Added README.md file for Implement Queue using Stacks

Author: mrprashantkumar

232-implement-queue-using-stacks/README.md

@@ -0,0 +1,48 @@
+<h2><a href="https://leetcode.com/problems/implement-queue-using-stacks">Implement Queue using Stacks</a></h2> <img src='https://img.shields.io/badge/Difficulty-Easy-brightgreen' alt='Difficulty: Easy' /><hr><p>Implement a first in first out (FIFO) queue using only two stacks. The implemented queue should support all the functions of a normal queue (<code>push</code>, <code>peek</code>, <code>pop</code>, and <code>empty</code>).</p>
+
+<p>Implement the <code>MyQueue</code> class:</p>
+
+<ul>
+	<li><code>void push(int x)</code> Pushes element x to the back of the queue.</li>
+	<li><code>int pop()</code> Removes the element from the front of the queue and returns it.</li>
+	<li><code>int peek()</code> Returns the element at the front of the queue.</li>
+	<li><code>boolean empty()</code> Returns <code>true</code> if the queue is empty, <code>false</code> otherwise.</li>
+</ul>
+
+<p><strong>Notes:</strong></p>
+
+<ul>
+	<li>You must use <strong>only</strong> standard operations of a stack, which means only <code>push to top</code>, <code>peek/pop from top</code>, <code>size</code>, and <code>is empty</code> operations are valid.</li>
+	<li>Depending on your language, the stack may not be supported natively. You may simulate a stack using a list or deque (double-ended queue) as long as you use only a stack&#39;s standard operations.</li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input</strong>
+[&quot;MyQueue&quot;, &quot;push&quot;, &quot;push&quot;, &quot;peek&quot;, &quot;pop&quot;, &quot;empty&quot;]
+[[], [1], [2], [], [], []]
+<strong>Output</strong>
+[null, null, null, 1, 1, false]
+
+<strong>Explanation</strong>
+MyQueue myQueue = new MyQueue();
+myQueue.push(1); // queue is: [1]
+myQueue.push(2); // queue is: [1, 2] (leftmost is front of the queue)
+myQueue.peek(); // return 1
+myQueue.pop(); // return 1, queue is [2]
+myQueue.empty(); // return false
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= x &lt;= 9</code></li>
+	<li>At most <code>100</code>&nbsp;calls will be made to <code>push</code>, <code>pop</code>, <code>peek</code>, and <code>empty</code>.</li>
+	<li>All the calls to <code>pop</code> and <code>peek</code> are valid.</li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong>Follow-up:</strong> Can you implement the queue such that each operation is <strong><a href="https://en.wikipedia.org/wiki/Amortized_analysis" target="_blank">amortized</a></strong> <code>O(1)</code> time complexity? In other words, performing <code>n</code> operations will take overall <code>O(n)</code> time even if one of those operations may take longer.</p>