remove-all-occurrences-of-a-substring

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Algorithms

Given two strings <code>s</code> and <code>part</code>, perform the following operation on <code>s</code> until all occurrences of the substring <code>part</code> are removed:

<ul>
	<li>Find the leftmost occurrence of the substring <code>part</code> and remove it from <code>s</code>.</li>
</ul>

Return <code>s</code> after removing all occurrences of <code>part</code>.

A substring is a contiguous sequence of characters in a string.

&nbsp;
Example 1:

<pre>
Input: s = &quot;daabcbaabcbc&quot;, part = &quot;abc&quot;
Output: &quot;dab&quot;
Explanation: The following operations are done:
- s = &quot;daabcbaabcbc&quot;, remove &quot;abc&quot; starting at index 2, so s = &quot;dabaabcbc&quot;.
- s = &quot;dabaabcbc&quot;, remove &quot;abc&quot; starting at index 4, so s = &quot;dababc&quot;.
- s = &quot;dababc&quot;, remove &quot;abc&quot; starting at index 3, so s = &quot;dab&quot;.
Now s has no occurrences of &quot;abc&quot;.
</pre>

Example 2:

<pre>
Input: s = &quot;axxxxyyyyb&quot;, part = &quot;xy&quot;
Output: &quot;ab&quot;
Explanation: The following operations are done:
- s = &quot;axxxxyyyyb&quot;, remove &quot;xy&quot; starting at index 4 so s = &quot;axxxyyyb&quot;.
- s = &quot;axxxyyyb&quot;, remove &quot;xy&quot; starting at index 3 so s = &quot;axxyyb&quot;.
- s = &quot;axxyyb&quot;, remove &quot;xy&quot; starting at index 2 so s = &quot;axyb&quot;.
- s = &quot;axyb&quot;, remove &quot;xy&quot; starting at index 1 so s = &quot;ab&quot;.
Now s has no occurrences of &quot;xy&quot;.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
	<li><code>1 &lt;= part.length &lt;= 1000</code></li>
	<li><code>s</code>​​​​​​ and <code>part</code> consists of lowercase English letters.</li>
</ul>

removeOccurrences

Remove All Occurrences of a Substring

Maximum Deletions on a String

maximum-deletions-on-a-string

对字母串可执行的最大删除数

Stack

String

Simulation

给你两个字符串 <code>s</code> 和 <code>part</code> ，请你对 <code>s</code> 反复执行以下操作直到 所有 子字符串 <code>part</code> 都被删除：

<ul>
	<li>找到 <code>s</code> 中 最左边 的子字符串 <code>part</code> ，并将它从 <code>s</code> 中删除。</li>
</ul>

请你返回从 <code>s</code> 中删除所有 <code>part</code> 子字符串以后得到的剩余字符串。

一个 子字符串 是一个字符串中连续的字符序列。

 

示例 1：

<pre>输入：s = "daabcbaabcbc", part = "abc"
输出："dab"
解释：以下操作按顺序执行：
- s = "daabcbaabcbc" ，删除下标从 2 开始的 "abc" ，得到 s = "dabaabcbc" 。
- s = "dabaabcbc" ，删除下标从 4 开始的 "abc" ，得到 s = "dababc" 。
- s = "dababc" ，删除下标从 3 开始的 "abc" ，得到 s = "dab" 。
此时 s 中不再含有子字符串 "abc" 。
</pre>

示例 2：

<pre>输入：s = "axxxxyyyyb", part = "xy"
输出："ab"
解释：以下操作按顺序执行：
- s = "axxxxyyyyb" ，删除下标从 4 开始的 "xy" ，得到 s = "axxxyyyb" 。
- s = "axxxyyyb" ，删除下标从 3 开始的 "xy" ，得到 s = "axxyyb" 。
- s = "axxyyb" ，删除下标从 2 开始的 "xy" ，得到 s = "axyb" 。
- s = "axyb" ，删除下标从 1 开始的 "xy" ，得到 s = "ab" 。
此时 s 中不再含有子字符串 "xy" 。
</pre>

 

提示：

<ul>
	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
	<li><code>1 &lt;= part.length &lt;= 1000</code></li>
	<li><code>s</code>​​​​​​ 和 <code>part</code> 只包小写英文字母。</li>
</ul>