Given an array with both +ive and -ive integers, return a pair with the 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}
A Simple Solution is to consider every pair and keep track of the maximum product. Below is the implementation of this simple solution.
C++
// 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; } |
Java
// JAVA Code to Find a pair with maximum // product in array of Integers import java.util.*; class GFG { // Function to find maximum product pair // in arr[0..n-1] static void maxProduct(int arr[], int n) { if (n < 2) { System.out.println("No pairs exists"); 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]; } System.out.println("Max product pair is {" + a + ", " + b + "}"); } /* Driver program to test above function */ public static void main(String[] args) { int arr[] = {1, 4, 3, 6, 7, 0}; int n = arr.length; maxProduct(arr, n); } } // This code is contributed by Arnav Kr. Mandal. |
Python3
# A simple Python3 program to find max # product pair in an array of integers # Function to find maximum # product pair in arr[0..n-1] def maxProduct(arr, n): if (n < 2): print("No pairs exists") return # Initialize max product pair a = arr[0]; b = arr[1] # Traverse through every possible pair # and keep track of max product for i in range(0, n): for j in range(i + 1, n): if (arr[i] * arr[j] > a * b): a = arr[i]; b = arr[j] print("Max product pair is {", a, ",", b, "}", sep = "") # Driver Code arr = [1, 4, 3, 6, 7, 0] n = len(arr) maxProduct(arr, n) # This code is contributed by Smitha Dinesh Semwal. |
C#
// C# Code to Find a pair with maximum // product in array of Integers using System; class GFG { // Function to find maximum // product pair in arr[0..n-1] static void maxProduct(int []arr, int n) { if (n < 2) { Console.Write("No pairs exists"); 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]; } Console.Write("Max product pair is {" + a + ", " + b + "}"); } // Driver Code public static void Main() { int []arr = {1, 4, 3, 6, 7, 0}; int n = arr.Length; maxProduct(arr, n); } } // This code is contributed by nitin mittal. |
PHP
<?php // A simple PHP program to // find max product pair in // an array of integers // Function to find maximum // product pair in arr[0..n-1] function maxProduct( $arr, $n) { if ($n < 2) { echo "No pairs exists\n"; return; } // Initialize max product pair $a = $arr[0]; $b = $arr[1]; // Traverse through every possible pair // and keep track of max product for ($i = 0; $i < $n; $i++) for ($j = $i + 1; $j < $n; $j++) { if ($arr[$i] * $arr[$j] > $a * $b) { $a = $arr[$i]; $b = $arr[$j]; } } echo "Max product pair is {" , $a , ", "; echo $b , "}"; } // Driver Code $arr = array(1, 4, 3, 6, 7, 0); $n = count($arr); maxProduct($arr, $n); // This code is contributed by anuj_67. ?> |
Javascript
<script> // A simple Javascript program to find max product pair in // an array of integers // Function to find maximum product pair in arr[0..n-1] function maxProduct(arr, n) { if (n < 2) { document.write("No pairs exists" + "<br>"); return; } // Initialize max product pair let a = arr[0], b = arr[1]; // Traverse through every possible pair // and keep track of max product for (let i=0; i<n; i++) for (let j=i+1; j<n; j++) if (arr[i]*arr[j] > a*b) a = arr[i], b = arr[j]; document.write("Max product pair is {" + a + ", " + b + "}"); } // Driver program to test let arr = [1, 4, 3, 6, 7, 0]; let n = arr.length; maxProduct(arr, n); // This code is contributed by Mayank Tyagi </script> |
Output
Max product pair is {6, 7}
Time Complexity: O(n2)
Auxiliary Space: O(1)
Better Approach:
A Better Solution is to use sorting. Below are detailed steps.
- Sort input array in increasing order.
- If all elements are positive, then return the product of the last two numbers.
- Else return a maximum of products of the first two and last two numbers.
Thanks to Rahul Jain for suggesting this method.
