Given three numbers, our task is to find the largest number by which when given 3 numbers are divided leads to same remainder. It may be assumed that all given numbers are given in increasing order.
Examples:
Input : a = 62, b = 132, c = 237 Output : 35 35 leads to same remainder 27 when divides 62, 132 and 237. Input : a = 74, b = 272, c = 584 Output : 6
Brute Force Approach:
The brute force approach to solve this problem is to start from the largest number among the three (i.e., c) and decrement it until we find a number that leaves the same remainder when divided by a and b.
We can find the remainder of a, b, and c when divided by any number n using the modulo operator (%). Then, we can check if the remainders of a, b, and c when divided by a candidate number n are equal. If they are equal, then n is a valid solution.
Below is the implementation of above approach:
// C++ program to find the largest numbers that // leads to same remainder when divides given // three sorted numbers #include <bits/stdc++.h> using namespace std;
// function return number which divides these // three number and leaves same remainder . int sameRemainder( int a, int b, int c)
{ // start from largest number and decrement until we find
// a solution
for ( int n = c; n >= 1; n--) {
int remainderA = a % n;
int remainderB = b % n;
int remainderC = c % n;
if (remainderA == remainderB
&& remainderB == remainderC) {
return n;
}
}
// no solution found
return -1;
} // driver program int main()
{ int a = 62, b = 132, c = 237;
cout << sameRemainder(a, b, c) << endl;
return 0;
} |
// Java program to find the largest numbers that // leads to same remainder when divides given // three sorted numbers import java.util.*;
public class Main {
// function return number which divides these
// three number and leaves same remainder .
static int sameRemainder( int a, int b, int c)
{
// start from largest number and decrement until we
// find a solution
for ( int n = c; n >= 1 ; n--) {
int remainderA = a % n;
int remainderB = b % n;
int remainderC = c % n;
if (remainderA == remainderB
&& remainderB == remainderC) {
return n;
}
}
// no solution found
return - 1 ;
}
// driver program
public static void main(String[] args)
{
int a = 62 , b = 132 , c = 237 ;
System.out.println(sameRemainder(a, b, c));
}
} |
# Python program to find the largest numbers that # leads to same remainder when divides given # three sorted numbers def same_remainder(a, b, c):
# start from largest number and decrement until we find a solution
for n in range (c, 0 , - 1 ):
remainder_a = a % n
remainder_b = b % n
remainder_c = c % n
if remainder_a = = remainder_b = = remainder_c:
return n
# no solution found
return - 1
a = 62
b = 132
c = 237
print (same_remainder(a, b, c))
|
using System;
class Program
{ // function return number which divides these
// three number and leaves same remainder .
static int SameRemainder( int a, int b, int c)
{
// start from largest number and decrement until we
// find a solution
for ( int n = c; n >= 1; n--) {
int remainderA = a % n;
int remainderB = b % n;
int remainderC = c % n;
if (remainderA == remainderB
&& remainderB == remainderC) {
return n;
}
}
// no solution found
return -1;
}
static void Main( string [] args)
{
int a = 62, b = 132, c = 237;
Console.WriteLine(SameRemainder(a, b, c));
}
} |
function sameRemainder(a, b, c) {
// start from largest number and decrement until we find a solution
for (let n = c; n >= 1; n--) {
let remainderA = a % n;
let remainderB = b % n;
let remainderC = c % n;
if (remainderA == remainderB && remainderB == remainderC) {
return n;
}
}
// no solution found
return -1;
} // driver program let a = 62, b = 132, c = 237; console.log(sameRemainder(a, b, c)); |
35
Time Complexity: O(c)
Auxiliary Space: O(1)
Approach:
The idea is based on the fact that if a number leaves same remainder with a, b and c, then it would divide their differences. Let us understand assuming that x is our result. Let a = x*d1 + r where r is the remainder when a is divided by x. Similarly we can write b = x*d2 + r and b = x*d3 + r. So the logic is here we first find differences of all three pairs and after that, we find greatest common divisor of differences to maximize result.
Below is the implementation of above idea.
