Sync LeetCode submission Runtime - 47 ms (9.05%), Memory - 16.5 MB (31.46%)

Author: jimit105

3055-maximum-odd-binary-number/README.md

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+<p>You are given a <strong>binary</strong> string <code>s</code> that contains at least one <code>&#39;1&#39;</code>.</p>
+
+<p>You have to <strong>rearrange</strong> the bits in such a way that the resulting binary number is the <strong>maximum odd binary number</strong> that can be created from this combination.</p>
+
+<p>Return <em>a string representing the maximum odd binary number that can be created from the given combination.</em></p>
+
+<p><strong>Note </strong>that the resulting string <strong>can</strong> have leading zeros.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;010&quot;
+<strong>Output:</strong> &quot;001&quot;
+<strong>Explanation:</strong> Because there is just one &#39;1&#39;, it must be in the last position. So the answer is &quot;001&quot;.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;0101&quot;
+<strong>Output:</strong> &quot;1001&quot;
+<strong>Explanation: </strong>One of the &#39;1&#39;s must be in the last position. The maximum number that can be made with the remaining digits is &quot;100&quot;. So the answer is &quot;1001&quot;.
+</pre>
+
+<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 <code>&#39;0&#39;</code> and <code>&#39;1&#39;</code>.</li>
+	<li><code>s</code> contains at least one <code>&#39;1&#39;</code>.</li>
+</ul>

3055-maximum-odd-binary-number/solution.py

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+# Approach 2: Greedy Bit Manipulation (Counting Ones)
+# Time: O(n)
+# Space: O(n)
+
+class Solution:
+    def maximumOddBinaryNumber(self, s: str) -> str:
+        n = len(s)
+        ones_cnt = s.count('1')
+
+        return '1' * (ones_cnt - 1) + '0' * (n - ones_cnt) + '1'
+