# Maximum sum of pairwise product in an array with negative allowed

Given an array of n elements. Find maximum sum of pairwise multiplications. Sum can be larger so take mod with 10^9+7. If there are odd elements, then we can add any one element (without forming a pair) to the sum.

Examples:

Input : arr[] = {-1, 4, 5, -7, -4, 9, 0}
Output : 77
So to get the maximum sum, the arrangement will
be {-7, -4}, {-1, 0}, {9, 5} and {4}.
So the answer is (-7*(-4))+((-1)*0)+(9*5)+(4) ={77}.

Input : arr[] = {8, 7, 9}
Output : 79
Answer is (9*8) +(7) = 79.

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

1- Sort the given array.
2- First, multiply the negative numbers pairwise from the starting and add to the total_sum.
3- Second, multiply the positive numbers pairwise from the last and to the total_sum.
4- Check if negative and positive both counts are odd, then add the product of last pair
i.e. last negative and positive left.
5- Or if any of the one counts is odd, then add that element left.
6- Return sum.

 // C++ program for above implementation #include #define Mod 1000000007 using namespace std;    // Function to find the maximum sum long long int findSum(int arr[], int n) {     long long int sum = 0;        // Sort the array first     sort(arr, arr + n);        // First multiply negative numbers pairwise     // and sum up from starting as to get maximum      // sum.      int i = 0;     while (i < n && arr[i] < 0) {         if (i != n - 1 && arr[i + 1] <= 0) {             sum = (sum + (arr[i] * arr[i + 1]) % Mod) % Mod;             i += 2;         }         else             break;     }        // Second multiply positive numbers pairwise     // and summed up from the last as to get maximum      // sum.     int j = n - 1;     while (j >= 0 && arr[j] > 0) {         if (j != 0 && arr[j - 1] > 0) {             sum = (sum + (arr[j] * arr[j - 1]) % Mod) % Mod;             j -= 2;         }         else             break;     }        // To handle case if positive and negative     // numbers both are odd in counts.     if (j > i)         sum = (sum + (arr[i] * arr[j]) % Mod) % Mod;        // If one of them occurs odd times     else if (i == j)         sum = (sum + arr[i]) % Mod;        return sum; }    // Drivers code int main() {     int arr[] = { -1, 9, 4, 5, -4, 7 };     int n = sizeof(arr) / sizeof(arr[0]);     cout << findSum(arr, n);     return 0; }

 // Java program for above implementation import java.io.*; import java.util.*;    class GFG {    static int Mod = 1000000007;    // Function to find the maximum sum static long findSum(int arr[], int n) {     long sum = 0;        // Sort the array first     Arrays.sort(arr);        // First multiply negative numbers      // pairwise and sum up from starting      // as to get maximum sum.     int i = 0;     while (i < n && arr[i] < 0) {     if (i != n - 1 && arr[i + 1] <= 0) {         sum = (sum + (arr[i] * arr[i + 1]) % Mod) % Mod;         i += 2;     }      else         break;     }        // Second multiply positive numbers      // pairwise and summed up from the      // last as to get maximum sum.     int j = n - 1;     while (j >= 0 && arr[j] > 0) {     if (j != 0 && arr[j - 1] > 0) {         sum = (sum + (arr[j] * arr[j - 1]) % Mod) % Mod;         j -= 2;     } else         break;     }        // To handle case if positive and negative     // numbers both are odd in counts.     if (j > i)     sum = (sum + (arr[i] * arr[j]) % Mod) % Mod;        // If one of them occurs odd times     else if (i == j)     sum = (sum + arr[i]) % Mod;        return sum; }    // Drivers code public static void main(String args[]) {     int arr[] = {-1, 9, 4, 5, -4, 7};     int n = arr.length;     System.out.println(findSum(arr, n)); } }    /*This code is contributed by Nikita Tiwari.*/

 # Python3 code for above implementation Mod= 1000000007    # Function to find the maximum sum def findSum(arr, n):     sum = 0            # Sort the array first     arr.sort()            # First multiply negative numbers      # pairwise and sum up from starting     # as to get maximum sum.     i = 0     while i < n and arr[i] < 0:         if i != n - 1 and arr[i + 1] <= 0:             sum = (sum + (arr[i] * arr[i + 1])                                   % Mod) % Mod             i += 2         else:             break                # Second multiply positive numbers     # pairwise and summed up from the      # last as to get maximum sum.     j = n - 1     while j >= 0 and arr[j] > 0:         if j != 0 and arr[j - 1] > 0:             sum = (sum + (arr[j] * arr[j - 1])                                  % Mod) % Mod             j -= 2         else:             break                # To handle case if positive      # and negative numbers both     # are odd in counts.     if j > i:         sum = (sum + (arr[i] * arr[j]) % Mod)                                         % Mod                # If one of them occurs odd times     elif i == j:         sum = (sum + arr[i]) % Mod            return sum    # Driver code arr = [ -1, 9, 4, 5, -4, 7 ] n = len(arr)  print(findSum(arr, n))    # This code is contributed by "Sharad_Bhardwaj".

 // C# program for above implementation using System;    class GFG {        static int Mod = 1000000007;        // Function to find the maximum sum     static long findSum(int[] arr, int n)     {         long sum = 0;            // Sort the array first         Array.Sort(arr);            // First multiply negative numbers         // pairwise and sum up from starting         // as to get maximum sum.         int i = 0;         while (i < n && arr[i] < 0) {             if (i != n - 1 && arr[i + 1] <= 0) {                 sum = (sum + (arr[i] * arr[i + 1]) % Mod) % Mod;                 i += 2;             }             else                 break;         }            // Second multiply positive numbers         // pairwise and summed up from the         // last as to get maximum sum.         int j = n - 1;         while (j >= 0 && arr[j] > 0) {             if (j != 0 && arr[j - 1] > 0) {                 sum = (sum + (arr[j] * arr[j - 1]) % Mod) % Mod;                 j -= 2;             }             else                 break;         }            // To handle case if positive and negative         // numbers both are odd in counts.         if (j > i)             sum = (sum + (arr[i] * arr[j]) % Mod) % Mod;            // If one of them occurs odd times         else if (i == j)             sum = (sum + arr[i]) % Mod;            return sum;     }        // Drivers code     public static void Main()     {         int[] arr = { -1, 9, 4, 5, -4, 7 };         int n = arr.Length;         Console.WriteLine(findSum(arr, n));     } }    /*This code is contributed by vt_m.*/

 = 0 && \$arr[\$j] > 0)     {         if (\$j != 0 && \$arr[\$j - 1] > 0)         {             \$sum = (\$sum + (\$arr[\$j] *                      \$arr[\$j - 1]) % \$Mod) % \$Mod;             \$j -= 2;         }         else             break;     }        // To handle case if positive and negative     // numbers both are odd in counts.     if (\$j > \$i)         \$sum = (\$sum + (\$arr[\$i] *                  \$arr[\$j]) % \$Mod) % \$Mod;        // If one of them occurs odd times     else if (\$i == \$j)         \$sum = (\$sum + \$arr[\$i]) % Mod;        return \$sum; }    // Driver code \$arr = array (-1, 9, 4, 5, -4, 7 ); \$n = sizeof(\$arr); echo findSum(\$arr, \$n);    // This code is contributed by ita_c ?>

Output:
87

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