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Author: koralkulacoglu

LeetCode/1520-number-of-steps-to-reduce-a-number-in-binary-representation-to-one/README.md

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+<p>Given the binary representation of an integer as a string <code>s</code>, return <em>the number of steps to reduce it to </em><code>1</code><em> under the following rules</em>:</p>
+
+<ul>
+	<li>
+	<p>If the current number is even, you have to divide it by <code>2</code>.</p>
+	</li>
+	<li>
+	<p>If the current number is odd, you have to add <code>1</code> to it.</p>
+	</li>
+</ul>
+
+<p>It is guaranteed that you can always reach one for all test cases.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;1101&quot;
+<strong>Output:</strong> 6
+<strong>Explanation:</strong> &quot;1101&quot; corressponds to number 13 in their decimal representation.
+Step 1) 13 is odd, add 1 and obtain 14.&nbsp;
+Step 2) 14 is even, divide by 2 and obtain 7.
+Step 3) 7 is odd, add 1 and obtain 8.
+Step 4) 8 is even, divide by 2 and obtain 4.&nbsp; 
+Step 5) 4 is even, divide by 2 and obtain 2.&nbsp;
+Step 6) 2 is even, divide by 2 and obtain 1.&nbsp; 
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;10&quot;
+<strong>Output:</strong> 1
+<strong>Explanation:</strong> &quot;10&quot; corresponds to number 2 in their decimal representation.
+Step 1) 2 is even, divide by 2 and obtain 1.&nbsp; 
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;1&quot;
+<strong>Output:</strong> 0
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length&nbsp;&lt;= 500</code></li>
+	<li><code>s</code> consists of characters &#39;0&#39; or &#39;1&#39;</li>
+	<li><code>s[0] == &#39;1&#39;</code></li>
+</ul>

LeetCode/1520-number-of-steps-to-reduce-a-number-in-binary-representation-to-one/solution.py

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+class Solution:
+    def numSteps(self, s: str) -> int:
+        ans = 0
+        n = int(s, 2)
+        while n > 1:
+            if n & 1:
+                n += 1
+            else:
+                n >>= 1
+            ans += 1
+        return ans