Sync LeetCode submission Runtime - 16 ms (18.84%), Memory - 18 MB (32.79%)

Author: jimit105

2021-remove-all-occurrences-of-a-substring/README.md

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+<p>Given two strings <code>s</code> and <code>part</code>, perform the following operation on <code>s</code> until <strong>all</strong> occurrences of the substring <code>part</code> are removed:</p>
+
+<ul>
+	<li>Find the <strong>leftmost</strong> occurrence of the substring <code>part</code> and <strong>remove</strong> it from <code>s</code>.</li>
+</ul>
+
+<p>Return <code>s</code><em> after removing all occurrences of </em><code>part</code>.</p>
+
+<p>A <strong>substring</strong> is a contiguous sequence of characters in a string.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;daabcbaabcbc&quot;, part = &quot;abc&quot;
+<strong>Output:</strong> &quot;dab&quot;
+<strong>Explanation</strong>: The following operations are done:
+- s = &quot;da<strong><u>abc</u></strong>baabcbc&quot;, remove &quot;abc&quot; starting at index 2, so s = &quot;dabaabcbc&quot;.
+- s = &quot;daba<strong><u>abc</u></strong>bc&quot;, remove &quot;abc&quot; starting at index 4, so s = &quot;dababc&quot;.
+- s = &quot;dab<strong><u>abc</u></strong>&quot;, remove &quot;abc&quot; starting at index 3, so s = &quot;dab&quot;.
+Now s has no occurrences of &quot;abc&quot;.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;axxxxyyyyb&quot;, part = &quot;xy&quot;
+<strong>Output:</strong> &quot;ab&quot;
+<strong>Explanation</strong>: The following operations are done:
+- s = &quot;axxx<strong><u>xy</u></strong>yyyb&quot;, remove &quot;xy&quot; starting at index 4 so s = &quot;axxxyyyb&quot;.
+- s = &quot;axx<strong><u>xy</u></strong>yyb&quot;, remove &quot;xy&quot; starting at index 3 so s = &quot;axxyyb&quot;.
+- s = &quot;ax<strong><u>xy</u></strong>yb&quot;, remove &quot;xy&quot; starting at index 2 so s = &quot;axyb&quot;.
+- s = &quot;a<strong><u>xy</u></strong>b&quot;, remove &quot;xy&quot; starting at index 1 so s = &quot;ab&quot;.
+Now s has no occurrences of &quot;xy&quot;.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<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>

2021-remove-all-occurrences-of-a-substring/solution.py

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+# Approach 2: Stack
+
+# Time: O(n * m)
+# Space: O(n + m)
+
+class Solution:
+    def removeOccurrences(self, s: str, part: str) -> str:
+        stack = []
+        part_length = len(part)
+
+        for char in s:
+            stack.append(char)
+
+            if len(stack) >= part_length and self._check_match(stack, part, part_length):
+                for _ in range(part_length):
+                    stack.pop()
+
+        return ''.join(stack)
+
+    def _check_match(self, stack, part, part_length):
+        return ''.join(stack[-part_length:]) == part
+