How to find Maximum Pairwise Product in Java using Naive approach

In this Java tutorial, we will learn how to find the maximum pairwise product using Java. Maximum Pairwise Product in easy words you can say that it is the product of those two numbers present in an array from which we get the highest value possible.

What is the meaning of Maximum Pairwise Product?

It means that we have to find the maximum product of two distinct numbers in a given array of non-negative numbers.

Input: A sequence of non-negative numbers.

Output: The maximum value that can be obtained by multiplying two different numbers from the input sequence.

=>, For example, I’m giving you a sequence of non-negative numbers: a1 . . . . an

Input format: The first line contains a single number whose value is equal to n. The next line contains n non-negative numbers a1 . . . . an.

Output format: The maximum pairwise product.

Sample 1

Input

:
3
1 2 3

Output

:
6

Sample 2

Input:

10
7 5 14 2 8 8 10 1 2 3

Output:

140

Naive Method :

A naive approach is to solve the Maximum Pairwise Product Question is to find all the possible pairs from the sequence which is inputted by the user. A[1 . . . . n] =[a1 . . . . an] and then we have to find the largest product value.

Logic :

First, we have to find the two largest values from the inputted sequence. Because we know that the product of the largest value is the maximum product we can get.

Note: All inputted numbers in the sequence is must be non-negative numbers.

Find Maximum Pairwise Product in Java using Native Approach

import java.util.*;
import java.io.*;

public class MaxPairwiseProduct {
static long getMaxPairwiseProductFast(int[] numbers) {
        int n_Size = numbers.length;

        int max_index1 = -1;
        for (int p = 0; p < n_Size; p++) {
            if ((max_index1 == -1) || (numbers[p] > numbers[max_index1]))
                max_index1 = i;
        }

        int max_index2 = -1;
        for (int k = 0; k < n_Size; k++) {
            if ((k != max_index1) && ((max_index2 == -1) || (numbers[k] > numbers[max_index2])))
                max_index2 = j;
        }

        return (long)numbers[max_index1] * numbers[max_index2];
    }

    public static void main(String[] args) {
        
        FastScanner scanner = new FastScanner(System.in);
        int n = scanner.nextInt();
        int[] numbers = new int[n];
        for (int i = 0; i < n; i++) {
            numbers[i] = scanner.nextInt();
        }
        
        System.out.println(getMaxPairwiseProductFast(numbers));
     
    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(InputStream stream) {
            try {
                br = new BufferedReader(new InputStreamReader(stream));
            } catch (Exception e) {
                e.printStackTrace();
            }
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
    }

}

We have used StringTokenizer class here.

You can also read,

• Frequency of Repeated words in a string in Java
• Working with HashMap (Insertion, Deletion, Iteration) in Java

 

One response to “How to find Maximum Pairwise Product in Java using Naive approach”

1. STUTI DEY says:

   October 6, 2021 at 9:59 pm

   After compiling it is showing illegal start of type ,I can’t solve this problem ….please help

   Reply