1910

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 = "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.

Companies:
Twitter, Zoho

Related Topics:
String

Solution 1. Brute Force

// OJ: https://leetcode.com/problems/remove-all-occurrences-of-a-substring/
// Author: github.com/lzl124631x
// 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;
    }
};