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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

  • class Solution { public String removeOccurrences(String s, String part) { while (s.contains(part)) { s = s.replaceFirst(part, ""); } return s; } } 
  • 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; } }; 
  • class Solution: def removeOccurrences(self, s: str, part: str) -> str: while part in s: s = s.replace(part, '', 1) return s 
  • func removeOccurrences(s string, part string) string { for strings.Contains(s, part) { s = strings.Replace(s, part, "", 1) } return s } 
  • function removeOccurrences(s: string, part: string): string { while (s.includes(part)) { s = s.replace(part, ''); } return s; } 

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