Given the count of digits 1, 2, 3, 4. Using these digits you are allowed to only form numbers 234 and 12. The task is to find the maximum possible sum that can be obtained after forming the numbers.
Note: The aim is only to maximize the sum, even if some of the digits left unused.
Examples:
Input : c1 = 5, c2 = 2, c3 = 3, c4 = 4 Output : 468 Explanation : We can form two 234s Input : c1 = 5, c2 = 3, c3 = 1, c4 = 5 Output : 258 Explanation : We can form one 234 and two 12s
Approach : An efficient approach is to first try to make 234’s. The possible number of 234s are minimum of c2, c3, c4. After this, with remaining 1’s and 2’s try to form 12s.
Below is the implementation of the above approach :
C++
// CPP program to maximum possible sum #include <bits/stdc++.h> using namespace std;
// Function to find the maximum possible sum int Maxsum( int c1, int c2, int c3, int c4)
{ // To store required sum
int sum = 0;
// Number of 234's can be formed
int two34 = min(c2, min(c3, c4));
// Sum obtained with 234s
sum = two34 * 234;
// Remaining 2's
c2 -= two34;
// Sum obtained with 12s
sum += min(c2, c1) * 12;
// Return the required sum
return sum;
} // Driver code int main()
{ int c1 = 5, c2 = 2, c3 = 3, c4 = 4;
cout << Maxsum(c1, c2, c3, c4);
return 0;
} |
Java
// Java program to maximum possible sum class GFG
{ // Function to find the maximum possible sum static int Maxsum( int c1, int c2, int c3, int c4)
{ // To store required sum
int sum = 0 ;
// Number of 234's can be formed
int two34 = Math.min(c2,Math.min(c3, c4));
// Sum obtained with 234s
sum = two34 * 234 ;
// Remaining 2's
c2 -= two34;
// Sum obtained with 12s
sum +=Math.min(c2, c1) * 12 ;
// Return the required sum
return sum;
} // Driver code public static void main(String[] args)
{ int c1 = 5 , c2 = 2 , c3 = 3 , c4 = 4 ;
System.out.println(Maxsum(c1, c2, c3, c4));
} } // This code is contributed by Code_Mech. |
Python3
# Python3 program to maximum possible sum # Function to find the maximum # possible sum def Maxsum(c1, c2, c3, c4):
# To store required sum
sum = 0
# Number of 234's can be formed
two34 = min (c2, min (c3, c4))
# Sum obtained with 234s
sum = two34 * 234
# Remaining 2's
c2 - = two34
sum + = min (c2, c1) * 12
# Return the required sum
return sum
# Driver Code c1 = 5 ; c2 = 2 ; c3 = 3 ; c4 = 4
print (Maxsum(c1, c2, c3, c4))
# This code is contributed by Shrikant13 |
C#
// C# program to maximum possible sum using System;
class GFG
{ // Function to find the maximum possible sum static int Maxsum( int c1, int c2, int c3, int c4)
{ // To store required sum
int sum = 0;
// Number of 234's can be formed
int two34 = Math.Min(c2, Math.Min(c3, c4));
// Sum obtained with 234s
sum = two34 * 234;
// Remaining 2's
c2 -= two34;
// Sum obtained with 12s
sum +=Math.Min(c2, c1) * 12;
// Return the required sum
return sum;
} // Driver code public static void Main()
{ int c1 = 5, c2 = 2, c3 = 3, c4 = 4;
Console.WriteLine(Maxsum(c1, c2, c3, c4));
} } // This code is contributed // by Akanksha Rai |
PHP
<?php // PHP program to maximum possible sum // Function to find the maximum possible sum function Maxsum( $c1 , $c2 , $c3 , $c4 )
{ // To store required sum
$sum = 0;
// Number of 234's can be formed
$two34 = min( $c2 , min( $c3 , $c4 ));
// Sum obtained with 234s
$sum = $two34 * 234;
// Remaining 2's
$c2 -= $two34 ;
// Sum obtained with 12s
$sum += min( $c2 , $c1 ) * 12;
// Return the required sum
return $sum ;
} // Driver code $c1 = 5; $c2 = 2;
$c3 = 3; $c4 = 4;
echo Maxsum( $c1 , $c2 , $c3 , $c4 );
// This code is contributed by Ryuga ?> |
Javascript
<script> // Java Script program to maximum possible sum // Function to find the maximum possible sum function Maxsum(c1,c2,c3,c4)
{ // To store required sum
let sum = 0;
// Number of 234's can be formed
let two34 = Math.min(c2,Math.min(c3, c4));
// Sum obtained with 234s
sum = two34 * 234;
// Remaining 2's
c2 -= two34;
// Sum obtained with 12s
sum +=Math.min(c2, c1) * 12;
// Return the required sum
return sum;
} // Driver code let c1 = 5, c2 = 2, c3 = 3, c4 = 4;
document.write(Maxsum(c1, c2, c3, c4));
// This code is contributed by sravan kumar G </script> |
Output:
468
Time Complexity: O(1)
Auxiliary Space: O(1)