Sum of two large numbers

Given two numbers as strings. The numbers may be very large (may not fit in long long int), the task is to find sum of these two numbers.

Examples:

Input  : str1 = "3333311111111111", 
         str2 =   "44422222221111"
Output : 3377733333332222

Input  : str1 = "7777555511111111", 
         str2 =    "3332222221111"
Output : 7780887733332222

Approach 1: Using BigInteger class

The simplest approach is use BigInteger class. BigInteger provides analogues to all of Java’s primitive integer operators, and all relevant methods from java.lang.Math. In this approach we will create 2 objects of BigInteger and pass string in it then add simply add those objects using add() method.

Java

import java.util.*;
import java.math.*;
public class GFG{

     public static void main(String []args){
        String str="7777555511111111";
        String str1="3332222221111";

        BigInteger a=new BigInteger(str); // creating obj of biginteger and pass str in it
        BigInteger b=new BigInteger(str1); // creating obj of biginteger and pass str1 in it

        System.out.println(a.add(b)); //add

     }
}

Python3

# Define the input strings
str = "7777555511111111"
str1 = "3332222221111"

# Create int objects with arbitrary precision
a = int(str)
b = int(str1)

# Add the int objects
result = a + b

# Print the result
print(result)

Javascript

// Define the input strings
let str = "7777555511111111";
let str1 = "3332222221111";

// Create BigInt objects with arbitrary precision
let a = BigInt(str);
let b = BigInt(str1);

// Add the BigInt objects
let result = a + b;

// Print the result
console.log(result.toString());

C++

#include <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <functional>
#include <iomanip>
#include <iostream>
#include <numeric>
#include <sstream>
#include <string>
#include <vector>

using namespace std;

// Define a struct to represent BigInteger
struct BigInteger {
    string str;

    // Constructor to initialize
    // BigInteger with a string
    BigInteger(string s) { str = s; }

    // Overload + operator to add
    // two BigInteger objects
    BigInteger operator+(const BigInteger& b)
    {
        string a = str;
        string c = b.str;
        int alen = a.length(), clen = c.length();
        int n = max(alen, clen);
        if (alen > clen)
            c.insert(0, alen - clen, '0');
        else if (alen < clen)
            a.insert(0, clen - alen, '0');
        string res(n + 1, '0');
        int carry = 0;
        for (int i = n - 1; i >= 0; i--) {
            int digit = (a[i] - '0') + (c[i] - '0') + carry;
            carry = digit / 10;
            res[i + 1] = digit % 10 + '0';
        }
        if (carry == 1) {
            res[0] = '1';
            return BigInteger(res);
        }
        else {
            return BigInteger(res.substr(1));
        }
    }

    // Overload << operator to output
    // BigInteger object
    friend ostream& operator<<(ostream& out,
                               const BigInteger& b)
    {
        out << b.str;
        return out;
    }
};

// Driver Code
int main()
{
    string str = "7777555511111111";
    string str1 = "3332222221111";

    // Create BigInteger objects
    // and add them
    BigInteger a(str);
    BigInteger b(str1);
    BigInteger sum = a + b;

    // Output the result
    cout << sum << endl;
    return 0;
}

C#

// C# code addition

using System;
using System.Collections.Generic;

// Implementing biginteger class.
public class BigInteger {
    private List<int> digits;

    // Constructor
    public BigInteger(string value) {
        digits = new List<int>();

        for (int i = value.Length - 1; i >= 0; i--) {
            digits.Add(int.Parse(value[i].ToString()));
        }
    }

    // Add function  in BigInteger.
    public static BigInteger Add(BigInteger a, BigInteger b) {
        List<int> sum = new List<int>();
        int carry = 0;
        int i = 0;

        while (i < a.digits.Count || i < b.digits.Count) {
            int x = i < a.digits.Count ? a.digits[i] : 0;
            int y = i < b.digits.Count ? b.digits[i] : 0;
            int s = x + y + carry;

            sum.Add(s % 10);
            carry = s / 10;
            i++;
        }

        if (carry > 0) {
            sum.Add(carry);
        }

        BigInteger result = new BigInteger("");

        for (int j = sum.Count - 1; j >= 0; j--) {
            result.digits.Add(sum[j]);
        }

        return result;
    }

    // An override function to convert them to string.
    public override string ToString() {
        string result = "";

        foreach (int digit in digits) {
            result += digit.ToString();
        }

        return result;
    }
}

public class GFG {
    public static void Main() {
        string str = "7777555511111111";
        string str1 = "3332222221111";

