# 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

 `// C++ program to implement  ` `// the above approach ` `#include ` `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; ` `}`

## Java

 `// 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 addStack(Stack N1, ` `                                   ``Stack N2) ` `    ``{ ` `        ``Stack res = ``new` `Stack(); ` `        ``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 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 N1 = ``new` `Stack(); ` `        ``N1.add(``5``); ` `        ``N1.add(``8``); ` `        ``N1.add(``7``); ` `        ``N1.add(``4``); ` `          ``Stack N2 = ``new` `Stack(); ` `        ``N2.add(``2``); ` `        ``N2.add(``1``); ` `        ``N2.add(``3``); ` `          ``Stack res = addStack(N1, N2); ` `          ``display(res); ` `    ``} ` `} ` ` `  `// This code is contributed by shikhasingrajput`

## Python3

 `# 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`

## C#

 `// 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 `

Output:

```6087
```

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

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