2017-03-24 00:02:30 -05:00
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// Author: Rodney Shaghoulian
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2018-07-27 16:45:56 -07:00
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// Github: github.com/RodneyShag
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// HackerRank: hackerrank.com/RodneyShag
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2017-03-24 00:02:30 -05:00
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import java.util.Scanner;
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2017-04-05 18:42:35 -05:00
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// For an element in A, if there's a matching element in B, this creates a "beautiful pair".
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// Each element can only be used once to create a beautiful pair.
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2017-03-24 00:02:30 -05:00
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2017-04-05 18:42:35 -05:00
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// Additionaly, We MUST change exactly 1 element in B. We attempt to change it to create
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// 1 more beautiful pair. In the special case where we already have the max number of
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// beautiful pairs, being forced to change it gives us 1 less beautiful pair.
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2017-03-24 00:02:30 -05:00
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// Time Complexity: O(n)
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// Space Complexity: O(1)
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public class Solution {
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private static final int MAX_NUM = 1000; // max value of any number in array
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2017-09-11 09:32:17 -07:00
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public static void main(String[] args) {
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2017-03-24 00:02:30 -05:00
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Scanner scan = new Scanner(System.in);
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int N = scan.nextInt();
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int [] bucketA = new int[MAX_NUM + 1];
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for (int i = 0; i < N; i++) {
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bucketA[scan.nextInt()]++;
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}
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int beautifulPairs = 0;
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for (int i = 0; i < N; i++) {
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int num = scan.nextInt();
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if (bucketA[num] > 0) {
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bucketA[num]--;
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beautifulPairs++;
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}
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}
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scan.close();
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/* Accounts for changing 1 element in B */
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if (beautifulPairs == N) {
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beautifulPairs--;
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} else {
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beautifulPairs++;
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}
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System.out.println(beautifulPairs);
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}
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}
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