Formatted question description: https://leetcode.ca/all/1910.html

# 1910. Remove All Occurrences of a Substring

Medium

## Description

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

Use indexOf to find the index of the occurrence of part in s, and remove the corresponding substring from s. Repeat this step until s does not contain part. Finally, return s.

class Solution {
public String removeOccurrences(String s, String part) {
int partLength = part.length();
while (s.indexOf(part) >= 0) {
int index = s.indexOf(part);
s = s.substring(0, index) + s.substring(index + partLength);
}
return s;
}
}
`