Sync LeetCode submission Runtime - 15 ms (91.77%), Memory - 19 MB (76.88%)

Author: jimit105

1903-design-most-recently-used-queue/README.md

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+<p>Design a queue-like data structure that moves the most recently used element to the end of the queue.</p>
+
+<p>Implement the <code>MRUQueue</code> class:</p>
+
+<ul>
+	<li><code>MRUQueue(int n)</code> constructs the <code>MRUQueue</code> with <code>n</code> elements: <code>[1,2,3,...,n]</code>.</li>
+	<li><code>int fetch(int k)</code> moves the <code>k<sup>th</sup></code> element <strong>(1-indexed)</strong> to the end of the queue and returns it.</li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong>
+[&quot;MRUQueue&quot;, &quot;fetch&quot;, &quot;fetch&quot;, &quot;fetch&quot;, &quot;fetch&quot;]
+[[8], [3], [5], [2], [8]]
+<strong>Output:</strong>
+[null, 3, 6, 2, 2]
+
+<strong>Explanation:</strong>
+MRUQueue mRUQueue = new MRUQueue(8); // Initializes the queue to [1,2,3,4,5,6,7,8].
+mRUQueue.fetch(3); // Moves the 3<sup>rd</sup> element (3) to the end of the queue to become [1,2,4,5,6,7,8,3] and returns it.
+mRUQueue.fetch(5); // Moves the 5<sup>th</sup> element (6) to the end of the queue to become [1,2,4,5,7,8,3,6] and returns it.
+mRUQueue.fetch(2); // Moves the 2<sup>nd</sup> element (2) to the end of the queue to become [1,4,5,7,8,3,6,2] and returns it.
+mRUQueue.fetch(8); // The 8<sup>th</sup> element (2) is already at the end of the queue so just return it.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 2000</code></li>
+	<li><code>1 &lt;= k &lt;= n</code></li>
+	<li>At most <code>2000</code> calls will be made to <code>fetch</code>.</li>
+</ul>
+
+<p>&nbsp;</p>
+<strong>Follow up:</strong> Finding an <code>O(n)</code> algorithm per <code>fetch</code> is a bit easy. Can you find an algorithm with a better complexity for each <code>fetch</code> call?

1903-design-most-recently-used-queue/solution.py

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+# Time:
+# Constructor: O(n)
+# fetch: O(n)
+
+class MRUQueue:
+
+    def __init__(self, n: int):
+        self.queue = list(range(1, n + 1))
+        
+
+    def fetch(self, k: int) -> int:
+        element = self.queue[k - 1]
+        self.queue.pop(k - 1)
+        self.queue.append(element)
+        return element
+
+        
+
+
+# Your MRUQueue object will be instantiated and called as such:
+# obj = MRUQueue(n)
+# param_1 = obj.fetch(k)