Package g1901_2000.s1910_remove_all_occurrences_of_a_substring

Class Solution

java.lang.Object

g1901_2000.s1910_remove_all_occurrences_of_a_substring.Solution

public class Solution
extends Object

1910 - Remove All Occurrences of a Substring\. Medium 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 = "da**abc**baabcbc", remove "abc" starting at index 2, so s = "dabaabcbc". - s = "daba**abc**bc", remove "abc" starting at index 4, so s = "dababc". - s = "dab**abc**", 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 = "axxx**xy**yyyb", remove "xy" starting at index 4 so s = "axxxyyyb". - s = "axx**xy**yyb", remove "xy" starting at index 3 so s = "axxyyb". - s = "ax**xy**yb", remove "xy" starting at index 2 so s = "axyb". - s = "a**xy**b", 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.

Constructor Summary

Constructors

Constructor	Description
Solution()

Method Summary

Modifier and Type	Method	Description
String	removeOccurrences(String s, String part)

Methods inherited from class java.lang.Object

clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait

Constructor Details

Solution

public Solution()

Method Details

removeOccurrences

public String removeOccurrences(String s,
 String part)