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Maximum value of (arr[i] * arr[j]) + (arr[j] – arr[i])) possible for any pair in an array
  • Last Updated : 04 May, 2021

Given an array arr[] consisting of N integers, the task is to find the maximum possible value of the expression (arr[i] * arr[j]) + (arr[j] – arr[i])) for any pair (i, j), such that i ≠ j and 0 ≤ (i, j) < N.

Examples:

Input: arr[] = {-2, -8, 0, 1, 2, 3, 4}
Output: 22
Explanation:
For the pair (-8, -2) the value of the expression (arr[i] * arr[j]) + (arr[j] – arr[i])) = ((-2)*(-8) + (-2 – (-8))) = 22, which is maximum among all possible pairs.

Input: arr[] = {-47, 0, 12}
Output: 47

Naive Approach: The simplest approach to solve the given problem is to generate all possible pairs of the array and find the value of the given expression (arr[i] * arr[j]) + (arr[j] – arr[i])) for each pair and print the maximum value obtained for any pair. 



Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be optimized, which is based on the observation that the maximum value of the expression will be obtained by selecting the pairs of maximum and second maximum or a minimum and second minimum of the array. Follow the steps below to solve the problem:

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the value of
// the expression a * b + (b - a)
int calc(int a, int b)
{
    return a * b + (b - a);
}
 
// Function to find the maximum value
// of the expression a * b + (b - a)
// possible for any pair (a, b)
int findMaximum(vector<int> arr, int N)
{
    // Sort the vector in ascending order
    sort(arr.begin(), arr.end());
 
    // Stores the maximum value
    int ans = -1e9;
 
    // Update ans by choosing the pair
    // of the minimum and 2nd minimum
    ans = max(ans, calc(arr[0], arr[1]));
 
    // Update ans by choosing the pair
    // of maximum and 2nd maximum
    ans = max(ans, calc(arr[N - 2],
                        arr[N - 1]));
 
    // Return the value of ans
    return ans;
}
 
// Driver Code
int main()
{
    vector<int> arr = { 0, -47, 12 };
    int N = (int)arr.size();
    cout << findMaximum(arr, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG {
     
// Function to find the value of
// the expression a * b + (b - a)
static int calc(int a, int b)
{
    return a * b + (b - a);
}
 
// Function to find the maximum value
// of the expression a * b + (b - a)
// possible for any pair (a, b)
static int findMaximum(int[] arr, int N)
{
     
    // Sort the vector in ascending order
    Arrays.sort(arr);
 
    // Stores the maximum value
    int ans = (int)-1e9;
 
    // Update ans by choosing the pair
    // of the minimum and 2nd minimum
    ans = Math.max(ans, calc(arr[0], arr[1]));
 
    // Update ans by choosing the pair
    // of maximum and 2nd maximum
    ans = Math.max(ans, calc(arr[N - 2],
                             arr[N - 1]));
 
    // Return the value of ans
    return ans;
}   
 
// Driver Code
public static void main(String[] args)
{
     
    // Given inputs
    int[] arr = { 0, -47, 12 };
    int N = arr.length;
     
    System.out.println(findMaximum(arr, N));
}
}
 
// This code is contributed by offbeat

Python3




# Python3 program for the above approach
 
# Function to find the value of
# the expression a * b + (b - a)
def calc(a, b):
    return a * b + (b - a)
 
# Function to find the maximum value
# of the expression a * b + (b - a)
# possible for any pair (a, b)
def findMaximum(arr, N):
   
    # Sort the vector in ascending order
    arr = sorted(arr)
 
    # Stores the maximum value
    ans = -10**9
 
    # Update ans by choosing the pair
    # of the minimum and 2nd minimum
    ans = max(ans, calc(arr[0], arr[1]))
 
    # Update ans by choosing the pair
    # of maximum and 2nd maximum
    ans = max(ans, calc(arr[N - 2],arr[N - 1]))
 
    # Return the value of ans
    return ans
 
# Driver Code
if __name__ == '__main__':
    arr = [0, -47, 12]
    N = len(arr)
    print (findMaximum(arr, N))
 
# This code is contributed by mohit kumar 29.

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
   
    // Function to find the value of
    // the expression a * b + (b - a)
    static int calc(int a, int b)
    {
        return a * b + (b - a);
    }
 
    // Function to find the maximum value
    // of the expression a * b + (b - a)
    // possible for any pair (a, b)
    static int findMaximum(List<int> arr, int N)
    {
       
        // Sort the vector in ascending order
        arr.Sort();
 
        // Stores the maximum value
        int ans = -1000000000;
 
        // Update ans by choosing the pair
        // of the minimum and 2nd minimum
        ans = Math.Max(ans, calc(arr[0], arr[1]));
 
        // Update ans by choosing the pair
        // of maximum and 2nd maximum
        ans = Math.Max(ans, calc(arr[N - 2], arr[N - 1]));
 
        // Return the value of ans
        return ans;
    }
 
    // Driver Code
    public static void Main()
    {
        List<int> arr = new List<int>{ 0, -47, 12 };
        int N = (int)arr.Count;
        Console.Write(findMaximum(arr, N));
    }
}
 
// This code is contributed by ukasp..
Output: 
47

 

Time Complexity: O(N)
Auxiliary Space: O(1)

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