You have an unlimited amount of banknotes worth A and B dollars (A not equals to B). You want to pay a total of S dollars using exactly N notes. The task is to find the number of notes worth A dollars you need. If there is no solution return -1.
Examples:
Input: A = 1, B = 2, S = 7, N = 5
Output: 3
3 * A + 2 * B = S
3 * 1 + 2 * 2 = 7Input: A = 2, B = 1, S = 7, N = 5
Output: 2Input: A = 2, B = 1, S = 4, N = 5
Output: -1Input: A = 2, B = 3, S = 20, N = 8
Output: 4
Approach: Let x be the number of notes of value A required then the rest of the notes i.e. N – x must of value B. Since, their sum is required be S then the following equation must be satisfied:
S = (A * x) + (B * (N – x))
Solving the equation further,
S = (A * x) + (B * N) – (B * x)
S – (B * N) = (A – B) * x
x = (S – (B * N)) / (A – B)
After solving for x, it is the number of notes of value A required
only if the value of x is an integer else the result is not possible.
Below is the implementation of the above approach:
// C++ implementation of the approach #include<bits/stdc++.h> using namespace std;
// Function to return the amount of notes // with value A required int bankNotes( int A, int B, int S, int N)
{ int numerator = S - (B * N);
int denominator = A - B;
// If possible
if (numerator % denominator == 0)
return (numerator / denominator);
return -1;
} // Driver code int main()
{ int A = 1, B = 2, S = 7, N = 5;
cout << bankNotes(A, B, S, N) << endl;
} // This code is contributed by mits |
// Java implementation of the approach class GFG {
// Function to return the amount of notes
// with value A required
static int bankNotes( int A, int B, int S, int N)
{
int numerator = S - (B * N);
int denominator = A - B;
// If possible
if (numerator % denominator == 0 )
return (numerator / denominator);
return - 1 ;
}
// Driver code
public static void main(String[] args)
{
int A = 1 , B = 2 , S = 7 , N = 5 ;
System.out.print(bankNotes(A, B, S, N));
}
} |
# Python3 implementation of the approach # Function to return the amount of notes # with value A required def bankNotes(A, B, S, N):
numerator = S - (B * N)
denominator = A - B
# If possible
if (numerator % denominator = = 0 ):
return (numerator / / denominator)
return - 1
# Driver code A, B, S, N = 1 , 2 , 7 , 5
print (bankNotes(A, B, S, N))
# This code is contributed # by mohit kumar |
// C# implementation of the approach using System;
class GFG
{ // Function to return the amount of notes
// with value A required
static int bankNotes( int A, int B,
int S, int N)
{
int numerator = S - (B * N);
int denominator = A - B;
// If possible
if (numerator % denominator == 0)
return (numerator / denominator);
return -1;
}
// Driver code
public static void Main()
{
int A = 1, B = 2, S = 7, N = 5;
Console.Write(bankNotes(A, B, S, N));
}
} // This code is contributed by Akanksha Rai |
<?php // PHP implementation of the approach // Function to return the amount of notes // with value A required function bankNotes( $A , $B , $S , $N )
{ $numerator = $S - ( $B * $N );
$denominator = $A - $B ;
// If possible
if ( $numerator % $denominator == 0)
return ( $numerator / $denominator );
return -1;
} // Driver code $A = 1; $B = 2; $S = 7; $N = 5;
echo (bankNotes( $A , $B , $S , $N ));
// This code is contributed by Code_Mech ?> |
<script> // Javascript implementation of the approach // Function to return the amount of notes // with value A required function bankNotes(A, B, S, N)
{ let numerator = S - (B * N);
let denominator = A - B;
// If possible
if (numerator % denominator == 0)
return (Math.floor(numerator /
denominator));
return -1;
} // Driver Code let A = 1, B = 2, S = 7, N = 5; document.write(bankNotes(A, B, S, N) + "</br>" );
// This code is contributed by jana_sayantan </script> |
3
Time Complexity: O(1)
Auxiliary Space: O(1)