Sales by Match HackerRank Solution | Java Solution

HackerRank Sales by Match problem solution in Java

 

Problem Description :

Alex works at a clothing store. There is a large pile of socks that must be paired by color for sale. Given an array of integers representing the color of each sock, determine how many pairs of socks with matching colors there are.

For example, there are n=7 socks with colors socks = [1,2,1,2,1,3,2]. There is one pair of color 1 and one of color 2. There are three odd socks left, one of each color. The number of pairs is 2.  

Example 1 :

Input :
n = 6
arr = [1, 2, 3, 4, 5, 6]

Output :
0

Explanation : We have 6 socks with all different colors, So print 0.

Example 2 :

Input :
n = 10
arr = [1, 2, 3, 4, 1, 4, 2, 7, 9, 9]

Output :
4

Explanation : We have 10 socks. There is pair of color 1, 2, 4 and 9, So print 4.

This problem easily solved by HashMap. Store all pair of socks one by one in Map and check if any pair is present in Map or not. If pair is present then increment ans variable by 1 and at last print total pair present in given array of socks. 

Solution 1 : Java Code for Sales By Match Problem

import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class SalesbyMatch {

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        System.out.println("Enter number of total socks");

        int numberOfSocks = sc.nextInt();
        int[] arrayOfSocks = new int[numberOfSocks];
        
        System.out.println("Enter all socks ");
        for (int i = 0; i < arrayOfSocks.length; i++) {
            arrayOfSocks[i] = sc.nextInt();
        }
        
        Map<Integer, Integer> map = new HashMap<>();
        
        int numberOfMatchingPairs = 0;
        /*
         If map does not contains pair of sock then put into map and 
         set value to 1. If map contains then we remove that sock 
         means we found 1 pair of socks. Delete that socks and 
         increment the value of numberOfMatchingPairs
        */
        for (int i : arrayOfSocks) {
            if (!map.containsKey(i)) {
                map.put(i, 1);
            } else {
                map.remove(i);
                ++numberOfMatchingPairs;
            }
        }
        
        System.out.println(numberOfMatchingPairs);
    }

}
Output :-
Enter number of total socks
9
Enter all socks 
10 20 20 10 10 30 50 10 20
3

Explanation :

• In above solution we are using HashMap data structure. HashMap contains key - value pair. HashMap does not contains any duplicate key but it can contains duplicate value. So we are taking advantage of Map in above code.
• Traverse through given arrayOfSocks array and check, if map does not contains any socks then we are storing into map with socks as key and 1 as value. If map already contains socks we are removing that sock and increment numberOfMatchingPairs by 1.
• Last print numberOfMatchingPairs, it will return all matching pair number.
  

 

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