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Author: jimit105

2221-check-if-a-parentheses-string-can-be-valid/README.md

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+<p>A parentheses string is a <strong>non-empty</strong> string consisting only of <code>&#39;(&#39;</code> and <code>&#39;)&#39;</code>. It is valid if <strong>any</strong> of the following conditions is <strong>true</strong>:</p>
+
+<ul>
+	<li>It is <code>()</code>.</li>
+	<li>It can be written as <code>AB</code> (<code>A</code> concatenated with <code>B</code>), where <code>A</code> and <code>B</code> are valid parentheses strings.</li>
+	<li>It can be written as <code>(A)</code>, where <code>A</code> is a valid parentheses string.</li>
+</ul>
+
+<p>You are given a parentheses string <code>s</code> and a string <code>locked</code>, both of length <code>n</code>. <code>locked</code> is a binary string consisting only of <code>&#39;0&#39;</code>s and <code>&#39;1&#39;</code>s. For <strong>each</strong> index <code>i</code> of <code>locked</code>,</p>
+
+<ul>
+	<li>If <code>locked[i]</code> is <code>&#39;1&#39;</code>, you <strong>cannot</strong> change <code>s[i]</code>.</li>
+	<li>But if <code>locked[i]</code> is <code>&#39;0&#39;</code>, you <strong>can</strong> change <code>s[i]</code> to either <code>&#39;(&#39;</code> or <code>&#39;)&#39;</code>.</li>
+</ul>
+
+<p>Return <code>true</code> <em>if you can make <code>s</code> a valid parentheses string</em>. Otherwise, return <code>false</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/11/06/eg1.png" style="width: 311px; height: 101px;" />
+<pre>
+<strong>Input:</strong> s = &quot;))()))&quot;, locked = &quot;010100&quot;
+<strong>Output:</strong> true
+<strong>Explanation:</strong> locked[1] == &#39;1&#39; and locked[3] == &#39;1&#39;, so we cannot change s[1] or s[3].
+We change s[0] and s[4] to &#39;(&#39; while leaving s[2] and s[5] unchanged to make s valid.</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;()()&quot;, locked = &quot;0000&quot;
+<strong>Output:</strong> true
+<strong>Explanation:</strong> We do not need to make any changes because s is already valid.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;)&quot;, locked = &quot;0&quot;
+<strong>Output:</strong> false
+<strong>Explanation:</strong> locked permits us to change s[0]. 
+Changing s[0] to either &#39;(&#39; or &#39;)&#39; will not make s valid.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>n == s.length == locked.length</code></li>
+	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
+	<li><code>s[i]</code> is either <code>&#39;(&#39;</code> or <code>&#39;)&#39;</code>.</li>
+	<li><code>locked[i]</code> is either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
+</ul>

2221-check-if-a-parentheses-string-can-be-valid/solution.py

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+# Approach 1: Stack
+
+# Time: O(n)
+# Space: O(n)
+
+class Solution:
+    def canBeValid(self, s: str, locked: str) -> bool:
+        length = len(s)
+
+        # If length of string is odd, return false.
+        if length % 2 == 1:
+            return False
+
+        open_brackets = []
+        unlocked = []
+
+        # Iterate through the string to handle '(' and ')'
+        for i in range(length):
+            if locked[i] == '0':
+                unlocked.append(i)
+            elif s[i] == '(':
+                open_brackets.append(i)
+            elif s[i] == ')':
+                if open_brackets:
+                    open_brackets.pop()
+                elif unlocked:
+                    unlocked.pop()
+                else:
+                    return False
+
+        # Match remaining open brackets and the unlocked characters
+        while open_brackets and unlocked and open_brackets[-1] < unlocked[-1]:
+            open_brackets.pop()
+            unlocked.pop()
+
+        if open_brackets:
+            return False
+
+        return True
+