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1910. Remove All Occurrences of a Substring

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.

Solutions

Solution 1

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

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class Solution {
    public String removeOccurrences(String s, String part) {
        while (s.contains(part)) {
            s = s.replaceFirst(part, "");
        }
        return s;
    }
}

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class Solution {
public:
    string removeOccurrences(string s, string part) {
        int m = part.size();
        while (s.find(part) != -1) {
            s = s.erase(s.find(part), m);
        }
        return s;
    }
};

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func removeOccurrences(s string, part string) string {
    for strings.Contains(s, part) {
        s = strings.Replace(s, part, "", 1)
    }
    return s
}

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function removeOccurrences(s: string, part: string): string {
    while (s.includes(part)) {
        s = s.replace(part, '');
    }
    return s;
}

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