C++14
// C++ code to find a pair with maximum // product in array of Integers #include<bits/stdc++.h> using namespace std; void maxProduct(vector<int>arr, int n) { // Sort the array sort(arr.begin(), arr.end()); int num1, num2; // Calculate product of two smallest numbers int sum1 = arr[0] * arr[1]; // Calculate product of two largest numbers int sum2 = arr[n - 1] * arr[n - 2]; // Print the pairs whose product is greater if (sum1 > sum2) { num1 = arr[0]; num2 = arr[1]; } else { num1 = arr[n - 2]; num2 = arr[n - 1]; } cout << ("Max product pair = ") << num1 << "," << num2; } // Driver Code int main() { vector<int>arr = { 1, 4, 3, 6, 7, 0 }; int n = arr.size(); maxProduct(arr, n); } // This code is contributed by Stream_Cipher |
Java
// JAVA Code to Find a pair with maximum // product in array of Integers import java.util.*; public class GFG { static void maxProduct(int arr[], int n) { // Sort the array Arrays.sort(arr); int num1, num2; // Calculate product of two smallest numbers int sum1 = arr[0] * arr[1]; // Calculate product of two largest numbers int sum2 = arr[n - 1] * arr[n - 2]; // print the pairs whose product is greater if (sum1 > sum2) { num1 = arr[0]; num2 = arr[1]; } else { num1 = arr[n - 2]; num2 = arr[n - 1]; } System.out.println("Max product pair = " + "{" + num1 + "," + num2 + "}"); } // Driver Code public static void main(String[] args) { int arr[] = { 1, 4, 3, 6, 7, 0 }; int n = arr.length; maxProduct(arr, n); } } // Contributed by Navtika Kumar |
Python3
# Python code to find a pair with maximum # product in array of Integers def maxProduct(arr, n): # Sort the array arr.sort() num1 = num2 = 0 # Calculate product of two smallest numbers sum1 = arr[0] * arr[1] # Calculate product of two largest numbers sum2 = arr[n - 1] * arr[n - 2] # Print the pairs whose product is greater if (sum1 > sum2): num1 = arr[0] num2 = arr[1] else: num1 = arr[n - 2] num2 = arr[n - 1] print("Max product pair = {", num1, ",", num2, "}", sep="") # Driver Code arr = [1, 4, 3, 6, 7, 0] n = len(arr) maxProduct(arr, n) # This code is contributed by subhammahato348. |
C#
// C# code to Find a pair with maximum // product in array of Integers using System; class GFG{ static void maxProduct(int []arr, int n) { // Sort the array Array.Sort(arr); int num1, num2; // Calculate product of two // smallest numbers int sum1 = arr[0] * arr[1]; // Calculate product of two // largest numbers int sum2 = arr[n - 1] * arr[n - 2]; // Print the pairs whose // product is greater if (sum1 > sum2) { num1 = arr[0]; num2 = arr[1]; } else { num1 = arr[n - 2]; num2 = arr[n - 1]; } Console.Write("Max product pair = " + "{" + num1 + "," + num2 + "}"); } // Driver Code public static void Main(String[] args) { int []arr = { 1, 4, 3, 6, 7, 0 }; int n = arr.Length; maxProduct(arr, n); } } // This code is contributed by shivanisinghss2110 |
Javascript
<script> // Javascript code to Find a pair with maximum // product in array of Integers function maxProduct(arr, n) { // Sort the array arr.sort(function(a, b){return a - b}); let num1, num2; // Calculate product of two smallest numbers let sum1 = arr[0] * arr[1]; // Calculate product of two largest numbers let sum2 = arr[n - 1] * arr[n - 2]; // print the pairs whose product is greater if (sum1 > sum2) { num1 = arr[0]; num2 = arr[1]; } else { num1 = arr[n - 2]; num2 = arr[n - 1]; } document.write("Max product pair = " + "{" + num1 + "," + num2 + "}"); } // Driver Code let arr = [ 1, 4, 3, 6, 7, 0 ]; let n = arr.length; maxProduct(arr, n); // This code is contributed by avanitrachhadiya2155 </script> |
Output
Max product pair = 6,7
Time Complexity: O(nlog n)
Auxiliary Space: O(1)
Efficient Approach:
An Efficient Solution can solve the above problem in a single traversal of the input array. The idea is to traverse the input array and keep track of the following four values.
- Maximum positive value
- Second maximum positive value
- Maximum negative value i.e., a negative value with maximum absolute value
- Second maximum negative value.
At the end of the loop, compare the products of the first two and last two and print the maximum of two products. Below is the implementation of this idea.
