Maximize the value of the given expression
Given three non-zero integers a, b and c. The task is to find the maximum value possible by putting addition and multiplication signs between them in any order.
Note: Rearrangement of integers is allowed but addition and multiplication sign must be used once. Braces can also be placed between equations as per your need.
Examples:
Input: a = 2, b = 1, c = 4
Output: 12
(1 + 2) * 4 = 3 * 4 = 12Input: a = 2, b = 2, c = 2
Output: 8
(2 + 2) * 2 = 4 * 2 = 8
Approach: To solve this problem one can opt the method of generating all the possibilities and calculate them to get the maximum value but this approach is not efficient. Take the advantage of given conditions that integers may got rearranged and mandatory use of each mathematical sign (+, *). There are total of four cases to solve which are listed below:
- All three integers are non-negative: For this simply add two smaller one and multiply their result by largest integer.
- One integer is negative and rest two positive : Multiply the both positive integer and add their result to negative integer.
- Two integers are negative and one is positive: As the product of two negative numbers is positive multiply both negative integers and then add their result to positive integer.
- All three are negative integers: Multiply the two smallest integers and add them to largest one.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the maximum result int maximumResult(int a, int b, int c) { // To store the count of negative integers int countOfNegative = 0; // Sum of all the three integers int sum = a + b + c; // Product of all the three integers int product = a * b * c; // To store the smallest and the largest // among all the three integers int largest = (a > b) ? ((a > c) ? a : c) : ((b > c) ? b : c); int smallest; // Calculate the count of negative integers if (a < 0) countOfNegative++; if (b < 0) countOfNegative++; if (c < 0) countOfNegative++; // Depending upon count of negatives switch (countOfNegative) { // When all three are positive integers case 0: return (sum - largest) * largest; // For single negative integer case 1: return (product / smallest) + smallest; // For two negative integers case 2: return (product / largest) + largest; // For three negative integers case 3: return (product / largest) + largest; } } // Driver Code int main() { int a = 2, b = 1, c = 4; cout << maximumResult(a, b, c); return 0; } |
chevron_right
filter_none
Java
// Java implementation of the approach class GFG { // Function to return the maximum result static int maximumResult(int a, int b, int c) { // To store the count of negative integers int countOfNegative = 0; // Sum of all the three integers int sum = a + b + c; // Product of all the three integers int product = a * b * c; // To store the smallest and the largest // among all the three integers int largest = (a > b) ? ((a > c) ? a : c) : ((b > c) ? b : c); int smallest=1; // Calculate the count of negative integers if (a < 0) countOfNegative++; if (b < 0) countOfNegative++; if (c < 0) countOfNegative++; // Depending upon count of negatives switch (countOfNegative) { // When all three are positive integers case 0: return (sum - largest) * largest; // For single negative integer case 1: return (product / smallest) + smallest; // For two negative integers case 2: return (product / largest) + largest; // For three negative integers case 3: return (product / largest) + largest; } return -1; } // Driver Code public static void main(String[] args) { int a = 2, b = 1, c = 4; System.out.print(maximumResult(a, b, c)); } } // This code contributed by Rajput-Ji |
chevron_right
filter_none
Python3
# Python3 implementation of the approach # Function to return the maximum result def maximumResult(a, b, c): # To store the count of negative integers countOfNegative = 0 # Sum of all the three integers Sum = a + b + c # Product of all the three integers product = a * b * c # To store the smallest and the # largest among all the three integers largest = max(a, b, c) smallest = min(a, b, c) # Calculate the count of negative integers if a < 0: countOfNegative += 1 if b < 0: countOfNegative += 1 if c < 0: countOfNegative += 1 # When all three are positive integers if countOfNegative == 0: return (Sum - largest) * largest # For single negative integer elif countOfNegative == 1: return (product // smallest) + smallest # For two negative integers elif countOfNegative == 2: return (product // largest) + largest # For three negative integers elif countOfNegative == 3: return (product // largest) + largest # Driver Code if __name__ == "__main__": a, b, c = 2, 1, 4 print(maximumResult(a, b, c)) # This code is contributed by Rituraj Jain |
chevron_right
filter_none
C#
// C# implementation of the approach using System; class GFG { // Function to return the maximum result static int maximumResult(int a, int b, int c) { // To store the count of negative integers int countOfNegative = 0; // Sum of all the three integers int sum = a + b + c; // Product of all the three integers int product = a * b * c; // To store the smallest and the largest // among all the three integers int largest = (a > b) ? ((a > c) ? a : c) : ((b > c) ? b : c); int smallest=1; // Calculate the count of negative integers if (a < 0) countOfNegative++; if (b < 0) countOfNegative++; if (c < 0) countOfNegative++; // Depending upon count of negatives switch (countOfNegative) { // When all three are positive integers case 0: return (sum - largest) * largest; // For single negative integer case 1: return (product / smallest) + smallest; // For two negative integers case 2: return (product / largest) + largest; // For three negative integers case 3: return (product / largest) + largest; } return -1; } // Driver Code static void Main() { int a = 2, b = 1, c = 4; Console.WriteLine(maximumResult(a, b, c)); } } // This code is contributed by mits |
chevron_right
filter_none
PHP
Output:
12
Recommended Posts:
- Expression Evaluation
- Maximize the value of x + y + z such that ax + by + cz = n
- Maximize the sum of arr[i]*i
- Find all possible outcomes of a given expression
- Rearrange an array to maximize i*arr[i]
- Maximize a value for a semicircle of given radius
- Maximize array sum after K negations | Set 2
- Maximize array sum after K negations | Set 1
- Print the balanced bracket expression using given brackets
- Find a permutation of 2N numbers such that the result of given expression is exactly 2K
- Maximum and Minimum Values of an Algebraic Expression
- Maximize volume of cuboid with given sum of sides
- Maximize the product of four factors of a Number
- Maximize the profit by selling at-most M products
- Maximize sum of consecutive differences in a circular array
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
- Smallest multiple of N formed using the given set of digits
- Find the Kth smallest element in the sorted generated array
- Length of the longest substring with no consecutive same letters
- Select numbers in such way to maximize the amount of money
- Maximize the size of array by deleting exactly k sub-arrays to make array prime
