1910. Remove All Occurrences of a Substring

Leetcode Link

Problem Description

The task described in the LeetCode problem is about string manipulation. You are given two strings s and part. Your goal is to repeatedly find the leftmost occurrence of the string part in s and remove it. You should keep doing this operation until part can no longer be found within s. To clarify, a substring is a sequence of characters that appear in consecutive order within another string. The operation is only complete when there are no more sequences of characters in s that match part. The output should be the resulting string s after all possible removals of the substring part have been conducted.


To solve the problem, one can use the built-in string methods available in python. The intuition lies in searching for the substring part within the string s and removing the leftmost occurrence of it. This can be done iteratively using a while loop, which continues as long as part is found in s.

To implement this:

  1. Check if part is a substring of s using the in keyword.
  2. If it is, use the replace method of the string object, which replaces the first occurrence of part in s with an empty string (effectively removing it), but make sure to limit the replacement to just one occurrence by passing 1 as the second argument to replace.
  3. Return the modified string s once there are no more occurrences of part in it.

The key here is that the replace function is used in a controlled manner to only remove the first (leftmost) occurrence of part within each iteration, ensuring that the algorithm works as intended by the problem description.

Solution Approach

The implementation of the solution is straightforward and relies primarily on Python's string processing capabilities. Here's a step-by-step explanation of the approach using algorithms and data structures:

  • Algorithm: Iterative Removal
    • The algorithm uses a loop to repeatedly search and remove the substring part from s. It continues to do so until the part can no longer be found within s.
  • Data Structure: String
    • Strings are the primary data structure used in this problem. In Python, strings are immutable, meaning a new string is created each time you modify it.

Given the Python code snippet:

1class Solution:
2    def removeOccurrences(self, s: str, part: str) -> str:
3        while part in s:
4            s = s.replace(part, '', 1)
5        return s

Here's an explanation of the code:

  • While Loop:
    • The while loop checks whether part is still a substring of s. The condition part in s returns a boolean value - True if part is found in s and False otherwise.
  • String Replacement:
    • Inside the loop, the replace method is called on s. The first argument is the substring part we are looking for, the second argument is an empty string to which the found part will be replaced, signifying its removal. The third argument 1 ensures that only the first instance of part is replaced, which corresponds to the "leftmost occurrence".
  • Return value:
    • Once the loop terminates (when part is not found in s), the final version of s is returned, which no longer contains any occurrences of part.

No complex patterns or sophisticated algorithms are needed beyond the basic iterative approach. The solution leverages Python's built-in string methods to achieve the result with a clean and easy-to-understand implementation.

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Example Walkthrough

To illustrate the solution approach, let's go through a small example. Imagine we have the string s = "axbxcx" and the substring part = "x". Our goal is to remove the leftmost occurrence of x from s until x can no longer be found in s.

Here's how the implementation works step by step:

  • Initial: s = "axbxcx", part = "x"
  • First Iteration:
    1. Check if part is in s: x is in "axbxcx"
    2. Replace the first occurrence of part with an empty string:
      • Before: s = "axbxcx"
      • After: s = "abxcx" (we removed the first x)
  • Second Iteration:
    1. Check if part is in s again: x is in "abxcx"
    2. Replace the first occurrence of part with an empty string:
      • Before: s = "abxcx"
      • After: s = "abcx" (we removed the second x)
  • Third Iteration:
    1. Check if part is in s again: x is in "abcx"
    2. Replace the first occurrence of part with an empty string:
      • Before: s = "abcx"
      • After: s = "abc" (we removed the last x)
  • Final Check:
    • part is no longer in s: x is not in "abc"

The resulting string s is "abc" because all instances of part have been removed. This gives us the output of the function. Using the provided code snippet, the final return statement return s would give us "abc" as the solution to our input example.

Python Solution

1class Solution:
2    def remove_occurrences(self, string: str, part: str) -> str:
3        # Repeatedly search for the 'part' in 'string' and remove its first occurrence
4        while part in string:
5            # Find the index of the first occurrence of 'part'
6            index = string.find(part)
7            # Remove 'part' by slicing the string before and after 'part'
8            string = string[:index] + string[index+len(part):]
10        # Return the modified string after removing all occurrences of 'part'
11        return string
13# Example usage:
14# sol = Solution()
15# new_string = sol.remove_occurrences("daabcbaabcbc", "abc")
16# print(new_string)  # Output would be "dab"

Java Solution

1class Solution {
3    /**
4     * Removes all occurrences of the substring 'part' from the string 's'.
5     *
6     * @param s    The original string from which occurrences of 'part' will be removed.
7     * @param part The substring to be removed from 's'.
8     * @return The modified string with all occurrences of 'part' removed.
9     */
10    public String removeOccurrences(String s, String part) {
11        // Keep removing 'part' from 's' while 's' contains 'part'
12        while (s.contains(part)) {
13            // Replace the first occurrence of 'part' in 's' with an empty string
14            s = s.replaceFirst(part, "");
15        }
16        return s;
17    }

C++ Solution

1class Solution {
3    // Function to remove all occurrences of a substring 'part' from the string 's'
4    string removeOccurrences(string s, string part) {
5        // Get the size of the substring 'part'
6        int partSize = part.size();
8        // Find the first occurrence of 'part' in 's'
9        size_t position = s.find(part);
11        // Continue looping as long as 'part' is found in 's'
12        while (position != string::npos) {
13            // Erase 'part' from 's'
14            s.erase(position, partSize);
15            // Find the next occurrence of 'part' in 's'
16            position = s.find(part);
17        }
18        // Return the modified string with all 'part' occurrences removed
19        return s;
20    }

Typescript Solution

2 * Removes all occurrences of a specified substring from the given string.
3 * @param {string} str - The original string from which to remove occurrences.
4 * @param {string} part - The substring to remove from the original string.
5 * @returns {string} The modified string with all occurrences of the substring removed.
6 */
7function removeOccurrences(str: string, part: string): string {
8    // Continue to look for the substring `part` in `str` until it cannot be found
9    while (str.includes(part)) {
10        // Replace the first occurrence of `part` in `str` with an empty string
11        str = str.replace(part, '');
12    }
13    // Return the modified string with all occurrences of `part` removed
14    return str;

Time and Space Complexity

Time Complexity

The provided function's time complexity can be analyzed based on the while loop and the str.replace() method used within it.

  • The while loop runs as long as the substring part is found within the string s.
  • str.replace() is called each time the part is found, and it has a complexity of O(n) in the worst case, where n is the length of string s, since in the worst case it has to scan the entire string to replace the occurrences.

If m is the length of the substring part, the worst-case scenario occurs when part is found in s multiple times and the positions of part in s are distributed such that most of the string has to be scanned for each replacement. Therefore, the complexity can approximately be O((n - m + 1) * n) because it could take up to (n - m + 1) searches through the string s.

Hence, the worst-case time complexity is O((n - m + 1) * n).

Space Complexity

The space complexity of the function arises from the storage required for the input string s and the additional strings created during the replacements.

  • The input string s has a space complexity of O(n), where n is its length.
  • Each time a replacement is performed, a new string is created, and this new string could have a length up to n - m.

Given that only one replacement string exists at a time (the previous is discarded when the new one is created), the additional space required for the algorithm as a result of replacements is also O(n).

Therefore, the overall space complexity of the algorithm is O(n).

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