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.
Approach:
- Sort the given array.
- First, multiply the negative numbers pairwise from the starting and add to the total_sum.
- Second, multiply the positive numbers pairwise from the last and to the total_sum.
- Check if negative and positive both counts are odd, then add the product of last pair
i.e. last negative and positive left. - Or if any of the one counts is odd, then add that element left.
- Return sum.
Implementation:
C++
// C++ program for above implementation#include <bits/stdc++.h>#define Mod 1000000007using namespace std;// Function to find the maximum sumlong 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 codeint main(){ int arr[] = { -1, 9, 4, 5, -4, 7 }; int n = sizeof(arr) / sizeof(arr[0]); cout << findSum(arr, n); return 0;} |
Java
// Java program for above implementationimport java.io.*;import java.util.*;class GFG {static int Mod = 1000000007;// Function to find the maximum sumstatic 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 codepublic 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
# Python3 code for above implementationMod= 1000000007# Function to find the maximum sumdef 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 codearr = [ -1, 9, 4, 5, -4, 7 ]n = len(arr) print(findSum(arr, n))# This code is contributed by "Sharad_Bhardwaj". |
C#
// C# program for above implementationusing 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.*/ |
PHP
<?php// PHP program for above implementation$Mod = 1000000007;// Function to find the maximum sumfunction findSum(&$arr, $n){ global $Mod; $sum = 0; // Sort the array first sort($arr); // First multiply negative numbers // pairwise and sum up from starting // as to get maximum sum. $i = 0; while ($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. $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 $sum;}// Driver code$arr = array (-1, 9, 4, 5, -4, 7 );$n = sizeof($arr);echo findSum($arr, $n);// This code is contributed by ita_c?> |
Javascript
<script>// JavaScript program for above implementation let Mod = 1000000007; // Function to find the maximum sumfunction findSum(arr, n) { let sum = 0; // Sort the array first arr.sort(); // First multiply negative numbers // pairwise and sum up from starting // as to get maximum sum. let 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. let 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;}// Driver code let arr = [-1, 9, 4, 5, -4, 7]; let n = arr.length; document.write(findSum(arr, n)); </script> |
Output
87
Time Complexity : O(N log(N))
Auxiliary Space: O(1)
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