Program to add two integers of given base (original) (raw)

Last Updated : 3 Oct, 2022

Given three integers X, Y, and B, where X and Y are Base-B integers. The task is to find the sum of integers X and Y.

Examples:

Input: X = 123, Y = 234, B = 6 Output: 401 Explanation: Sum of two integers in base 6 - 1 1
1 2 3 + 2 3 4

 4 0 1

Input: X = 546, Y = 248 B = 9 Output: 805 Explanation: Sum of two integers in base 9 - 1 1 5 4 6 + 2 4 8

 8 0 5   

Approach: The idea is to use the fact that whenever two digits of the numbers are added, then the place value will be the modulo of the sum of digits by the base whereas carry will be the integer division of the sum of digits by base. i.e.

Let two digits of the number be D1 and D2 - Place Value = (D1 + D2) % B

Carry = (D1 + D2) / B

Similarly, Add every digit from the last to get the desired result.

Below is the implementation of the above approach:

C++ `

// C++ implementation to find the // sum of two integers of base B

#include <bits/stdc++.h>

using namespace std;

// Function to find the sum of // two integers of base B string sumBaseB(string a, string b, int base) { int len_a, len_b;

len_a = a.size();
len_b = b.size();

string sum, s;
s = "";
sum = "";

int diff;
diff = abs(len_a - len_b);

// Padding 0 in front of the 
// number to make both numbers equal
for (int i = 1; i <= diff; i++)
    s += "0";

// Condition to check if the strings
// have lengths mis-match
if (len_a < len_b)
    a = s + a;
else
    b = s + b;

int curr, carry = 0;

// Loop to find the find the sum 
// of two integers of base B
for (int i = max(len_a, len_b) - 1; 
                       i > -1; i--) {
    
    // Current Place value for 
    // the resultant sum
    curr = carry + (a[i] - '0') + 
                   (b[i] - '0');

    // Update carry
    carry = curr / base;

    // Find current digit
    curr = curr % base;

    // Update sum result
    sum = (char)(curr + '0') + sum;
}
if (carry > 0)
    sum = (char)(carry + '0') + sum;
return sum;

}

// Driver Code int main() { string a, b, sum; int base; a = "123"; b = "234"; base = 6;

// Function Call
sum = sumBaseB(a, b, base);
cout << sum << endl;
return 0;

}

Java

// Java implementation to find the // sum of two integers of base B class GFG {

// Function to find the sum of 
// two integers of base B 
static String sumBaseB(String a, String b, int base) 
{ 
    int len_a, len_b; 

    len_a = a.length(); 
    len_b = b.length(); 

    String sum, s; 
    s = ""; 
    sum = ""; 

    int diff; 
    diff = Math.abs(len_a - len_b); 
    
    // Padding 0 in front of the 
    // number to make both numbers equal 
    for (int i = 1; i <= diff; i++) 
        s += "0"; 
    
    // Condition to check if the strings 
    // have lengths mis-match 
    if (len_a < len_b) 
        a = s + a; 
    else
        b = s + b; 

    int curr, carry = 0; 
    
    // Loop to find the find the sum 
    // of two integers of base B 
    for (int i = Math.max(len_a, len_b) - 1; 
                        i > -1; i--) { 
        
        // Current Place value for 
        // the resultant sum 
        curr = carry + (a.charAt(i) - '0') + 
                    (b.charAt(i) - '0'); 

        // Update carry 
        carry = curr / base; 

        // Find current digit 
        curr = curr % base; 

        // Update sum result 
        sum = (char)(curr + '0') + sum; 
    } 
    if (carry > 0) 
        sum = (char)(carry + '0') + sum; 
    return sum; 
} 

// Driver Code 
public static void main (String[] args) 
{ 
    String a, b, sum; 
    int base; 
    a = "123"; 
    b = "234"; 
    base = 6; 
    
    // Function Call 
    sum = sumBaseB(a, b, base); 
    System.out.println(sum); 
} 

}

// This code is contributed by AnkitRai01

Python3

Python 3 implementation to find the

sum of two integers of base B

Function to find the sum of

two integers of base B

def sumBaseB(a,b,base):

len_a = len(a)
len_b = len(b)

s = "";
sum = "";

diff = abs(len_a - len_b);
 
# Padding 0 in front of the 
# number to make both numbers equal
for i in range(1,diff+1):
    s += "0"
 
# Condition to check if the strings
# have lengths mis-match
if (len_a < len_b):
    a = s + a
else:
    b = s + b;

carry = 0;
 
# Loop to find the find the sum 
# of two integers of base B
for i in range(max(len_a, len_b) - 1,-1,-1):
     
    # Current Place value for 
    # the resultant sum
    curr = carry + (ord(a[i]) -ord('0')) +( ord(b[i]) - ord('0'));

    # Update carry
    carry = curr // base

    # Find current digit
    curr = curr % base;

    # Update sum result
    sum = chr(curr + ord('0')) + sum
    
if (carry > 0):
    sum = chr(carry + ord('0')) + sum;
return sum

Driver Code

a = "123" b = "234" base = 6

Function Call

sum = sumBaseB(a, b, base); print(sum)

This code is contributed by atul_kumar_shrivastava

C#

// C# implementation to find the // sum of two integers of base B using System;

class GFG {

// Function to find the sum of 
// two integers of base B 
static string sumBaseB(string a, string b, int base_var) 
{ 
    int len_a, len_b; 

    len_a = a.Length; 
    len_b = b.Length; 

    string sum, s; 
    s = ""; 
    sum = ""; 

    int diff; 
    diff = Math.Abs(len_a - len_b); 
    
    // Padding 0 in front of the 
    // number to make both numbers equal 
    for (int i = 1; i <= diff; i++) 
        s += "0"; 
    
    // Condition to check if the strings 
    // have lengths mis-match 
    if (len_a < len_b) 
        a = s + a; 
    else
        b = s + b; 

    int curr, carry = 0; 
    
    // Loop to find the find the sum 
    // of two integers of base B 
    for (int i = Math.Max(len_a, len_b) - 1; 
                        i > -1; i--) { 
        
        // Current Place value for 
        // the resultant sum 
        curr = carry + (a[i] - '0') + 
                    (b[i] - '0'); 

        // Update carry 
        carry = curr / base_var; 

        // Find current digit 
        curr = curr % base_var; 

        // Update sum result 
        sum = (char)(curr + '0') + sum; 
    } 
    if (carry > 0) 
        sum = (char)(carry + '0') + sum; 
    return sum; 
} 

// Driver Code 
public static void Main (string[] args) 
{ 
    string a, b, sum; 
    int base_var; 
    a = "123"; 
    b = "234"; 
    base_var = 6; 
    
    // Function Call 
    sum = sumBaseB(a, b, base_var); 
    Console.WriteLine(sum); 
} 

}

// This code is contributed by AnkitRai01

JavaScript

PHP

`

Time Complexity: O(max(len_a , len_b))
Auxiliary Space: O(max(len_a , len_b))