Sync LeetCode submission Runtime - 0 ms (100.00%), Memory - 17.6 MB (70.70%)

Author: jimit105

3447-clear-digits/README.md

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+<p>You are given a string <code>s</code>.</p>
+
+<p>Your task is to remove <strong>all</strong> digits by doing this operation repeatedly:</p>
+
+<ul>
+	<li>Delete the <em>first</em> digit and the <strong>closest</strong> <b>non-digit</b> character to its <em>left</em>.</li>
+</ul>
+
+<p>Return the resulting string after removing all digits.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">s = &quot;abc&quot;</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">&quot;abc&quot;</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>There is no digit in the string.<!-- notionvc: ff07e34f-b1d6-41fb-9f83-5d0ba3c1ecde --></p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">s = &quot;cb34&quot;</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">&quot;&quot;</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>First, we apply the operation on <code>s[2]</code>, and <code>s</code> becomes <code>&quot;c4&quot;</code>.</p>
+
+<p>Then we apply the operation on <code>s[1]</code>, and <code>s</code> becomes <code>&quot;&quot;</code>.</p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 100</code></li>
+	<li><code>s</code> consists only of lowercase English letters and digits.</li>
+	<li>The input is generated such that it is possible to delete all digits.</li>
+</ul>

3447-clear-digits/solution.py

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+# Approach 2: Stack
+
+# Time: O(n)
+# Space: O(n)
+
+class Solution:
+    def clearDigits(self, s: str) -> str:
+        stack = []
+
+        for char in s:
+            if char.isdigit() and stack:
+                stack.pop()
+            else:
+                stack.append(char)
+
+        return ''.join(stack)
+