Given two numbers N1and 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.
- Create a new stack, res to store the sum of the two stacks.
- Initialize variables rem and sum to store the carry generated and the sum of top elements respectively.
- Keep popping the top elements of both the stacks and push the sum % 10 to res and update rem as sum/10.
- Repeat the above step until the stacks are empty. If rem is greater than 0, insert rem into the stack.
- 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++ 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;
} |
// 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 |
# 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# 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 |
<script> // JavaScript program to implement // the above approach // Function to return the stack that // contains the sum of two numbers function addStack(N1, N2)
{ var res = [];
var sum = 0, rem = 0;
while (N1.length!=0 && N2.length!=0) {
// Calculate the sum of the top
// elements of both the stacks
sum = (rem + N1[N1.length-1] +N2[N2.length-1]);
// Push the sum into the stack
res.push(sum % 10);
// Store the carry
rem = parseInt(sum / 10);
// Pop the top elements
N1.pop();
N2.pop();
}
// If N1 is not empty
while (N1.length!=0) {
sum = (rem + N1[N1.length-1]);
res.push(sum % 10);
rem = parseInt(sum / 10);
N1.pop();
}
// If N2 is not empty
while (N2.length!=0) {
sum = (rem +N2[N2.length-1]);
res.push(sum % 10);
rem = parseInt(sum / 10);
N2.pop();
}
// If carry remains
while (rem > 0) {
res.push(rem);
rem = parseInt(rem/10);
}
// Reverse the stack.so that
// most significant digit is
// at the bottom of the stack
while (res.length!=0) {
N1.push(res[res.length-1]);
res.pop();
}
res = N1;
return res;
} // Function to display the // resultamt stack function display(res)
{ var N = res.length;
var s = "" ;
while (res.length!=0) {
s = (res[res.length-1].toString()) + s;
res.pop();
}
document.write( s );
} // Driver Code var N1 = [];
N1.push(5); N1.push(8); N1.push(7); N1.push(4); var N2 = [];
N2.push(2); N2.push(1); N2.push(3); var res = addStack(N1, N2);
display(res); </script> |
6087
Time Complexity: O(N)
Auxiliary Space: O(N)