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
<script> // Javascript implementation to find the // sum of two integers of base B // Function to find the sum of // two integers of base B function sumBaseB(a, b, base_var)
{ let len_a, len_b;
len_a = a.length;
len_b = b.length;
let sum, s;
s = "" ;
sum = "" ;
let diff;
diff = Math.abs(len_a - len_b);
// Padding 0 in front of the
// number to make both numbers equal
for (let 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;
let curr, carry = 0;
// Loop to find the find the sum
// of two integers of base B
for (let i = Math.max(len_a, len_b) - 1;
i > -1; i--)
{
// Current Place value for
// the resultant sum
curr = carry + (a[i].charCodeAt() -
'0' .charCodeAt()) +
(b[i].charCodeAt() -
'0' .charCodeAt());
// Update carry
carry = parseInt(curr / base_var, 10);
// Find current digit
curr = curr % base_var;
// Update sum result
sum = String.fromCharCode(
curr + '0' .charCodeAt()) + sum;
}
if (carry > 0)
sum = String.fromCharCode(
carry + '0' .charCodeAt()) + sum;
return sum;
} // Driver code let a, b, sum; let base_var; a = "123" ;
b = "234" ;
base_var = 6; // Function Call sum = sumBaseB(a, b, base_var); document.write(sum + "</br>" );
// This code is contributed by divyesh072019 </script> |
PHP
<?php // PHP implementation to find the // sum of two integers of base B // Function to find the sum of // two integers of base B function sumBaseB( $a , $b , $base )
{ $len_a = strlen ( $a );
$len_b = strlen ( $b );
$s = "" ;
$sum = "" ;
$diff = abs ( $len_a - $len_b );
// Padding 0 in front of the
// number to make both numbers equal
for ( $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 ;
$carry = 0;
// Loop to find the find the sum
// of two integers of base B
for ( $i = max( $len_a , $len_b ) - 1; $i > -1; $i --) {
// Current Place value for
// the resultant sum
$curr = $carry + (ord( $a [ $i ]) - ord( '0' )) +(ord( $b [ $i ]) - ord( '0' ));
// Update carry
$carry = intdiv( $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 );
echo $sum ;
return 0;
//This code is contributed by Susobhan Akhuli ?> |
Output
401
Time Complexity: O(max(len_a , len_b))
Auxiliary Space: O(max(len_a , len_b))
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