        BigInteger a = new BigInteger(str);// convert the string to a BigInteger
        BigInteger b = new BigInteger(str1);// convert the string to a BigInteger

        Console.WriteLine(BigInteger.Add(a, b)); // add the two BigIntegers and print the result
    }
}

// The code is  contributed by Nidhi goel.

Output

7780887733332222

Time Complexity: O(1)
Auxiliary Space: O(1)

Approach 2: Iterative approach

The idea is based on school mathematics. We traverse both strings from end, one by one add digits and keep track of carry. To simplify the process, we do following: 
1) Reverse both strings. 
2) Keep adding digits one by one from 0’th index (in reversed strings) to end of smaller string, append the sum % 10 to end of result and keep track of carry as sum/10. 
3) Finally reverse the result. 

C++

// C++ program to find sum of two large numbers.
#include<bits/stdc++.h>
using namespace std;

// Function for finding sum of larger numbers
string findSum(string str1, string str2)
{
    // Before proceeding further, make sure length
    // of str2 is larger.
    if (str1.length() > str2.length())
        swap(str1, str2);

    // Take an empty string for storing result
    string str = "";

    // Calculate length of both string
    int n1 = str1.length(), n2 = str2.length();

    // Reverse both of strings
    reverse(str1.begin(), str1.end());
    reverse(str2.begin(), str2.end());

    int carry = 0;
    for (int i=0; i<n1; i++)
    {
        // Do school mathematics, compute sum of
        // current digits and carry
        int sum = ((str1[i]-'0')+(str2[i]-'0')+carry);
        str.push_back(sum%10 + '0');

        // Calculate carry for next step
        carry = sum/10;
    }

    // Add remaining digits of larger number
    for (int i=n1; i<n2; i++)
    {
        int sum = ((str2[i]-'0')+carry);
        str.push_back(sum%10 + '0');
        carry = sum/10;
    }

    // Add remaining carry
    if (carry)
        str.push_back(carry+'0');

    // reverse resultant string
    reverse(str.begin(), str.end());

    return str;
}

// Driver code
int main()
{
    string str1 = "12";
    string str2 = "198111";
    cout << findSum(str1, str2);
    return 0;
}

Java

// Java program to find sum of two large numbers.
import java.util.*;
class GFG
{
// Function for finding sum of larger numbers
static String findSum(String str1, String str2)
{
    // Before proceeding further, make sure length
    // of str2 is larger.
    if (str1.length() > str2.length()){
        String t = str1;
        str1 = str2;
        str2 = t;
    }

    // Take an empty String for storing result
    String str = "";

    // Calculate length of both String
    int n1 = str1.length(), n2 = str2.length();

    // Reverse both of Strings
    str1=new StringBuilder(str1).reverse().toString();
    str2=new StringBuilder(str2).reverse().toString();

    int carry = 0;
    for (int i = 0; i < n1; i++)
    {
        // Do school mathematics, compute sum of
        // current digits and carry
        int sum = ((int)(str1.charAt(i) - '0') +
                    (int)(str2.charAt(i) - '0') + carry);
        str += (char)(sum % 10 + '0');

        // Calculate carry for next step
        carry = sum / 10;
    }

    // Add remaining digits of larger number
    for (int i = n1; i < n2; i++)
    {
        int sum = ((int)(str2.charAt(i) - '0') + carry);
        str += (char)(sum % 10 + '0');
        carry = sum / 10;
    }

    // Add remaining carry
    if (carry > 0)
        str += (char)(carry + '0');

