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 sumint 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 codeint main(){ int c1 = 5, c2 = 2, c3 = 3, c4 = 4; cout << Maxsum(c1, c2, c3, c4); return 0;} |
Java
// Java program to maximum possible sumclass GFG{ // Function to find the maximum possible sumstatic 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 codepublic 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 sumdef 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 Codec1 = 5; c2 = 2; c3 = 3; c4 = 4print(Maxsum(c1, c2, c3, c4))# This code is contributed by Shrikant13 |
C#
// C# program to maximum possible sumusing System;class GFG{ // Function to find the maximum possible sumstatic 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 codepublic 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 sumfunction 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)
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