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Interview Question for Senior Software Engineer at Google:
Jul 13, 2009
1 of
1 people found this helpful
Create a stack of numbers where the maximum number is always known.
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Answers & Comments (6)
1 of 1 people found this helpful
Jul 31, 2009
maintain a sorted stack, the insert would look like
insert(int p,stack s){
stack large;
int top = s.peek();
while(top>s){
large.push(s.pop());
top = s.peek();
}
s.push(p);
while(!large.empty()){
s.push(large.pop());
}
}
insert(int p,stack s){
stack large;
int top = s.peek();
while(top>s){
large.push(s.pop());
top = s.peek();
}
s.push(p);
while(!large.empty()){
s.push(large.pop());
}
}
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1 of 1 people found this helpful
Jul 31, 2009
sorry, typo
while(top>p)
while(top>p)
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Aug 21, 2009
I was asked of this question as well. I not sure how to implment a second stack in a way that the max number is always on top.
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Aug 21, 2009
I was asked of this question as well. I not sure how to implment a second stack in a way that the max number is always on top.
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5 of 5 people found this helpful
Aug 21, 2009
To Job Seeker:
The basic idea is that when a new number is being pushed onto the stack, you need to see if that number is greater than or equal to the current top of the maximums-stack. If it is, you push the number onto maximums-stack as well.
Also, when pop is called, you look to see if the number being popped is equal to the number on the top of the maximums-stack. If it it, pop that stack as well.
The basic idea is that when a new number is being pushed onto the stack, you need to see if that number is greater than or equal to the current top of the maximums-stack. If it is, you push the number onto maximums-stack as well.
Also, when pop is called, you look to see if the number being popped is equal to the number on the top of the maximums-stack. If it it, pop that stack as well.
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