Maximize sum of product and difference between any pair of array elements possible
Last Updated :
23 Jul, 2025
Given an array arr[] of size N, the task is to find the maximum value of arr[i] ∗ arr[j] + arr[i] − arr[j] for any pair (arr[i], arr[j]) from the given array, where i != j and 0 < i, j < N - 1.
Examples:
Input: arr[] = {1, 2, 3, 4, 5}
Output: 21
Explanation:
Among all the pairs of the array, the maximum value is obtained for the pair (arr[4], arr[3]), which is equal to
=> arr[4] * arr[3] + arr[4] - arr[3] = 5 * 4 + 5 - 4 = 20 + 1 = 21.
Input: {-4, -5, 0, 1, 3}
Output: 21
Explanation:
Among all the pairs of the array, the maximum value is obtained for the pair (arr[0], arr[1]), which is equal to
=> arr[0] * arr[1] + arr[0] - arr[1] = (-4) * (-5) + (-4) - (-5) = 20 + 1 = 21.
Naive Approach: The simplest approach to solve the problem is to traverse the array and generate all possible pairs (arr[i], arr[j]) (i != j) from the array and evaluate the expression for all the pairs. Finally, print the maximum of all the pairs.
Below is the implementation of the approach:
C++
// C++ code for the approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the maximum value of arr[i] * arr[j] + arr[i] - arr[j]
int maxProductSum(int arr[], int N) {
int maxSum = INT_MIN;
// Generate all possible pairs (arr[i], arr[j])
for (int i = 0; i < N - 1; i++) {
for (int j = i + 1; j < N; j++) {
// Evaluate the expression for each pair
// consider two permutation of pair for any 2 particular elements
int sum1 = (arr[i] * arr[j]) + (arr[i] - arr[j]);
int sum2 = (arr[j] * arr[i]) + (arr[j] - arr[i]);
// taking maximum expression value among two permutations
int sum = max(sum1, sum2);
// Update the maximum value
maxSum = max(maxSum, sum);
}
}
// Return the maximum value
return maxSum;
}
// Driver code
int main() {
int arr[] = { -4, -5, 0, 1, 3 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function call to find the maximum value
int maxSum = maxProductSum(arr, N);
// Print the maximum value
cout << maxSum << endl;
return 0;
}
// Java code for the approach
import java.util.*;
public class GFG {
// Function to find the maximum value of arr[i] * arr[j]
// + arr[i] - arr[j]
static int maxProductSum(int[] arr, int N)
{
int maxSum = Integer.MIN_VALUE;
// Generate all possible pairs (arr[i], arr[j])
for (int i = 0; i < N - 1; i++) {
for (int j = i + 1; j < N; j++) {
// Evaluate the expression for each pair
// consider two permutation of pair for any
// 2 particular elements
int sum1
= (arr[i] * arr[j]) + (arr[i] - arr[j]);
int sum2
= (arr[j] * arr[i]) + (arr[j] - arr[i]);
// taking maximum expression value among two
// permutations
int sum = Math.max(sum1, sum2);
// Update the maximum value
maxSum = Math.max(maxSum, sum);
}
}
// Return the maximum value
return maxSum;
}
// Driver code
public static void main(String[] args)
{
int[] arr = { -4, -5, 0, 1, 3 };
int N = arr.length;
// Function call to find the maximum value
int maxSum = maxProductSum(arr, N);
// Print the maximum value
System.out.println(maxSum);
}
}
# Python3 code for the approach
import sys
# Function to find the maximum value of arr[i] * arr[j] + arr[i] - arr[j]
def maxProductSum(arr, N):
maxSum = -sys.maxsize
# Generate all possible pairs (arr[i], arr[j])
for i in range(N - 1):
for j in range(i + 1, N):
# Evaluate the expression for each pair
sum1 = (arr[i] * arr[j]) + (arr[i] - arr[j])
sum2 = (arr[j] * arr[i]) + (arr[j] - arr[i])
# Taking maximum expression value among two permutations
sum = max(sum1, sum2)
# Update the maximum value
maxSum = max(maxSum, sum)
# Return the maximum value
return maxSum
# Driver code
if __name__ == '__main__':
# Input array
arr = [-4, -5, 0, 1, 3]
N = len(arr)
# Function call to find the maximum value
maxSum = maxProductSum(arr, N)
# Print the maximum value
print(maxSum)
using System;
public class GFG
{
// Function to find the maximum value of arr[i] * arr[j]
// + arr[i] - arr[j]
static int MaxProductSum(int[] arr, int N)
{
int maxSum = int.MinValue;
// Generate all possible pairs (arr[i], arr[j])
for (int i = 0; i < N - 1; i++)
{
for (int j = i + 1; j < N; j++)
{
// Evaluate the expression for each pair
// consider two permutation of pair for any
// 2 particular elements
int sum1 = (arr[i] * arr[j]) + (arr[i] - arr[j]);
int sum2 = (arr[j] * arr[i]) + (arr[j] - arr[i]);
// taking maximum expression value among two
// permutations
int sum = Math.Max(sum1, sum2);
// Update the maximum value
maxSum = Math.Max(maxSum, sum);
}
}
// Return the maximum value
return maxSum;
}
// Driver code
static void Main(string[] args)
{
int[] arr = { -4, -5, 0, 1, 3 };
int N = arr.Length;
// Function call to find the maximum value
int maxSum = MaxProductSum(arr, N);
// Print the maximum value
Console.WriteLine(maxSum);
}
}
// Function to find the maximum value of arr[i] * arr[j] + arr[i] - arr[j]
function maxProductSum(arr, N) {
let maxSum = -Infinity;
// Generate all possible pairs (arr[i], arr[j])
for (let i = 0; i < N - 1; i++) {
for (let j = i + 1; j < N; j++) {
// Evaluate the expression for each pair
const sum1 = arr[i] * arr[j] + arr[i] - arr[j];
const sum2 = arr[j] * arr[i] + arr[j] - arr[i];
// Taking maximum expression value among two permutations
const sum = Math.max(sum1, sum2);
// Update the maximum value
maxSum = Math.max(maxSum, sum);
}
}
// Return the maximum value
return maxSum;
}
// Driver code
const arr = [-4, -5, 0, 1, 3];
const N = arr.length;
// Function call to find the maximum value
const maxSum = maxProductSum(arr, N);
// Print the maximum value
console.log(maxSum);
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: The optimal idea is based on the observation that the value of the expression can be maximum for the following 2 cases:
Follow the steps below to solve the problem:
- Sort the array arr[] in ascending order.
