Sync LeetCode submission Runtime - 12 ms (92.46%), Memory - 18 MB (61.91%)

Author: jimit105

0301-remove-invalid-parentheses/README.md

@@ -0,0 +1,34 @@
+<p>Given a string <code>s</code> that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.</p>
+
+<p>Return <em>a list of <strong>unique strings</strong> that are valid with the minimum number of removals</em>. You may return the answer in <strong>any order</strong>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;()())()&quot;
+<strong>Output:</strong> [&quot;(())()&quot;,&quot;()()()&quot;]
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;(a)())()&quot;
+<strong>Output:</strong> [&quot;(a())()&quot;,&quot;(a)()()&quot;]
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;)(&quot;
+<strong>Output:</strong> [&quot;&quot;]
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 25</code></li>
+	<li><code>s</code> consists of lowercase English letters and parentheses <code>&#39;(&#39;</code> and <code>&#39;)&#39;</code>.</li>
+	<li>There will be at most <code>20</code> parentheses in <code>s</code>.</li>
+</ul>

0301-remove-invalid-parentheses/solution.py

@@ -0,0 +1,53 @@
+# Approach: Backtracking
+
+# Time: O(2^n)
+# Space: O(n)
+
+class Solution:
+    def removeInvalidParentheses(self, s: str) -> List[str]:
+        def isValid(s):
+            count = 0
+            for char in s:
+                if char == '(':
+                    count += 1
+                elif char == ')':
+                    count -= 1
+                if count < 0:
+                    return False
+            return count == 0
+
+        # Use backtracking to try different combinations of removals
+        def backtrack(s, start, left_count, right_count):
+            if left_count == 0 and right_count == 0:
+                if isValid(s):
+                    result.add(s)
+                return
+
+            for i in range(start, len(s)):
+                # Skip duplicates (Optimization)
+                if i > start and s[i] == s[i - 1]:
+                    continue
+
+                # Try removing left parenthesis
+                if left_count > 0 and s[i] == '(':
+                    backtrack(s[:i] + s[i+1 :], i, left_count - 1, right_count)
+
+                # Try removing right parenthesis
+                if right_count > 0 and s[i] == ')':
+                    backtrack(s[:i] + s[i+1 :], i, left_count, right_count - 1)
+
+        # Count invalid parentheses
+        left = right = 0
+        for char in s:
+            if char == '(':
+                left += 1
+            elif char == ')':
+                if left == 0:
+                    right += 1
+                else:
+                    left -= 1
+
+        result = set()
+        backtrack(s, 0, left, right)
+        return list(result)
+