    // reverse resultant String
    str = new StringBuilder(str).reverse().toString();

    return str;
}

// Driver code
public static void main(String[] args)
{
    String str1 = "12";
    String str2 = "198111";
    System.out.println(findSum(str1, str2));
}
}
// This code is contributed by mits

Python3

# Python3 program to find sum of
# two large numbers.

# Function for finding sum of
# larger numbers
def findSum(str1, str2):

    # Before proceeding further,
    # make sure length of str2 is larger.
    if (len(str1) > len(str2)):
        t = str1;
        str1 = str2;
        str2 = t;

    # Take an empty string for
    # storing result
    str = "";

    # Calculate length of both string
    n1 = len(str1);
    n2 = len(str2);

    # Reverse both of strings
    str1 = str1[::-1];
    str2 = str2[::-1];

    carry = 0;
    for i in range(n1):

        # Do school mathematics, compute
        # sum of current digits and carry
        sum = ((ord(str1[i]) - 48) +
              ((ord(str2[i]) - 48) + carry));
        str += chr(sum % 10 + 48);

        # Calculate carry for next step
        carry = int(sum / 10);

    # Add remaining digits of larger number
    for i in range(n1, n2):
        sum = ((ord(str2[i]) - 48) + carry);
        str += chr(sum % 10 + 48);
        carry = (int)(sum / 10);

    # Add remaining carry
    if (carry):
        str += chr(carry + 48);

    # reverse resultant string
    str = str[::-1];

    return str;

# Driver code
str1 = "12";
str2 = "198111";
print(findSum(str1, str2));

# This code is contributed by mits

C#

// C# program to find sum of two large numbers.
using System;
class GFG
{
// Function for finding sum of larger numbers
static string findSum(string str1, string str2)
{
    // Before proceeding further, make sure length
    // of str2 is larger.
    if (str1.Length > str2.Length){
        string t = str1;
        str1 = str2;
        str2 = t;
    }

    // Take an empty string for storing result
    string str = "";

    // Calculate length of both string
    int n1 = str1.Length, n2 = str2.Length;

    // Reverse both of strings
    char[] ch = str1.ToCharArray();
    Array.Reverse( ch );
    str1 = new string( ch );
    char[] ch1 = str2.ToCharArray();
    Array.Reverse( ch1 );
    str2 = new string( ch1 );

    int carry = 0;
    for (int i = 0; i < n1; i++)
    {
        // Do school mathematics, compute sum of
        // current digits and carry
        int sum = ((int)(str1[i] - '0') +
                (int)(str2[i] - '0') + carry);
        str += (char)(sum % 10 + '0');

        // Calculate carry for next step
        carry = sum/10;
    }

    // Add remaining digits of larger number
    for (int i = n1; i < n2; i++)
    {
        int sum = ((int)(str2[i] - '0') + carry);
        str += (char)(sum % 10 + '0');
        carry = sum/10;
    }

    // Add remaining carry
    if (carry > 0)
        str += (char)(carry + '0');

    // reverse resultant string
    char[] ch2 = str.ToCharArray();
    Array.Reverse( ch2 );
    str = new string( ch2 );

    return str;
}

// Driver code
static void Main()
{
    string str1 = "12";
    string str2 = "198111";
    Console.WriteLine(findSum(str1, str2));
}
}

// This code is contributed by mits

PHP

<?php
// PHP program to find sum of two large numbers.


// Function for finding sum of larger numbers
function findSum($str1, $str2)
{
    // Before proceeding further, make sure length
    // of str2 is larger.
    if (strlen($str1) > strlen($str2)) {
        $t=$str1;
        $str1=$str2;
        $str2=$t;
    }

    // Take an empty string for storing result
    $str = "";

    // Calculate length of both string
    $n1 = strlen($str1);
    $n2 = strlen($str2);

    // Reverse both of strings
    $str1 = strrev($str1);
    $str2 = strrev($str2);

    $carry = 0;
    for ($i=0; $i<$n1; $i++)
    {
        // Do school mathematics, compute sum of
        // current digits and carry
        $sum = ((ord($str1[$i])-48)+((ord($str2[$i])-48)+$carry));
        $str.=chr($sum%10 + 48);

