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: add the two largest integers and multiply them to smallest one. case 3-: (sum – smallest) * smallest
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 = max(a,max(b,c)); int smallest = min(a,min(b,c) ); // 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 (sum - smallest) * smallest; } } // Driver Code int main() { int a=-2,b=-1,c=-4; cout << maximumResult(a, b, c); return 0; } // This code contributed by Nikhil |
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= (a<b)?((a<c)? a : c):((b<c) ? b : c); // 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 (sum - smallest) * smallest; } 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 Nikhil |
Python3
# Python3 implementation of the approach # Function to return the maximum result # 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 (Sum - smallest) * smallest# Driver Code if __name__ == "__main__": a, b, c = -2, -1, -4 print(maximumResult(a, b, c)) |
C#
// C# implementation of the approachusing System;class GFG{ // Function to return the maximum resultstatic 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=(a<b)?((a<c)? a : c):((b<c) ? b : c); // 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 (sum - smallest) * smallest; } return -1;}// Driver Codestatic void Main(){ int a = -2, b = -1, c = -4; Console.WriteLine(maximumResult(a, b, c));}}// This code is contributed by mits & Nikhil |
PHP
<?php// PHP implementation of the approach// Function to return the maximum resultfunction 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) $countOfNegative++; if ($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) + $largest; // For three negative integers case 3: return ($sum - $smallest) * $smallest; }}// Driver Code$a = -2;$b = -1;$c = -4;echo maximumResult($a, $b, $c);// This code is contributed by ihritik?> |
Javascript
<script>// JavaScript implementation of the approach // Function to return the maximum result function maximumResult(a, b, c) { // To store the count of negative integers let countOfNegative = 0; // Sum of all the three integers let sum = a + b + c; // Product of all the three integers let product = a * b * c; // To store the smallest and the largest // among all the three integers let largest = Math.max(a,Math.max(b,c)); let smallest = Math.min(a,Math.min(b,c) ); // 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 (sum - smallest) * smallest; } } // Driver Code let a = -2, b = -1, c = -4; document.write(maximumResult(a, b, c)); // This code is contributed by Surbhi Tyagi.</script> |
Output:
12
Time Complexity: O(1)
Auxiliary Space: O(1)
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