- Evaluate the expression for the pair arr[N - 1] and arr[N - 2] and store it in a variable, say max1.
- Similarly, evaluate the expression for the pair arr[1] and arr[0] and store it in a variable, say max2.
- Store the maximum of max1 and max2 in a variable, say ans.
- Print the value of ans as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to evaluate given expression
int compute(int a, int b)
{
// Store the result
int ans = a * b + a - b;
return ans;
}
// Function to find the maximum value of
// the given expression possible for any
// unique pair from the given array
void findMaxValue(int arr[], int N)
{
// Sort the array in ascending order
sort(arr, arr + N);
// Evaluate the expression for
// the two largest elements
int maxm = compute(arr[N - 1], arr[N - 2]);
// Evaluate the expression for
// the two smallest elements
maxm = max(maxm, compute(arr[1], arr[0]));
// Print the maximum
cout << maxm;
}
// Driver Code
int main()
{
// Given array
int arr[] = { -4, -5, 0, 1, 3 };
// Store the size of the array
int N = sizeof(arr) / sizeof(arr[0]);
findMaxValue(arr, N);
return 0;
}
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG
{
// Function to evaluate given expression
static int compute(int a, int b)
{
// Store the result
int ans = a * b + a - b;
return ans;
}
// Function to find the maximum value of
// the given expression possible for any
// unique pair from the given array
static void findMaxValue(int arr[], int N)
{
// Sort the array in ascending order
Arrays.sort(arr);
// Evaluate the expression for
// the two largest elements
int maxm = compute(arr[N - 1], arr[N - 2]);
// Evaluate the expression for
// the two smallest elements
maxm = Math.max(maxm, compute(arr[1], arr[0]));
// Print the maximum
System.out.print(maxm);
}
// Driver Code
public static void main(String[] args)
{
// Given array
int arr[] = { -4, -5, 0, 1, 3 };
// Store the size of the array
int N = arr.length;
findMaxValue(arr, N);
}
}
// This code is contributed by saanjoy_62
# Python Program for the above approach
# Function to evaluate given expression
def compute(a, b):
# Store the result
res = (a * b) + (a - b)
return res
# Function to find the maximum value of
# the given expression possible for any
# unique pair from the given array
def findMaxValue(arr, N):
# Sort the list in ascending order
arr.sort()
# Evaluate the expression for
# the two largest elements
maxm = compute(arr[N - 1], arr[N - 2])
# Evaluate the expression for
# the two smallest elements
maxm = max(maxm, compute(arr[1], arr[0]));
print(maxm)
# Driver code
# given list
arr = [-4, -5, 0, 1, 3]
# store the size of the list
N = len(arr)
findMaxValue(arr, N)
# This code is contributed by santhoshcharan.
// C# program for above approach
using System;
public class GFG
{
// Function to evaluate given expression
static int compute(int a, int b)
{
// Store the result
int ans = a * b + a - b;
return ans;
}
// Function to find the maximum value of
// the given expression possible for any
// unique pair from the given array
static void findMaxValue(int[] arr, int N)
{
// Sort the array in ascending order
Array.Sort(arr);
// Evaluate the expression for
// the two largest elements
int maxm = compute(arr[N - 1], arr[N - 2]);
// Evaluate the expression for
// the two smallest elements
maxm = Math.Max(maxm, compute(arr[1], arr[0]));
// Print the maximum
Console.WriteLine(maxm);
}
// Driver code
public static void Main(String[] args)
{
// Given array
int[] arr = { -4, -5, 0, 1, 3 };
// Store the size of the array
int N = arr.Length;
findMaxValue(arr, N);
}
}
// This code is contributed by susmitakundugoaldanga.
<script>
// Javascript implementation of the above approach
// Function to evaluate given expression
function compute(a, b)
{
// Store the result
var ans = a * b + a - b;
return ans;
}
// Function to find the maximum value of
// the given expression possible for any
// unique pair from the given array
function findMaxValue(arr, N)
{
// Sort the array in ascending order
arr.sort(function(a,b){return a - b});
// Evaluate the expression for
// the two largest elements
var maxm = compute(arr[N - 1], arr[N - 2]);
// Evaluate the expression for
// the two smallest elements
maxm = Math.max(maxm, compute(arr[1], arr[0]));
// Print the maximum
document.write(maxm);
}
// Driver Code
var arr = [-4, -5, 0, 1, 3];
var N = arr.length;
findMaxValue(arr, N);
// This code is contributed by Shubhamsingh10
</script>
Time Complexity: O(N*log(N))
Auxiliary Space: O(1)
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