        // Calculate carry for next step
        $carry = (int)($sum/10);
    }

    // Add remaining digits of larger number
    for ($i=$n1; $i<$n2; $i++)
    {
        $sum = ((ord($str2[$i])-48)+$carry);
        $str.=chr($sum%10 + 48);
        $carry = (int)($sum/10);
    }

    // Add remaining carry
    if ($carry)
        $str.=chr($carry+48);

    // reverse resultant string
    $str=strrev($str);

    return $str;
}

// Driver code

    $str1 = "12";
    $str2 = "198111";
    echo findSum($str1, $str2);

// This code is contributed by mits
?>

Javascript

<script>

// Javascript program to find sum of
// two large numbers.

// Function for finding sum of larger numbers
function findSum(str1, str2)
{

    // Before proceeding further, make
    // sure length of str2 is larger.
    if (str1.length > str2.length)
    {
        let t = str1;
        str1 = str2;
        str2 = t;
    }

    // Take an empty String for storing result
    let str = "";

    // Calculate length of both String
    let n1 = str1.length, n2 = str2.length;

    // Reverse both of Strings
    str1 = str1.split("").reverse().join("");
    str2 = str2.split("").reverse().join("");

    let carry = 0;
    for(let i = 0; i < n1; i++)
    {

        // Do school mathematics, compute sum of
        // current digits and carry
        let sum = ((str1[i].charCodeAt(0) -
                        '0'.charCodeAt(0)) +
                   (str2[i].charCodeAt(0) -
                        '0'.charCodeAt(0)) + carry);
        str += String.fromCharCode(sum % 10 +
                        '0'.charCodeAt(0));

        // Calculate carry for next step
        carry = Math.floor(sum / 10);
    }

    // Add remaining digits of larger number
    for(let i = n1; i < n2; i++)
    {
        let sum = ((str2[i].charCodeAt(0) -
                        '0'.charCodeAt(0)) + carry);
        str += String.fromCharCode(sum % 10 +
                        '0'.charCodeAt(0));
        carry = Math.floor(sum / 10);
    }

    // Add remaining carry
    if (carry > 0)
        str += String.fromCharCode(carry +
                       '0'.charCodeAt(0));

    // reverse resultant String
    str = str.split("").reverse().join("");

    return str;
}

// Driver code
let str1 = "12";
let str2 = "198111";

document.write(findSum(str1, str2))

// This code is contributed by rag2127

</script>

Output:

198123

Time Complexity: O(n1+n2) where n1 and n2 are lengths of two input strings representing numbers.

Auxiliary Space: O(max(n1, n2))

Approach 3 : Optimization

We can avoid the first two string reverse operations by traversing them from the end. Below is the optimized solution. 

C++

// C++ program to find sum of two large numbers.
#include<bits/stdc++.h>
using namespace std;

// Function for finding sum of larger numbers
string findSum(string str1, string str2)
{
    // Before proceeding further, make sure length
    // of str2 is larger.
    if (str1.length() > str2.length())
        swap(str1, str2);

    // Take an empty string for storing result
    string str = "";

    // Calculate length of both string
    int n1 = str1.length(), n2 = str2.length();
    int diff = n2 - n1;

    // Initially take carry zero
    int carry = 0;

    // Traverse from end of both strings
    for (int i=n1-1; i>=0; i--)
    {
        // Do school mathematics, compute sum of
        // current digits and carry
        int sum = ((str1[i]-'0') +
                   (str2[i+diff]-'0') +
                   carry);
        str.push_back(sum%10 + '0');
        carry = sum/10;
    }

    // Add remaining digits of str2[]
    for (int i=n2-n1-1; i>=0; i--)
    {
        int sum = ((str2[i]-'0')+carry);
        str.push_back(sum%10 + '0');
        carry = sum/10;
    }

    // Add remaining carry
    if (carry)
        str.push_back(carry+'0');

