Problem: Given two integer sequences, one of which is the push sequence of a stack, please check whether the other sequence is a corresponding pop sequence or not.
For example, if 1, 2, 3, 4, 5 is a push sequence, 4, 5, 3, 2, 1 is a corresponding pop sequence, but the sequence 4, 3, 5, 1, 2 is not.
Analysis: An intuitive thought on this problem is to create an auxiliary stack. We push the numbers in the first sequence one by one, and try to pop them out according to the order in the second sequence.
Take the sequence 4, 5, 3, 2, 1 as an example to analyze the process to push and pop. The first number to be popped is 4, so we need to push it into a stack. The pushing order is defined in the first sequence, where there are numbers 1, 2 and 3 prior to 4. Therefore, numbers 1, 2, and 3 are pushed into a stack before 4 is pushed. At this time, there are 4 numbers in a stack, which are 1, 2, 3 and 4, with 4 on top. When 4 is popped, numbers 1, 2 and 3 are left. The next number to be popped is 5, which is not on top of stack, so we have to push numbers in the first sequence into stack until 5 is pushed. When number 5 is on top of a stack, we can pop it. The next three numbers to be popped are 3, 2 and 1. Since these numbers are on top of a stack before pop operations, they can be popped directly. The whole process to push and pop is summarized in Table 1.
Step

Operation

Stack Status

Popped

Step

Operation

Stack Status

Popped

1

Push 1

1

6

Push 5

1, 2, 3, 5
 
2

Push 2

1, 2

7

Pop

1, 2, 3

5
 
3

Push 3

1, 2, 3

8

Pop

1, 2

3
 
4

Push 4

1, 2, 3, 4

9

Pop

1

2
 
5

Pop

1, 2, 3

4

10

Pop

1

Table 1: The process to push and pop with a push sequence 1, 2, 3, 4, 5 and pop sequence 4, 5, 3, 2, 1
Let us continue to analyze another pop sequence 4, 3, 5, 1, 2. The process to pop the first number 4 is similar to the process above. After the number 4 is popped, 3 is on the top of stack and it can be popped. The next number to be popped is 5. Since it is not on top, we have to push numbers in the first sequence until the number 5 is pushed. The number 5 can be popped when it is pushed onto the top of a stack. After 5 is popped out, there are only two numbers 1 and 2 left in stack. The next number to be popped is 1, but it is not on the top of stack. We have to push numbers in the first sequence until 1 is pushed. However, all numbers in the first sequence have been pushed. Therefore, the sequence 4, 3, 5, 1, 2 is not a pop sequence of the stack with push sequence 1, 2, 3, 4, 5. The whole process to push and pop is summarized in Table 2.
Step

Operation

Stack Status

Popped

Step

Operation

Stack Status

Popped

1

Push 1

1

6

Pop

1, 2

3
 
2

Push 2

1, 2

7

Push 5

1, 2, 5
 
3

Push 3

1, 2, 3

8

Pop

1, 2

5
 
4

Push 4

1, 2, 3, 4

The next number to be popped is 1, which is neither on the top of stack, nor in the remaining numbers of push sequence.
 
5

Pop

1, 2, 3

4

Table 1: The process to push and pop with a push sequence 1, 2, 3, 4, 5 and pop sequence 4, 3, 5, 1, 2
According to the analysis above, we get a solution to check whether a sequence is a pop sequence of a stack or not. If the number to be popped is currently on top of stack, just pop it. If it is not on the top of stack, we have to push remaining numbers into the auxiliary stack until we meet the number to be popped. If the next number to be popped is not remaining in the push sequence, it is not a pop sequence of a stack. The following is some sample code based on this solution:
bool IsPopOrder(const int* pPush, const int* pPop, int nLength)
{
bool bPossible = false;
if(pPush != NULL && pPop != NULL && nLength > 0)
{
const int* pNextPush = pPush;
const int* pNextPop = pPop;
std::stack<int> stackData;
while(pNextPop  pPop < nLength)
{
// When the number to be popped is not on top of stack,
// push some numbers in the push sequence into stack
while(stackData.empty()  stackData.top() != *pNextPop)
{
// If all numbers have been pushed, break
if(pNextPush  pPush == nLength)
break;
stackData.push(*pNextPush);
pNextPush ++;
}
if(stackData.top() != *pNextPop)
break;
stackData.pop();
pNextPop ++;
}
if(stackData.empty() && pNextPop  pPop == nLength)
bPossible = true;
}
return bPossible;
}
The discussion about this problem is included in my book <Coding Interviews: Questions, Analysis & Solutions>, with some revisions. You may find the details of this book on Amazon.com, or Apress.
The author Harry He owns all the rights of this post. If you are going to use part of or the whole of this ariticle in your blog or webpages, please add a reference to http://codercareer.blogspot.com/. If you are going to use it in your books, please contact me (zhedahht@gmail.com) . Thanks.
another method using divid and conquer with constant space and worst case O(nlogn) time (I guess O(n) expected time but I didn't prove it):
ReplyDeletepush: 1, 2, 3, 4, 5
pop: 4, 5, 3, 2, 1
without loss of generality, let push seq is sorted, do a binary search of first element of pop from push sequence, in this case search for 4, then we have two subproblems which are:
push: 123
pop: 321
and
push: 5
pop:5
Your solution does not work on push={1,4,3,4,5} and pop={4,5,3,4,1};
ReplyDeleteProve: http://ideone.com/SQtdV
This comment has been removed by the author.
DeleteThis algorithm wont give good results if sequence has duplicate numbers.
ReplyDeleteworks only in case of unique numbers
ReplyDeleteHi here's my solution
ReplyDeleteThe concept comes from that the inorder pop sequence means we need to start from some interior point of push array and then sequentially move right(push new number) or left(pop current stack) and clean the push array until we reach the two end of the push array. So, just reverse the procedure and start from the tail of pop sequence, we could have the following code, which could handle duplicate numbers
bool IsPopOrder(const int* pPush, const int* pPop, int nLength)
{
int s = 0;
int e = nLength1;
for(int i=nLength1;i>=0;i)
{
if(pPop[i]==pPush[e])
e;
else if(pPop[i]==pPush[s])
s++;
else
return false;
}
return true;
}
The solution doesn't work when sequence has duplicated values
ReplyDelete服了老gay
ReplyDeletecan you extend this for duplicate elements?
ReplyDelete