fixing formating on 451. Sort Character by Frequency

Author: YonashPetit

problems/451. Sort Characters By Frequency.md

@@ -26,11 +26,12 @@ How to reverse a linked list given start, end node?
 
 * `hold`: This is a dictionary (hash map) used to store the frequency of each character in the string. The keys represent the characters, and the values represent the number of times they appear.
 
-```
+```python
 biggest = 0
 hold = {}
 ```
 
+
 2. Building the Frequency Dictionary
 
 * For each character (`char`) in the string `s`:
@@ -41,19 +42,21 @@ hold = {}
 
 * `biggest` keeps track of the highest frequency of any character in `s` by updating it using `max(biggest, hold[char])` for each character.
 
-```
+```python
 for char in s:
     if char not in hold:
         hold[char] = 1
     else:
         hold[char] += 1
-    biggest = max(biggest, hold[char])
-```
-Example:
-For a string like s = "abbccc", this step would produce:
+    biggest = max(biggest, hold[char])```
 
-hold = {'a': 1, 'b': 2, 'c': 3}
-biggest = 3 (since the character c appears the most with a frequency of 3).
++ Example:
+
+* For a string like s = "abbccc", this step would produce:
+
+* `hold = {'a': 1, 'b': 2, 'c': 3}`
+
+* `biggest = 3 (since the character c appears the most with a frequency of 3).`
 
 
 3. Creating an Array to Hold Characters by Frequency
@@ -62,23 +65,27 @@ biggest = 3 (since the character c appears the most with a frequency of 3).
 
 * For a string like s = "abbccc", this step would produce:
 
-'''
+
+```python
 array = [""] * (biggest + 1)
-'''
+```
+
 
 4. Populating the Frequency Array
 
 * This loop iterates over each character (key) and its frequency (value) in the hold dictionary.
 
 * For each character, it places that character (repeated value times) at the index corresponding to its frequency in the array. The string of characters is concatenated at the corresponding index in array.
 
-'''
+
+```python
 for key, value in hold.items():
-    array[value] += key * value
-'''
+    array[value] += key * value```
 
-Example:
-Continuing with `s = "abbccc"`:
+
++ Example:
+
+* Continuing with `s = "abbccc"`:
 
 * `hold = {'a': 1, 'b': 2, 'c': 3}`
 * After this step, array will look like this:
@@ -87,6 +94,7 @@ Continuing with `s = "abbccc"`:
   - `array[2]` has `"bb"` because `b` appears twice.
   - `array[3]` has `"ccc"` because `c` appears three times.
 
+  
 5. Building the Answer String in Decreasing Order of Frequency
 
 * The answer variable is initialized as an empty string. It will store the final result.
@@ -95,15 +103,16 @@ Continuing with `s = "abbccc"`:
 
 * Since array.pop() removes elements from the end, it effectively appends characters to answer in decreasing order of their frequency.
 
-'''
+```python
 answer = ""
 for i in range(len(array)):
     answer += array.pop()
-return answer
-'''
+return answer```
 
-Example:
-With array = ['', 'a', 'bb', 'ccc'], the popping process will occur as follows:
+
++ Example:
+
++ With array = ['', 'a', 'bb', 'ccc'], the popping process will occur as follows:
 
 * Pop "ccc" → answer = "ccc"
 * Pop "bb" → answer = "cccbb"