    // reverse resultant string
    reverse(str.begin(), str.end());

    return str;
}

// Driver code
int main()
{
    string str1 = "12";
    string str2 = "198111";
    cout << findSum(str1, str2);
    return 0;
}

Java

// Java program to find sum of two large numbers.
import java.util.*;

class GFG{

 // Function for finding sum of larger numbers
static String findSum(String str1, String str2)
{
    // Before proceeding further, make sure length
    // of str2 is larger.
    if (str1.length() > str2.length()){
        String t = str1;
        str1 = str2;
        str2 = t;
    }

    // Take an empty String for storing result
    String str = "";

    // Calculate length of both String
    int n1 = str1.length(), n2 = str2.length();
    int diff = n2 - n1;

    // Initially take carry zero
    int carry = 0;

    // Traverse from end of both Strings
    for (int i = n1 - 1; i>=0; i--)
    {
        // Do school mathematics, compute sum of
        // current digits and carry
        int sum = ((int)(str1.charAt(i)-'0') +
            (int)(str2.charAt(i+diff)-'0') + carry);
        str += (char)(sum % 10 + '0');
        carry = sum / 10;
    }

    // Add remaining digits of str2[]
    for (int i = n2 - n1 - 1; i >= 0; i--)
    {
        int sum = ((int)(str2.charAt(i) - '0') + carry);
        str += (char)(sum % 10 + '0');
        carry = sum / 10;
    }

    // Add remaining carry
    if (carry > 0)
        str += (char)(carry + '0');

    // reverse resultant String
    return new StringBuilder(str).reverse().toString();
}

// Driver code
public static void main(String[] args)
{
    String str1 = "12";
    String str2 = "198111";
    System.out.println(findSum(str1, str2));
}
}

// This code is contributed by mits

Python3

# python 3 program to find sum of two large numbers.

# Function for finding sum of larger numbers
def findSum(str1, str2):

    # Before proceeding further, make sure length
    # of str2 is larger.
    if len(str1)> len(str2):
        temp = str1
        str1 = str2
        str2 = temp

    # Take an empty string for storing result
    str3 = ""

    # Calculate length of both string
    n1 = len(str1)
    n2 = len(str2)
    diff = n2 - n1

    # Initially take carry zero
    carry = 0

    # Traverse from end of both strings
    for i in range(n1-1,-1,-1):

        # Do school mathematics, compute sum of
        # current digits and carry

        sum = ((ord(str1[i])-ord('0')) +
                   int((ord(str2[i+diff])-ord('0'))) + carry)

        str3 = str3+str(sum%10 )


        carry = sum//10

    # Add remaining digits of str2[]
    for i in range(n2-n1-1,-1,-1):

        sum = ((ord(str2[i])-ord('0'))+carry)
        str3 = str3+str(sum%10 )
        carry = sum//10

    # Add remaining carry
    if (carry):
        str3+str(carry+'0')

    # reverse resultant string
    str3 = str3[::-1]

    return str3

# Driver code
if __name__ == "__main__":
    str1 = "12"
    str2 = "198111"
    print(findSum(str1, str2))

# This code is contributed by ChitraNayal

C#

// C# program to find sum of two large numbers.
using System;

class GFG{

// Function for finding sum of larger numbers
static string findSum(string str1, string str2)
{
    // Before proceeding further, make sure length
    // of str2 is larger.
    if (str1.Length > str2.Length)
    {
        string t = str1;
        str1 = str2;
        str2 = t;
    }

    // Take an empty string for storing result
    string str = "";

    // Calculate length of both string
    int n1 = str1.Length, n2 = str2.Length;
    int diff = n2 - n1;

    // Initially take carry zero
    int carry = 0;

    // Traverse from end of both strings
    for (int i = n1 - 1; i >= 0; i--)
    {
        // Do school mathematics, compute sum of
        // current digits and carry
        int sum = ((int)(str1[i] - '0') +
                (int)(str2[i + diff]-'0') + carry);
        str += (char)(sum % 10 + '0');
        carry = sum / 10;
    }

