Chapter 三 Stacks and Queues - 3.3

Chapter 3 Stacks and Queues - 3.3
Problem 3.3: Imagine a (literal) stack of plates. If the stack gets too high, it might topple. Therefore, in real life, we would likely start a new stack when the previous stack exceeds some threshold. Implement a data structure SetOfStacks that mimics this. SetOfStacks should be composed of several stacks, and should create a new stack once the previous one exceeds capacity. SetOfStacks.push() and SetOfStacks.pop() should behave identically to a single stack (that is, pop() should return the same values as it would if there were just a single stack).
FOLLOW UP
Implement a function popAt(int index) which performs a pop operation on a specific sub-stack.

It is a quite easy one. However, the requirement of follow up question is not clear and we should discuss details with interviewers.
from stack import *MAX_CAPACITY = 3class SetOfStacks: def __init__(self): self.stacks = [stack()] self.last = 0 def push(self, value): if self.stacks[self.last].size() >= MAX_CAPACITY: self.stacks.append(stack()) self.last = self.last + 1 self.stacks[self.last].push(value) def pop(self): if self.stacks[self.last].size() <= 0: if len(self.stacks) <= 1: return False del self.stacks[self.last] self.last = self.last - 1 return self.stacks[self.last].pop() def popAt(self, index): if (index < 0) or (index > self.last): return False to_return = self.stacks[index].pop() if self.stacks[index].size() <= 0: del self.stacks[index] self.last = self.last - 1 return to_return# Test casesif __name__ == "__main__": stacks_set = SetOfStacks() for i in range(0, 10): print "push", i stacks_set.push(i) for i in range(0, 3): print "pop", stacks_set.pop() for i in range(0, 5): print "push", i stacks_set.push(i) for i in range(0, 13): print "popAt(0)", stacks_set.popAt(0)