1910. Remove All Occurrences of a Substring
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
partand 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 <= 10001 <= part.length <= 1000sandpartconsists 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`.
JAVA
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; } }
C++
// Time: O()
// Space: O()
class Solution {
public:
string removeOccurrences(string s, string t) {
auto f = s.find(t);
while (f != string::npos) {
s.erase(f, t.size());
f = s.find(t);
}
return s;
}
};PYTHON
class Solution: def removeOccurrences(self, s: str, part: str) -> str: while part in s: s = s.replace(part, "", 1) # Replace only the first occurrence in each iteration return s
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