    // Add remaining digits of str2[]
    for (int i = n2 - n1 - 1; i >= 0; i--)
    {
        int sum = ((int)(str2[i] - '0') + carry);
        str += (char)(sum % 10 + '0');
        carry = sum / 10;
    }

    // Add remaining carry
    if (carry > 0)
        str += (char)(carry + '0');

    // reverse resultant string
    char[] ch2 = str.ToCharArray();
    Array.Reverse(ch2);
    return new string(ch2);
}

// Driver code
static void Main()
{
    string str1 = "12";
    string str2 = "198111";
    Console.WriteLine(findSum(str1, str2));
}
}

// This code is contributed by mits

PHP

<?php
// PHP program to find sum of two large numbers.

// Function for finding sum of larger numbers
function findSum($str1, $str2)
{
    // Before proceeding further, make
    // sure length of str2 is larger.
    if(strlen($str1)> strlen($str2))
    {
        $temp = $str1;
        $str1 = $str2;
        $str2 = $temp;
    }

    // Take an empty string for storing result
    $str3 = "";

    // Calculate length of both string
    $n1 = strlen($str1);
    $n2 = strlen($str2);
    $diff = $n2 - $n1;

    // Initially take carry zero
    $carry = 0;

    // Traverse from end of both strings
    for ($i = $n1 - 1; $i >= 0; $i--)
    {
        // Do school mathematics, compute sum
        // of current digits and carry
        $sum = ((ord($str1[$i]) - ord('0')) +
               ((ord($str2[$i + $diff]) -
                 ord('0'))) + $carry);

        $str3 .= chr($sum % 10 + ord('0'));


        $carry = (int)($sum / 10);
    }

    // Add remaining digits of str2[]
    for ($i = $n2 - $n1 - 1; $i >= 0; $i--)
    {
        $sum = ((ord($str2[$i]) - ord('0')) + $carry);
        $str3 .= chr($sum % 10 + ord('0'));
        $carry = (int)($sum / 10);
    }

    // Add remaining carry
    if ($carry)
        $str3 .= chr($carry + ord('0'));

    // reverse resultant string
    return strrev($str3);
}

// Driver code
$str1 = "12";
$str2 = "198111";
print(findSum($str1, $str2));

// This code is contributed by mits
?>

Javascript

<script>

// JavaScript program to find sum of two large numbers.

// Function for finding sum of larger numbers
function findSum(str1,str2)
{
    // Before proceeding further, make sure length
    // of str2 is larger.
    if (str1.length > str2.length){

        let temp = str1
        str1 = str2
        str2 = temp

    }

    // Take an empty string for storing result
    let str = "";

    // Calculate length of both string
    let n1 = str1.length, n2 = str2.length;
    let diff = n2 - n1;
    // Initially take carry zero
    let carry = 0;

    // Traverse from end of both strings
    for (let i=n1-1; i>=0; i--)
    {
        // Do school mathematics, compute sum of
        // current digits and carry
        // console.log((str1.charCodeAt(i)-48),(str2.charCodeAt(i+diff)-48))
        let sum = ((str1.charCodeAt(i)-48) +
                (str2.charCodeAt(i+diff)-48) +
                carry);
        str+=(sum%10);
        carry = Math.floor(sum/10);
    }

    // // Add remaining digits of str2[]
    for (let i=n2-n1-1; i>=0; i--)
    {
        let sum = ((str2.charCodeAt(i)-48)+carry);
        str+=(sum%10);
        carry = Math.floor(sum/10);
    }

    // Add remaining carry
    if (carry)
        str+=(carry+'0');

    // reverse resultant string

    str = str.split("").reverse().join("");

    return str;
}

// Driver code
let str1 = "12";
let str2 = "198111";
document.write(findSum(str1, str2),"</br>");

// This code is contributed by shinjanpatra.
</script>

Output:

198123

Time Complexity: O(max(n1, n2)) where n1 and n2 are lengths of two input strings representing numbers.

Auxiliary Space: O(max(n1, n2))

This article is contributed by DANISH_RAZA . If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.