The Increasing Decreasing String problem on LeetCode is a fascinating and somewhat quirky problem that tests your ability to manipulate strings and arrays effectively. This problem provides an opportunity to explore various aspects of Python’s string and list capabilities, along with improving your overall problem-solving skills.

In this comprehensive article, we will delve deep into the problem, discuss its requirements, and explore multiple approaches to solve it. We will examine the performance of each solution and provide detailed explanations.

## Problem Statement

The task is to build a string `result`

using the characters from the input string `s`

according to the following rules:

- Pick the smallest character from
`s`

and append it to`result`

. - Pick the smallest character from
`s`

that is greater than the last appended character and append it to`result`

. - Repeat this process until you cannot pick more characters.
- Pick the largest character from
`s`

and append it to`result`

. - Pick the largest character from
`s`

that is smaller than the last appended character and append it to`result`

. - Repeat this process until you cannot pick more characters.
- Repeat steps 1-6 until you pick every character from
`s`

.

### Constraints

`1 <= s.length <= 500`

`s`

consists of only lowercase English letters.

### Example

```
Input: s = "aaaabbbbcccc"
Output: "abccbaabccba"
```

## Approaches to Solve the Problem

### Sorting and List Manipulation

The idea is to first sort the input string and then iteratively build the result string based on the sorted characters.

Here’s the Python code snippet:

```
def sortString(s):
s = sorted(s)
result = []
while s:
for char in sorted(set(s)):
result.append(char)
s.remove(char)
for char in sorted(set(s), reverse=True):
result.append(char)
s.remove(char)
return ''.join(result)
```

Time Complexity: O(n log n)

Space Complexity: O(n)

### Frequency Counting Approach

Another efficient approach is to use a frequency counter to count occurrences of each character in `s`

. Then build the `result`

string by following the rules.

Here’s the Python code snippet:

```
from collections import Counter
def sortString(s):
freq = Counter(s)
keys = sorted(freq.keys())
result = []
while len(result) < len(s):
for key in keys:
if freq[key]:
result.append(key)
freq[key] -= 1
for key in reversed(keys):
if freq[key]:
result.append(key)
freq[key] -= 1
return ''.join(result)
```

Time Complexity: O(n)

Space Complexity: O(n)

## Comparative Analysis

### Sorting and List Manipulation

**Pros**: Intuitive, easy to understand.**Cons**: Slower due to sorting and list manipulation. Removing elements from a list is not efficient.

### Frequency Counting Approach

**Pros**: More efficient, uses constant space for storing frequencies.**Cons**: Slightly harder to understand for newcomers.

## Edge Cases and Optimization

### Edge Cases

**Single character**: If`s`

contains a single unique character, the output will be the same as the input.**No repeating characters**: If`s`

contains no repeating characters, the output will also not have any repeating characters.

### Optimization

The frequency counting approach already optimizes the problem well. However, you can micro-optimize by breaking the loop if the result string’s length matches the input string’s length, thus avoiding any unnecessary iterations.

## Conclusion

The Increasing Decreasing String problem serves as a good example to explore different techniques in string and list manipulation. By solving it using different approaches, one can understand the importance of algorithmic optimization and gain a deeper insight into how string manipulations work in Python. Both approaches have their merits, but the frequency counting approach offers better performance and should be preferred for large inputs.