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Reversing a Stack with the help of another empty Stack
  • Difficulty Level : Easy
  • Last Updated : 05 May, 2021

Given a Stack consisting of N elements, the task is to reverse the Stack using an extra stack.

Examples:

Input: stack = {1, 2, 3, 4, 5} 
Output: 





Explanation: 
Input Stack: 





Reversed Stack: 




5

Input: stack = {1, 3, 5, 4, 2} 
Output: 




2

Approach: Follow the steps below to solve the problem:



Below is the implementation of the above approach.

C++




// C++ program to reverse a stack
// by using an extra stack
#include <bits/stdc++.h>
using namespace std;
 
// Function to transfer elements of
// the stack s1 to the stack s2
void transfer(stack<int>& s1,
              stack<int>& s2, int n)
{
    for (int i = 0; i < n; i++) {
 
        // Store the top element
        // in a temporary variable
        int temp = s1.top();
 
        // Pop out of the stack
        s1.pop();
 
        // Push it into s2
        s2.push(temp);
    }
}
 
// Function to reverse a stack using another stack
void reverse_stack_by_using_extra_stack(stack<int>& s,
                                        int n)
{
    stack<int> s2;
 
    for (int i = 0; i < n; i++) {
 
        // Store the top element
        // of the given stack
        int x = s.top();
 
        // Pop that element
        // out of the stack
        s.pop();
 
        transfer(s, s2, n - i - 1);
        s.push(x);
        transfer(s2, s, n - i - 1);
    }
}
 
// Driver Code
int main()
{
    int n = 5;
 
    stack<int> s;
    s.push(1);
    s.push(2);
    s.push(3);
    s.push(4);
    s.push(5);
 
    reverse_stack_by_using_extra_stack(s, n);
 
    for (int i = 0; i < n; i++) {
        cout << s.top() << " ";
        s.pop();
    }
    return 0;
}

Python3




# Python3 program to reverse a stack
# by using an extra stack
 
# Function to transfer elements of
# the stack s1 to the stack s2
def transfer(s1, s2, n):
     
    for i in range(n):
 
        # Store the top element
        # in a temporary variable
        temp = s1[-1]
 
        # Pop out of the stack
        s1.pop()
 
        # Push it into s2
        s2.append(temp)
 
# Function to reverse a stack using another stack
def reverse_stack_by_using_extra_stack(s, n):
 
    s2 = []
     
    for i in range(n):
 
        # Store the top element
        # of the given stack
        x = s[-1]
 
        # Pop that element
        # out of the stack
        s.pop()
 
        transfer(s, s2, n - i - 1)
        s.append(x)
        transfer(s2, s, n - i - 1)
 
# Driver Code
if __name__ == "__main__":
 
    n = 5
 
    s = []
    s.append(1)
    s.append(2)
    s.append(3)
    s.append(4)
    s.append(5)
 
    reverse_stack_by_using_extra_stack(s, n)
 
    for i in range(n):
        print(s[-1], end = " ")
        s.pop()
         
# This code is contributed by ukasp
Output: 
1 2 3 4 5

 

Time Complexity: O(N2
Auxiliary Space: O(N)

 

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