Sum of two large numbers

Difficulty Level : Medium
Last Updated : 04 Jun, 2019

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

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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 
?> 

Output:

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

Python 3

# 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  
?> 

Output:

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

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My Personal Notes

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Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python 3

C#

PHP