C++
// 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; } // Initialize maximum and second maximum int posa = INT_MIN, posb = INT_MIN; // Initialize 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; } |
Java
// JAVA Code to Find a pair with maximum // product in array of Integers import java.util.*; class GFG { // Function to find maximum product pair // in arr[0..n-1] static void maxProduct(int arr[], int n) { if (n < 2) { System.out.println("No pairs exists"); return; } if (n == 2) { System.out.println(arr[0] + " " + arr[1]); return; } // Initialize maximum and second maximum int posa = Integer.MIN_VALUE, posb = Integer.MIN_VALUE; // Initialize minimum and second minimum int nega = Integer.MIN_VALUE, negb = Integer.MIN_VALUE; // 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 && Math.abs(arr[i]) > Math.abs(nega)) { negb = nega; nega = arr[i]; } else if(arr[i] < 0 && Math.abs(arr[i]) > Math.abs(negb)) negb = arr[i]; } if (nega * negb > posa * posb) System.out.println("Max product pair is {" + nega + ", " + negb + "}"); else System.out.println("Max product pair is {" + posa + ", " + posb + "}"); } /* Driver program to test above function */ public static void main(String[] args) { int arr[] = {1, 4, 3, 6, 7, 0}; int n = arr.length; maxProduct(arr, n); } } // This code is contributed by Arnav Kr. Mandal. |
Python3
# A O(n) Python 3 program to find # maximum product pair in an array # Function to find maximum product # pair in arr[0..n-1] def maxProduct(arr, n): if (n < 2): print("No pairs exists") return if (n == 2): print(arr[0] ," " , arr[1]) return # Initialize maximum and # second maximum posa = 0 posb = 0 # Initialize minimum and # second minimum nega = 0 negb = 0 # Traverse given array for i in range(n): # Update maximum and second # maximum if needed if (arr[i] > posa): posb = posa posa = arr[i] elif (arr[i] > posb): posb = arr[i] # Update minimum and second # minimum if needed if (arr[i] < 0 and abs(arr[i]) > abs(nega)): negb = nega nega = arr[i] elif(arr[i] < 0 and abs(arr[i]) > abs(negb)): negb = arr[i] if (nega * negb > posa * posb): print("Max product pair is {" , nega ,", ", negb , "}") else: print( "Max product pair is {" , posa ,", " ,posb , "}") # Driver Code if __name__ =="__main__": arr = [1, 4, 3, 6, 7, 0] n = len(arr) maxProduct(arr, n) # This code is contributed # by ChitraNayal |
C#
// C# Code to Find a pair with maximum // product in array of Integers using System; class GFG { // Function to find maximum // product pair in arr[0..n-1] static void maxProduct(int []arr, int n) { if (n < 2) { Console.WriteLine("No pairs exists"); return; } if (n == 2) { Console.WriteLine(arr[0] + " " + arr[1]); return; } // Initialize maximum and // second maximum int posa = int.MinValue; int posb = int.MaxValue; // Initialize minimum and // second minimum int nega = int.MinValue; int negb = int.MaxValue; // 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 && Math.Abs(arr[i]) > Math.Abs(nega)) { negb = nega; nega = arr[i]; } else if(arr[i] < 0 && Math.Abs(arr[i]) > Math.Abs(negb)) negb = arr[i]; } if (nega * negb > posa * posb) Console.WriteLine("Max product pair is {" + nega + ", " + negb + "}"); else Console.WriteLine("Max product pair is {" + posa + ", " + posb + "}"); } // Driver Code public static void Main() { int []arr = {1, 4, 3, 6, 7, 0}; int n = arr.Length; maxProduct(arr, n); } } // This code is contributed by anuj_67. |
PHP
<?php // A O(n) PHP program to find maximum // product pair in an array // Function to find maximum product // pair in arr[0..n-1] function maxProduct(&$arr, $n) { if ($n < 2) { echo("No pairs exists\n"); return; } if ($n == 2) { echo ($arr[0]); echo (" "); echo($arr[1]); echo("\n"); return; } // Initialize maximum and // second maximum $posa = 0; $posb = 0; // Initialize minimum and // second minimum $nega = 0; $negb = 0; // Traverse given array for ($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) { echo("Max product pair is {"); echo $nega; echo(", "); echo ($negb); echo ("}"); } else { echo("Max product pair is {"); echo $posa; echo(", "); echo ($posb); echo ("}"); } } // Driver Code $arr = array(1, 4, 3, 6, 7, 0); $n = sizeof($arr); maxProduct($arr, $n); // This code is contributed // by Shivi_Aggarwal ?> |
Javascript
<script> // JAVAscript Code to Find a pair with maximum // product in array of Integers // Function to find maximum product pair // in arr[0..n-1] function maxProduct(arr,n) { if (n < 2) { document.write("No pairs exists"); return; } if (n == 2) { document.write(arr[0] + " " + arr[1]); return; } // Initialize maximum and second maximum let posa = Number.MIN_VALUE, posb = Number.MIN_VALUE; // Initialize minimum and second minimum let nega = Number.MIN_VALUE, negb = Number.MIN_VALUE; // Traverse given array for (let 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 && Math.abs(arr[i]) > Math.abs(nega)) { negb = nega; nega = arr[i]; } else if(arr[i] < 0 && Math.abs(arr[i]) > Math.abs(negb)) negb = arr[i]; } if (nega * negb > posa * posb) document.write("Max product pair is {" + nega + ", " + negb + "}"); else document.write("Max product pair is {" + posa + ", " + posb + "}"); } /* Driver program to test above function */ let arr=[1, 4, 3, 6, 7, 0]; let n = arr.length; maxProduct(arr, n); // This code is contributed by rag2127 </script> |
Output
Max product pair is {7, 6}
Time complexity: O(n)
Auxiliary Space: O(1)
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