Clone a stack without extra space

Given a source stack, copy the contents of the source stack to destination stack maintaining the same order without using extra space.

Examples:

Input : Source:- |3|
                 |2|
                 |1|

Output : Destination:- |3|
                       |2|
                       |1|

Input : Source:- |a|
                 |b|
                 |c|

Output : Destination:- |a|
                       |b|
                       |c|

Approach:

In order to solve this without using extra space, we first reverse the source stack, then pop the top elements of the source stack one by one and push it into the destination stack. We follow the below steps to reverse the source stack: 

1. Initialize a variable count to 0.
2. Pop the top element from the source stack and store it in variable topVal.
3. Now pop the elements from the source stack and push them into the dest stack until the length of the source stack is equal to count.
4. Push topVal into the source stack and then pop all the elements in the dest stack and push them into the source stack.
5. Increment the value of count.
6. If count is not equal to length of source stack – 1, repeat the process from step-2.

Below is the implementation of the above approach:

Python3

# Python3 program to copy the contents from source stack
# into destination stack without using extra space

# Define a class for Stack
class Stack(object):

    def __init__(self):
        self.stack = []

    def push(self, value):
        self.stack.append(value)

    def pop(self):
        return self.stack.pop()

    def length(self):
        return len(self.stack)

    def display(self):
        for i in range(len(self.stack)-1, -1, -1):
            print(self.stack[i])

        print()

source = Stack()     # Source Stack
dest = Stack()         # Destination Stack

source.push(1)
source.push(2)
source.push(3)

print("Source Stack:")
source.display()

count = 0

# Reverse the order of the values in source stack
while count != source.length() - 1:

    topVal = source.pop()
    while count != source.length():
        dest.push(source.pop())

    source.push(topVal)
    while dest.length() != 0:
        source.push(dest.pop())

    count += 1

# Pop the values from source and push into destination stack
while source.length() != 0:
    dest.push(source.pop())

print("Destination Stack:")
dest.display()

Output

Source Stack:
3
2
1

Destination Stack:
3
2
1

Complexity Analysis:

• Time Complexity: O(n²)
• Auxiliary Space: O(n)

Efficient Approach: A better approach would be to represent the stack as a linked list. Reverse the source stack in the same way we reverse a linked list, pop the top elements of the source stack one by one and push it into the destination stack.

Below is the implementation of the above approach:

Python3

# Python3 program to copy the contents from source stack
# into destination stack without using extra space
# in linear time using Linked List

class StackNode(object):

    def __init__(self, data):
        self.data = data
        self.next = None

# Class for Stack to represent it as Linked list
class Stack(object):

    def __init__(self):
        self.top = None

    def push(self, value):

        newVal = StackNode(value)

        if self.top == None:
            self.top = newVal

        else:
            newVal.next = self.top
            self.top = newVal

    def pop(self):

        val = self.top.data
        self.top = self.top.next
        return val

    def display(self):

        current = self.top
        while current != None:
            print(current.data)
            current = current.next

        print()

    def reverse(self):

        current, temp, prev = self.top, None, None

        while current != None:
            temp = current.next
            current.next = prev
            prev = current
            current = temp

        self.top = prev

    def isEmpty(self):
        return self.top == None

source = Stack()        # Source Stack
dest = Stack()          # Destination Stack

source.push(1)
source.push(2)
source.push(3)

print("Source Stack:")
source.display()

source.reverse()

# Pop the values from source and push into destination stack
while source.isEmpty() != True:
    dest.push(source.pop())

print("Destination Stack:")
dest.display()

Output

Source Stack:
3
2
1

Destination Stack:
3
2
1

Complexity Analysis:

• Time Complexity: O(n)
• Auxiliary Space: O(n)