Find a pair with maximum product in array of Integers

Given an array with both +ive and -ive integers, return a pair with highest product.

Examples:

Input: arr[] = {1, 4, 3, 6, 7, 0}  
Output: {6,7}  

Input: arr[] = {-1, -3, -4, 2, 0, -5} 
Output: {-4,-5} 

Source: http://qa.geeksforgeeks.org/446/return-a-pair-with-maximum-product-in-array-of-integers/

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A Simple Solution is to consider every pair and keep track maximum product. Below is C++ implementation of this simple solution.

// A simple C++ program to find max product pair in // an array of integers #include<bits/stdc++.h> using namespace std; // Function to find maximum product pair in arr[0..n-1] void maxProduct(int arr[], int n) {     if (n < 2)     {         cout << "No pairs exists\n";         return;     }     // Initialize max product pair     int a = arr[0], b = arr[1];     // Traverse through every possible pair     // and keep track of max product     for (int i=0; i<n; i++)       for (int j=i+1; j<n; j++)          if (arr[i]*arr[j] > a*b)             a = arr[i], b = arr[j];     cout << "Max product pair is {" << a << ", "          << b << "}"; } // Driver program to test int main() {     int arr[] = {1, 4, 3, 6, 7, 0};     int n = sizeof(arr)/sizeof(arr[0]);     maxProduct(arr, n);     return 0; }

Output:

Max product pair is {6, 7}

Time Complexity: O(n2)

A Better Solution is to use sorting. Below are detailed steps.
1) Sort input array in increasing order.
2) If all elements are positive, then return product of last two numbers.
3) Else return maximum of products of first two and last two numbers.
Time complexity of this solution is O(nLog n). Thanks to Rahul Jain for suggesting this method.

An Efficient Solution can solve the above problem in single traversal of input array. The idea is to traverse the input array and keep track of following four values.
a) Maximum positive value
b) Second maximum positive value
c) Maximum negative value i.e., a negative value with maximum absolute value
d) Second maximum negative value.
At the end of the loop, compare the products of first two and last two and print the maximum of two products. Below is C++ implementation of this idea.

// A O(n) C++ program to find maximum product pair in an array #include<bits/stdc++.h> using namespace std; // Function to find maximum product pair in arr[0..n-1] void maxProduct(int arr[], int n) {     if (n < 2)     {         cout << "No pairs exists\n";         return;     }     if (n == 2)     {         cout << arr[0] << " " << arr[1] << endl;         return;     }     // Iniitialize maximum and second maximum     int posa = INT_MIN, posb = INT_MIN;     // Iniitialize minimum and second minimum     int nega = INT_MIN, negb = INT_MIN;     // Traverse given array     for (int i = 0; i < n; i++)     {         // Update maximum and second maximum if needed         if (arr[i] > posa)         {             posb = posa;             posa = arr[i];         }         else if (arr[i] > posb)             posb = arr[i];         // Update minimum and second minimum if needed         if (arr[i] < 0 && abs(arr[i]) > abs(nega))         {             negb = nega;             nega = arr[i];         }         else if(arr[i] < 0 && abs(arr[i]) > abs(negb))             negb = arr[i];     }     if (nega*negb > posa*posb)         cout << "Max product pair is {" << nega << ", "              << negb << "}";     else         cout << "Max product pair is {" << posa << ", "              << posb << "}"; } // Driver program to test above function int main() {     int arr[] = {1, 4, 3, 6, 7, 0};     int n = sizeof(arr)/sizeof(arr[0]);     maxProduct(arr, n);     return 0; }

Output:

Max product pair is {6, 7}

Time complexity: O(n)
Auxiliary Space: O(1)