Solution Explanation

The problem asks to repeatedly remove the leftmost occurrence of a substring part from a string s until no more occurrences of part exist. The solutions provided utilize a straightforward iterative approach. Let's break down the logic and complexity:

Approach:

The core idea is to repeatedly search for the substring part within s and remove it if found. This continues until part is no longer present in s. Different programming languages offer built-in functions to achieve this efficiently:

• replaceFirst() or replace(...,1): This method finds the first occurrence of part and removes it. The 1 in the python replace function limits replacement to the first occurence. If no occurrence is found, the string remains unchanged. This ensures that the leftmost occurrence is always targeted.

• erase(): (C++) Similar to replaceFirst(), but it directly manipulates the string.

• strings.Replace(s, part, "", 1): (Go) Again, this replaces the first occurrence only.

Code Explanation (Python Example):

class Solution:
    def removeOccurrences(self, s: str, part: str) -> str:
        while part in s:
            s = s.replace(part, '', 1) #Removes the first occurrence of 'part'
        return s

The Python code uses a while loop that continues as long as part is found within s. Inside the loop, s.replace(part, "", 1) replaces the first instance of part with an empty string. The loop terminates when no more occurrences of part are found.

The other languages follow a similar pattern, leveraging their respective string manipulation functions.

Time Complexity Analysis:

The time complexity is dominated by the repeated substring searches. In the worst-case scenario, the substring part might be found at the beginning of the string repeatedly, leading to O(nmk) complexity, where:

• n = length of s
• m = length of part
• k = Number of times the substring needs to be removed. In the worst case, k could be O(n/m).

However, in practice, the number of times the loop iterates is significantly less than n in most cases, making this solution efficient for many inputs. The replace operation itself has an average time complexity of O(n) where n is the length of the string in many implementations.

Space Complexity Analysis:

The space complexity depends on how the string manipulations are implemented. In most cases, the space used is proportional to the length of the string, resulting in O(n) space complexity where n is the length of string s. The algorithm does not create significant auxiliary data structures.

Improvements (Optional):

For extremely large strings, a more optimized approach might involve using a more efficient string searching algorithm (like Knuth-Morris-Pratt or Boyer-Moore) to reduce the search time in the while loop, although this is often not necessary unless dealing with very large strings. For average cases, the provided solutions are efficient.