Translate all code to English (#1836)

* Review the EN heading format.

* Fix pythontutor headings.

* Fix pythontutor headings.

* bug fixes

* Fix headings in **/summary.md

* Revisit the CN-to-EN translation for Python code using Claude-4.5

* Revisit the CN-to-EN translation for Java code using Claude-4.5

* Revisit the CN-to-EN translation for Cpp code using Claude-4.5.

* Fix the dictionary.

* Fix cpp code translation for the multipart strings.

* Translate Go code to English.

* Update workflows to test EN code.

* Add EN translation for C.

* Add EN translation for CSharp.

* Add EN translation for Swift.

* Trigger the CI check.

* Revert.

* Update en/hash_map.md

* Add the EN version of Dart code.

* Add the EN version of Kotlin code.

* Add missing code files.

* Add the EN version of JavaScript code.

* Add the EN version of TypeScript code.

* Fix the workflows.

* Add the EN version of Ruby code.

* Add the EN version of Rust code.

* Update the CI check for the English version  code.

* Update Python CI check.

* Fix cmakelists for en/C code.

* Fix Ruby comments

en/docs/chapter_greedy/fractional_knapsack_problem.md

@@ -1,4 +1,4 @@
-# Fractional knapsack problem
+# Fractional Knapsack Problem
 
 !!! question
 
@@ -15,7 +15,7 @@ The difference is that this problem allows selecting only a portion of an item.
 
 ![Value of items per unit weight](fractional_knapsack_problem.assets/fractional_knapsack_unit_value.png)
 
-### Greedy strategy determination
+### Greedy Strategy Determination
 
 Maximizing the total value of items in the knapsack **is essentially maximizing the value per unit weight of items**. From this, we can derive the greedy strategy shown in the figure below.
 
@@ -25,7 +25,7 @@ Maximizing the total value of items in the knapsack **is essentially maximizing
 
 ![Greedy strategy for the fractional knapsack problem](fractional_knapsack_problem.assets/fractional_knapsack_greedy_strategy.png)
 
-### Code implementation
+### Code Implementation
 
 We created an `Item` class to facilitate sorting items by unit value. We loop to make greedy selections, breaking when the knapsack is full and returning the solution:
 
@@ -39,7 +39,7 @@ Apart from sorting, in the worst case the entire item list needs to be traversed
 
 Since an `Item` object list is initialized, **the space complexity is $O(n)$**.
 
-### Correctness proof
+### Correctness Proof
 
 Using proof by contradiction. Suppose item $x$ has the highest unit value, and some algorithm yields a maximum value of `res`, but this solution does not include item $x$.