1910. Remove All Occurrences of a Substring

Problem

Given two strings s and part, perform the following operation on s until all occurrences of the substring part are removed:

• Find the leftmost occurrence of the substring part and remove it from s.

Return s** after removing all occurrences of **part.

A substring is a contiguous sequence of characters in a string.

  Example 1:

Input: s = "daabcbaabcbc", part = "abc"
Output: "dab"
Explanation: The following operations are done:
- s = "daabcbaabcbc", remove "abc" starting at index 2, so s = "dabaabcbc".
- s = "dabaabcbc", remove "abc" starting at index 4, so s = "dababc".
- s = "dababc", remove "abc" starting at index 3, so s = "dab".
Now s has no occurrences of "abc".

Example 2:

Input: s = "axxxxyyyyb", part = "xy"
Output: "ab"
Explanation: The following operations are done:
- s = "axxxxyyyyb", remove "xy" starting at index 4 so s = "axxxyyyb".
- s = "axxxyyyb", remove "xy" starting at index 3 so s = "axxyyb".
- s = "axxyyb", remove "xy" starting at index 2 so s = "axyb".
- s = "axyb", remove "xy" starting at index 1 so s = "ab".
Now s has no occurrences of "xy".

  Constraints:

• 1 <= s.length <= 1000

• 1 <= part.length <= 1000

• s​​​​​​ and part consists of lowercase English letters.

Solution (Java)

class Solution {
    public String removeOccurrences(String s, String part) {
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < s.length(); i++) {
            sb.append(s.charAt(i));
            if (sb.length() >= part.length()
                    && sb.substring(sb.length() - part.length()).equals(part)) {
                sb.setLength(sb.length() - part.length());
            }
        }
        return sb.toString();
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).