LeetCode 1910: Remove All Occurrences of a Substring Solution
Master LeetCode problem 1910 (Remove All Occurrences of a Substring), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
1910. Remove All Occurrences of a Substring
Problem Explanation
Explanation:
To solve this problem, we can iterate through the string s and keep removing occurrences of the substring part until there are no more occurrences left. We can achieve this by using the indexOf() method to find the index of the first occurrence of part. If the index is found, we remove the substring at that index. We repeat this process until no more occurrences of part are present in s.
Algorithm:
- Initialize a variable
indexto store the index of the first occurrence ofpartins. - While
indexis not equal to -1, remove the substringpartat the indexindexins. - Update the value of
indexto the index of the next occurrence ofpart. - Repeat steps 2 and 3 until no more occurrences of
partare found ins. - Return the modified string
s.
Time Complexity:
The time complexity of this approach is O(n*m), where n is the length of the string s and m is the length of the substring part. In the worst case, we may have to iterate through the entire string s multiple times to remove all occurrences of part.
Space Complexity:
The space complexity of this approach is O(n), where n is the length of the string s. This is because we are modifying the string s in-place without using any additional data structures.
:
Solution Code
class Solution {
public String removeOccurrences(String s, String part) {
int index = s.indexOf(part);
while (index != -1) {
s = s.substring(0, index) + s.substring(index + part.length());
index = s.indexOf(part);
}
return s;
}
}Try It Yourself
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