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 froms
.
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
andpart
consists of lowercase English letters.
class Solution:
def removeOccurrences(self, s: str, part: str) -> str:
while part in s:
s = s.replace(part, '', 1)
return s
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;
}
};
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;
}