maximum-frequency-stack

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Algorithms

Design a stack-like data structure to push elements to the stack and pop the most frequent element from the stack.

Implement the <code>FreqStack</code> class:

<ul>
	<li><code>FreqStack()</code> constructs an empty frequency stack.</li>
	<li><code>void push(int val)</code> pushes an integer <code>val</code> onto the top of the stack.</li>
	<li><code>int pop()</code> removes and returns the most frequent element in the stack.
	<ul>
		<li>If there is a tie for the most frequent element, the element closest to the stack&#39;s top is removed and returned.</li>
	</ul>
	</li>
</ul>

&nbsp;
Example 1:

<pre>
Input
[&quot;FreqStack&quot;, &quot;push&quot;, &quot;push&quot;, &quot;push&quot;, &quot;push&quot;, &quot;push&quot;, &quot;push&quot;, &quot;pop&quot;, &quot;pop&quot;, &quot;pop&quot;, &quot;pop&quot;]
[[], [5], [7], [5], [7], [4], [5], [], [], [], []]
Output
[null, null, null, null, null, null, null, 5, 7, 5, 4]

Explanation
FreqStack freqStack = new FreqStack();
freqStack.push(5); // The stack is [5]
freqStack.push(7); // The stack is [5,7]
freqStack.push(5); // The stack is [5,7,5]
freqStack.push(7); // The stack is [5,7,5,7]
freqStack.push(4); // The stack is [5,7,5,7,4]
freqStack.push(5); // The stack is [5,7,5,7,4,5]
freqStack.pop(); // return 5, as 5 is the most frequent. The stack becomes [5,7,5,7,4].
freqStack.pop(); // return 7, as 5 and 7 is the most frequent, but 7 is closest to the top. The stack becomes [5,7,5,4].
freqStack.pop(); // return 5, as 5 is the most frequent. The stack becomes [5,7,4].
freqStack.pop(); // return 4, as 4, 5 and 7 is the most frequent, but 4 is closest to the top. The stack becomes [5,7].
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>0 &lt;= val &lt;= 109</code></li>
	<li>At most <code>2 * 104</code> calls will be made to <code>push</code> and <code>pop</code>.</li>
	<li>It is guaranteed that there will be at least one element in the stack before calling <code>pop</code>.</li>
</ul>