**Mastering String Manipulation: Removing Occurrences with Efficiency**

### Thoughts and Talk

String manipulation tasks are ubiquitous in programming, and they often involve intricate logic and edge cases. The problem we're tackling today revolves around removing specific occurrences of a substring from a given string, but with a twist: we need to remove not only the substring but also the characters to its right until the next occurrence of the last character of the substring.

This problem might seem deceptively simple at first glance, but it demands a careful approach and a deep understanding of data structures and algorithms. By exploring various solutions, we can appreciate the nuances of problem-solving and the importance of optimizing our code for efficiency.

### Problem Statement

Given a string

`s`

and a substring `part`

, we need to remove all occurrences of `part`

from `s`

. However, the catch is that when we encounter an occurrence of `part`

, we should remove not only `part`

itself but also all the characters to its right until the next occurrence of the last character of `part`

. If there is no such occurrence, we remove all characters to the right of `part`

.### Approaches Explored

#### Approach 1: Brute Force

**Pros**:- Simple and easy to understand
- No complex data structures or algorithms required

**Cons**:- Inefficient, especially for large strings and frequent occurrences of the substring
- Time complexity: O(n * m), where n is the length of the string and m is the length of the substring

#### Approach 2: Stack-based Solution

**Pros**:- Efficient and optimized solution
- Time complexity: O(n), where n is the length of the string

**Cons**:- Slightly more complex implementation compared to the brute force approach
- Requires additional space for the stack data structure

### Finalized Approach

After evaluating the pros and cons of both approaches, we opted for the stack-based solution as it provides better time complexity and efficiency, especially for larger inputs. This approach leverages the power of a stack data structure to maintain the characters that need to be preserved while removing the occurrences of

`part`

and the characters to its right until the next occurrence of the last character of `part`

.### Executable Code

TheTechCruise.com Pyodide Terminal

```
class Solution:
def removeOccurrences(self, s: str, part: str) -> str:
st = [] # Initialize an empty stack
# Helper function to get a slice of a list or string
def get_part(m, n):
return m[-n+1:] if len(m) >= n else m
lp = len(part)
# Special case: If the length of 'part' is 1, we can use string replacement
if lp == 1:
return s.replace(part, "")
for i in range(len(s)):
# If the stack contains enough characters to potentially form 'part'
if len(st) + 1 >= lp and s[i] == part[-1]:
# Check if the characters in the stack, combined with the current character,
# form 'part'
if get_part(st, lp) + [s[i]] == list(part):
# Pop the characters from the stack that form 'part'
for i in range(lp-1):
st.pop()
else:
st.append(s[i])
else:
st.append(s[i])
return "".join(st) # Convert the stack back to a string and return it
sol = Solution()
# Test Case 1
print("Test Case 1:", sol.removeOccurrences("daabcbaabcbc", "abc")) # "dab"
# Test Case 2
print("Test Case 2:", sol.removeOccurrences("axxxxyyyyb", "xy")) # "axxxb"
# Test Case 3
print("Test Case 3:", sol.removeOccurrences("aabccbac", "abc")) # "bac"
# Test Case 4
print("Test Case 4:", sol.removeOccurrences("aabbaabb", "ab")) # ""
# Test Case 5
print("Test Case 5:", sol.removeOccurrences("xyz", "abc")) # "xyz"
```

### Code Analysis and Code Walk-through

Let's break down the code and understand how it works:

- We start by initializing an empty stack
`st`

.

- We define a helper function
`get_part`

that takes a list or string`m`

and an integer`n`

. It returns a slice of`m`

starting from the last`n`

characters. This function is used to check if the characters in the stack, combined with the current character, form`part`

.

- We store the length of the substring
`part`

in the variable`lp`

.

- If the length of
`part`

is 1, we can simply use the built-in`replace`

method to remove all occurrences of`part`

from`s`

. This is a special case optimization.

- We iterate over each character
`s[i]`

in the string`s`

: - If the length of the stack
`st`

plus 1 is greater than or equal to the length of`part`

(`lp`

), and the current character`s[i]`

matches the last character of`part`

, we check if the characters in the stack, combined with the current character, form`part`

. - If they form
`part`

, we pop`lp-1`

characters from the stack, effectively removing the occurrence of`part`

and the characters to its right until the next occurrence of the last character of`part`

. - If they don't form
`part`

, we push the current character`s[i]`

onto the stack. - If the length of the stack plus 1 is less than the length of
`part`

, we simply push the current character`s[i]`

onto the stack.

- After iterating over the entire string, the stack will contain the modified string with the desired occurrences of
`part`

removed.

- Finally, we convert the stack back to a string using the
`join`

method and return it as the result.

### Example with Explanation

Let's consider an example to better understand the code's execution:

TheTechCruise.com Pyodide Terminal

```
s = "daabcbaabcbc"
part = "abc"
```

Here's a step-by-step explanation of how the code works:

- The stack
`st`

is initialized as an empty list:`[]`

.

- The length of
`part`

is`lp = 3`

.

- The string
`s`

is iterated over character by character: `'d'`

is pushed onto the stack:`['d']`

`'a'`

is pushed onto the stack:`['d', 'a']`

`'a'`

is pushed onto the stack:`['d', 'a', 'a']`

`'b'`

is pushed onto the stack:`['d', 'a', 'a', 'b']`

`'c'`

matches the last character of`part`

, and the stack contains`['a', 'a', 'b']`

which, combined with`'c'`

, forms`part`

. So, we pop`'b'`

,`'a'`

, and`'a'`

from the stack, leaving`['d']`

.`'b'`

is pushed onto the stack:`['d', 'b']`

`'a'`

is pushed onto the stack:`['d', 'b', 'a']`

`'a'`

is pushed onto the stack:`['d', 'b', 'a', 'a']`

`'b'`

is pushed onto the stack:`['d', 'b', 'a', 'a', 'b']`

`'c'`

matches the last character of`part`

, and the stack contains`['a', 'a', 'b']`

which, combined with`'c'`

, forms`part`

. So, we pop`'b'`

,`'a'`

, and`'a'`

from the stack, leaving`['d', 'b']`

.`'b'`

is pushed onto the stack:`['d', 'b', 'b']`

`'c'`

is pushed onto the stack:`['d', 'b', 'b', 'c']`

- After iterating over the entire string, the stack contains
`['d', 'a', 'b']`

.

- The stack is converted back to a string
`"dab"`

and returned as the result.

### Discussion on Complexity

The time complexity of this solution is O(n), where n is the length of the input string

`s`

. This is because we iterate over the string once, performing constant-time operations (stack push and pop) for each character.The space complexity is O(n) in the worst case, where the entire input string needs to be stored in the stack. However, in practice, the space complexity is often much smaller, as we pop characters from the stack when we encounter occurrences of

`part`

.### Conclusion

In this blog post, we explored a problem involving string manipulation and the removal of specific occurrences of a substring, along with the characters to its right until the next occurrence of the last