Clone a stack without extra space

Difficulty Level : Medium
Last Updated : 03 Nov, 2018

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:

Initialize a variable count to 0.
Pop the top element from the source stack and store it in variable topVal.
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.
Push topVal ino the source stack and then pop all the elements in the dest stack and push them into the source stack.
Increment the value of count.
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 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

Time Complexity: $O(n^2)$

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 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

Time Complexity: $O(n)$

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