This question was asked in Round 1 of Code Jam 2008 contest.

The problem

You are given two vectors v1=(x1,x2,…,xn) and v2=(y1,y2,…,yn). The scalar product of these vectors is a single number, calculated as x1y1+x2y2+…+xnyn.

Suppose you are allowed to permute the coordinates of each vector as you wish. Choose two permutations such that the scalar product of your two new vectors is the smallest possible, and output that minimum scalar product.

Input

The first line of the input file contains integer number T – the number of test cases. For each test case, the first line contains integer number n. The next two lines contain n integers each, giving the coordinates of v1 and v2 respectively.Output

For each test case, output a line

Case #X: Ywhere X is the test case number, starting from 1, and Y is the minimum scalar product of all permutations of the two given vectors.

Limits

Small dataset

T = 1000

1 ≤ n ≤ 8

-1000 ≤ xi, yi ≤ 1000Large dataset

T = 10

100 ≤ n ≤ 800

-100000 ≤ xi, yi ≤ 100000

**Sample Input:**

2

3

1 3 -5

-2 4 1

5

1 2 3 4 5

1 0 1 0 1

**Sample Output:**

Case #1: -25

Case #2: 6

The simple approach would be to multiply the smaller numbers from one array with the larger numbers of the other array and then sum up their products.

So basically, we sort both the arrays and then multiply the smallest to the largest or we can sort and reverse one array and multiply the same indexes of both arrays. I have used the former one.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 |
/* * Author: Vlad - www.mycoding.net * This program is the solution to code jam's "Minimum Scalar Product" problem. */ import java.io.BufferedReader; import java.io.DataInputStream; import java.io.FileInputStream; import java.io.IOException; import java.io.InputStreamReader; import java.math.BigInteger; import java.util.Arrays; public class Solution { /** * @param args * */ public static long[] convertStringToLong(String array[]) { long longArray[] = new long[array.length]; for (int i = 0; i < array.length; i++) longArray[i] = Long.parseLong(array[i]); return longArray; } public static void main(String[] args) throws IOException { // TODO Auto-generated method stub FileInputStream fis = new FileInputStream("MinimumScalarProductB.in"); DataInputStream in = new DataInputStream(fis); BufferedReader br = new BufferedReader(new InputStreamReader(in)); int N = Integer.parseInt(br.readLine()); int x = 1; int totalVectors; BigInteger min; long arrX[], arrY[]; while(x <= N) { totalVectors = Integer.parseInt(br.readLine()); arrX = convertStringToLong(br.readLine().split(" ")); arrY = convertStringToLong(br.readLine().split(" ")); Arrays.sort(arrX); Arrays.sort(arrY); min= new BigInteger("0"); for (int i = 0; i< totalVectors; i++ ) { min = min.add(new BigInteger(Long.toString(arrX[i] * arrY[(totalVectors-1)-i]))); } System.out.println("Case #"+x+": "+min); x++; } br.close(); } } |

#### Vlad

#### Latest posts by Vlad (see all)

- Code jam “Tic-Tac-Toe-Tomek” solution in java - April 19, 2013
- Code jam “Minimum Scalar product” solution in java - March 18, 2013
- Code jam Store Credit solution in java - March 10, 2013

## Leave a Reply