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Use a queue which will have enqueue() and dequeue() operations.
- CoolGeek March 02, 2012Also, there would be a stack which would contain all max elements encountered till now.
Whenever an element is being enqueued, and it is greater than current max (top of stack), it will be pushed on stack also.
If a no. 'k' is current max and 'k' comes for enqueueing once again we will push 'k' once more on stack.
Whenever we dequeue an element, we check if it is same as element on top of stack. If it is same pop the element from stack as well.