Add two numbers represented by Stacks

Given two numbers N1 and N2 represented by two stacks, such that their most significant digits are present at the bottom of the stack, the task is to calculate and return the sum of the two numbers in the form of a stack.

Examples: 

Input: N1={5, 8, 7, 4}, N2={2, 1, 3} 
Output: {6, 0, 8, 7}
Explanation:
Step 1: Popped element from N1(= 4) +  Popped element from N2(= 3) = {7} and rem=0.
Step 2: Popped element from N1(= 7) +  Popped element from N2(= 1) = {7, 8} and rem=0.
Step 3: Popped element from N1(= 8) +  Popped element from N2(= 2) = {7, 8, 0} and rem=1.
Step 4: Popped element from N1(= 5) = {7, 8, 0, 6}
On reverse the stack, the desired arrangement {6,0,8,7} is obtained.

Input: N1={6,4,9,5,7}, N2={213} 
Output:{6, 5, 0, 0, 5}

Approach: The problem can be solved using the concept of Add two numbers represented by linked lists. Follow the below steps to solve the problem.



  1. Create a new stack, res to store the sum of the two stacks.
  2. Initialize variables rem and sum to store the carry generated and the sum of top elements respectively.
  3. Keep popping the top elements of both the stacks and push the sum % 10 to res and update rem as sum/10.
  4. Repeat the above step until the stacks are empty. If rem is greater than 0, insert rem into the stack.
  5. Reverse the res stack so that the most significant digit present at the bottom of the res stack.

Below is the implementation of the resultant approach:

C++14

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// C++ program to implement 
// the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the stack that
// contains the sum of two numbers
stack<int> addStack(stack<int> N1, 
                    stack<int> N2)
{
    stack<int> res;
    int sum = 0, rem = 0;
  
    while (!N1.empty() and !N2.empty()) {
        
      // Calculate the sum of the top
      // elements of both the stacks
        sum = (rem + N1.top() + N2.top());
        
      // Push the sum into the stack
        res.push(sum % 10);
        
      // Store the carry
        rem = sum / 10;
        
      // Pop the top elements
        N1.pop();
        N2.pop();
    }
    
    // If N1 is not empty
    while (!N1.empty()) {
        sum = (rem + N1.top());
        res.push(sum % 10);
        rem = sum / 10;
        N1.pop();
    }
    // If N2 is not empty
    while (!N2.empty()) {
        sum = (rem + N2.top());
        res.push(sum % 10);
        rem = sum / 10;
        N2.pop();
    }
  
    // If carry remains
    while (rem > 0) {
        res.push(rem);
        rem /= 10;
    }
  
    // Reverse the stack.so that
    // most significant digit is
    // at the bottom of the stack
    while (!res.empty()) {
        N1.push(res.top());
        res.pop();
    }
    res = N1;
    return res;
}
  
// Function to display the 
// resultamt stack
void display(stack<int>& res)
{
    int N = res.size();
    string s = "";
    while (!res.empty()) {
        s = to_string(res.top()) + s;
        res.pop();
    }
      
  cout << s << endl;
}
  
// Driver Code
int main()
{
    stack<int> N1;
    N1.push(5);
    N1.push(8);
    N1.push(7);
    N1.push(4);
  
    stack<int> N2;
    N2.push(2);
    N2.push(1);
    N2.push(3);
  
    stack<int> res = addStack(N1, N2);
  
    display(res);
    
    return 0;
}

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Java

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// Java program to implement
// the above approach
import java.util.Stack;
class GFG{
  
    // Function to return the stack that
    // contains the sum of two numbers
    static Stack<Integer> addStack(Stack<Integer> N1,
                                   Stack<Integer> N2)
    {
        Stack<Integer> res = new Stack<Integer>();
        int sum = 0, rem = 0;
        while (!N1.isEmpty() && !N2.isEmpty()) 
        {
  
            // Calculate the sum of the top
            // elements of both the stacks
            sum = (rem + N1.peek() + N2.peek());
  
            // Push the sum into the stack
            res.add(sum % 10);
  
            // Store the carry
            rem = sum / 10;
  
            // Pop the top elements
            N1.pop();
            N2.pop();
        }
  
        // If N1 is not empty
        while (!N1.isEmpty()) 
        {
            sum = (rem + N1.peek());
            res.add(sum % 10);
            rem = sum / 10;
            N1.pop();
        }
          
          // If N2 is not empty
        while (!N2.isEmpty()) 
        {
            sum = (rem + N2.peek());
            res.add(sum % 10);
            rem = sum / 10;
            N2.pop();
        }
  
        // If carry remains
        while (rem > 0
        {
            res.add(rem);
            rem /= 10;
        }
  