// C++ program to find the largest numbers that // leads to same remainder when divides given // three sorted numbers #include <bits/stdc++.h> using namespace std;
//__gcd function int gcd( int a, int b)
{ if (a == 0)
return b;
return gcd(b % a, a);
} // function return number which divides these // three number and leaves same remainder . int sameRemainder( int a, int b, int c)
{ // We find the differences of all three pairs
int a1 = (b - a), b1 = (c - b), c1 = (c - a);
// Return GCD of three differences.
return gcd(a1, gcd(b1, c1));
} // driver program int main()
{ int a = 62, b = 132, c = 237;
cout << sameRemainder(a, b, c) << endl;
return 0;
} |
// Java program to find the largest // numbers that leads to same // remainder when divides given // three sorted numbers class GFG {
//__gcd function
static int gcd( int a, int b)
{
if (a == 0 )
return b;
return gcd(b % a, a);
}
// function return number which divides these
// three number and leaves same remainder .
static int sameRemainder( int a, int b, int c)
{
// We find the differences of all three pairs
int a1 = (b - a), b1 = (c - b), c1 = (c - a);
// Return GCD of three differences.
return gcd(a1, gcd(b1, c1));
}
// Driver code
public static void main(String[] args)
{
int a = 62 , b = 132 , c = 237 ;
System.out.println(sameRemainder(a, b, c));
}
} // This code is contributed by Anant Agarwal. |
# Python program to find # the largest numbers that # leads to same remainder # when divides given # three sorted numbers # __gcd function def gcd(a, b):
if (a = = 0 ):
return b
return gcd(b % a, a)
# function return number # which divides these # three number and leaves # same remainder . def sameRemainder(a, b, c):
# We find the differences
# of all three pairs
a1 = (b - a)
b1 = (c - b)
c1 = (c - a)
# Return GCD of three differences.
return gcd(a1, gcd(b1, c1))
# Driver program a = 62
b = 132
c = 237
print (sameRemainder(a, b, c))
# This code is contributed # by Anant Agarwal. |
// C# program to find the largest // numbers that leads to same // remainder when divides given // three sorted numbers using System;
class GFG {
// gcd function
static int gcd( int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// function return number which divides
// these three number and leaves same
// remainder .
static int sameRemainder( int a, int b, int c)
{
// We find the differences of all three pairs
int a1 = (b - a), b1 = (c - b), c1 = (c - a);
// Return GCD of three differences.
return gcd(a1, gcd(b1, c1));
}
// Driver code
public static void Main()
{
int a = 62, b = 132, c = 237;
Console.WriteLine(sameRemainder(a, b, c));
}
} // This code is contributed by vt_m. |
<?php // PHP program to find the // largest numbers that leads // to same remainder when // divides given three sorted // numbers //__gcd function function gcd( $a , $b )
{ if ( $a == 0)
return $b ;
return gcd( $b % $a , $a );
} // function return number // which divides these // three number and // leaves same remainder . function sameRemainder( $a , $b , $c )
{ // We find the differences
// of all three pairs
$a1 = ( $b - $a );
$b1 = ( $c - $b );
$c1 = ( $c - $a );
// Return GCD of
// three differences.
return gcd( $a1 , gcd( $b1 , $c1 ));
} // Driver Code
$a = 62;
$b = 132;
$c = 237;
echo sameRemainder( $a , $b , $c ) ;
// This code is contributed by anuj_67. ?> |
<script> // Javascript program to find the largest // numbers that leads to same remainder // when divides given three sorted numbers //__gcd function function gcd(a, b)
{ if (a == 0)
return b;
return gcd(b % a, a);
} // Function return number which divides these // three number and leaves same remainder . function sameRemainder(a, b, c)
{ // We find the differences of all three pairs
var a1 = (b - a), b1 = (c - b), c1 = (c - a);
// Return GCD of three differences.
return gcd(a1, gcd(b1, c1));
} // Driver code var a = 62, b = 132, c = 237;
document.write(sameRemainder(a, b, c)); // This code is contributed by noob2000 </script> |
35
Time Complexity: O( log N )
Auxiliary Space: O(1)