

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class Solution {
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public String optimalDivision(int[] nums) {
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int n = nums.length;
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if (n == 1) {
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return nums[0] + "";
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}
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if (n == 2) {
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return nums[0] + "/" + nums[1];
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}
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StringBuilder ans = new StringBuilder(nums[0] + "/(");
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for (int i = 1; i < n - 1; ++i) {
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ans.append(nums[i] + "/");
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}
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ans.append(nums[n - 1] + ")");
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return ans.toString();
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}
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}

‎solution/0500-0599/0553.Optimal Division/Solution.py

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class Solution:
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def optimalDivision(self, nums: List[int]) -> str:
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n = len(nums)
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if n == 1:
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return str(nums[0])
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if n == 2:
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return f'{nums[0]}/{nums[1]}'
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return f'{nums[0]}/({"/".join(map(str, nums[1:]))})'

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