        // Reverse the stack.so that
        // most significant digit is
        // at the bottom of the stack
        while (!res.isEmpty()) 
        {
            N1.add(res.peek());
            res.pop();
        }
        res = N1;
        return res;
    }
  
    // Function to display the
    // resultamt stack
    static void display(Stack<Integer> res)
    {
        int N = res.size();
        String s = "";
        while (!res.isEmpty()) 
        {
            s = String.valueOf(res.peek()) + s;
            res.pop();
        }
  
        System.out.print(s + "\n");
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        Stack<Integer> N1 = new Stack<Integer>();
        N1.add(5);
        N1.add(8);
        N1.add(7);
        N1.add(4);
          Stack<Integer> N2 = new Stack<Integer>();
        N2.add(2);
        N2.add(1);
        N2.add(3);
          Stack<Integer> res = addStack(N1, N2);
          display(res);
    }
}
  
// This code is contributed by shikhasingrajput

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Python3

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# Python3 program to implement
# the above approach
  
# Function to return the stack that 
# contains the sum of two numbers
def addStack(N1, N2):
  
    res = []
    s = 0
    rem = 0
  
    while (len(N1) != 0 and len(N2) != 0):
  
        # Calculate the sum of the top
        # elements of both the stacks
        s = (rem + N1[-1] + N2[-1])
  
        # Push the sum into the stack
        res.append(s % 10)
  
        # Store the carry
        rem = s // 10
  
        # Pop the top elements
        N1.pop(-1)
        N2.pop(-1)
  
    # If N1 is not empty
    while(len(N1) != 0):
        s = rem + N1[-1]
        res.append(s % 10)
        rem = s // 10
        N1.pop(-1)
  
    # If N2 is not empty
    while(len(N2) != 0):
        s = rem + N2[-1]
        res.append(s % 10)
        rem = s // 10
        N2.pop(-1)
  
    # If carry remains
    while(rem > 0):
        res.append(rem)
        rem //= 10
  
    # Reverse the stack.so that
    # most significant digit is
    # at the bottom of the stack
    res = res[::-1]
  
    return res
  
# Function to display the
# resultamt stack
def display(res):
  
    s = ""
    for i in res:
        s += str(i)
  
    print(s)
  
# Driver Code
N1 = []
N1.append(5)
N1.append(8)
N1.append(7)
N1.append(4)
  
N2 = []
N2.append(2)
N2.append(1)
N2.append(3)
  
# Function call
res = addStack(N1, N2)
  
display(res)
  
# This code is contributed by Shivam Singh

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C#

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// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
  
class GFG{
  
// Function to return the stack that
// contains the sum of two numbers
static Stack<int> PushStack(Stack<int> N1,
                            Stack<int> N2)
{
    Stack<int> res = new Stack<int>();
    int sum = 0, rem = 0;
      
    while (N1.Count != 0 && N2.Count != 0) 
    {
  
        // Calculate the sum of the top
        // elements of both the stacks
        sum = (rem + N1.Peek() + N2.Peek());
  
        // Push the sum into the stack
        res.Push((int)sum % 10);
  
        // Store the carry
        rem = sum / 10;
  
        // Pop the top elements
        N1.Pop();
        N2.Pop();
    }
  
    // If N1 is not empty
    while (N1.Count != 0) 
    {
        sum = (rem + N1.Peek());
        res.Push(sum % 10);
        rem = sum / 10;
        N1.Pop();
    }
      
    // If N2 is not empty
    while (N2.Count != 0) 
    {
        sum = (rem + N2.Peek());
        res.Push(sum % 10);
        rem = sum / 10;
        N2.Pop();
    }
  
    // If carry remains
    while (rem > 0) 
    {
        res.Push(rem);
        rem /= 10;
    }
  
    // Reverse the stack.so that
    // most significant digit is
    // at the bottom of the stack
    while (res.Count != 0) 
    {
        N1.Push(res.Peek());
        res.Pop();
    }
      
    res = N1;
    return res;
}
  
// Function to display the
// resultamt stack
static void display(Stack<int> res)
{
    int N = res.Count;
    String s = "";
      
    while (res.Count != 0) 
    {
        s = String.Join("", res.Peek()) + s;
        res.Pop();
    }
  
    Console.Write(s + "\n");
}
  
// Driver Code
public static void Main(String[] args)
{
    Stack<int> N1 = new Stack<int>();
    N1.Push(5);
    N1.Push(8);
    N1.Push(7);
    N1.Push(4);
      
    Stack<int> N2 = new Stack<int>();
    N2.Push(2);
    N2.Push(1);
    N2.Push(3);
      
    Stack<int> res = PushStack(N1, N2);
      
    display(res);
}
}
  
// This code is contributed by Amit Katiyar

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Output: 

6087

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

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