Algorithms: find the maximum pairwise product

Let’s find out the maximum product of two biggest numbers in a given array. We will implement this in Java. Before that you need to understand how to find out biggest number in a given array, Please refer to the blog post below,

Example in Java

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.StringTokenizer;

public class MaxPairwiseProduct {

    static BigInteger getMaxPairwiseProductFast(int[] numbers) {
        int n = numbers.length;
        //System.out.println("size is: "+n);
        int index1 = 0;
        int index2;

        for (int i = 1; i < n; i++) {
            if (numbers[i] > numbers[index1]) {
                index1 = i;
            }
        }
        //System.out.println("Highest number is: "+numbers[index1]);
        if (index1 == 0) {
            index2 = 1;
        } else {
            index2 = 0;
        }
        for (int i = 1; i < n; i++) {
            if (i != index1 && numbers[i] > numbers[index2])
                index2 = i;
        }
        BigInteger x = new BigInteger(Integer.toString(numbers[index1]));
        BigInteger y = new BigInteger(Integer.toString(numbers[index2]));
        return x.multiply(y);
    }

    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());
        }
    }
}

Result

4
4 3 9 1
Highest number is: 9
36

Please observe the condition below,

if (index1 == 0) {
  index2 = 1;
} else {
  index2 = 0;
}

We are making sure, that if index1=0 which means that the highest number is at first place (index=0) then index2=1 otherwise the initial value of index2 will get be equal to index1 and will remain.

Explanation

For Example, if we are not using this condition,

consider the data set, a[4 2 3]

Say we have not used the condition so On first loop index1 will be 0

On the second loop index2 will have initial value of 0

a[0] > a[index2] && 0 != 0 —> FALSE —> index2 remains 0 (i.e; value 4)

a[1] > a[index2] && 1 != 0 —> FALSE —> index2 remains 0 (i.e; value 4)

a[2] > a[index2] && 2 != 0 —> FALSE —> index2 remains 0 (i.e; value 4)