#Problem Statement
Problem Statement
Input — The input will consist of one case. The first line of the input specifies the number of telephone numbers in the directory (up to 100,000) as a positive integer alone on the line. The remaining lines list the telephone numbers in the directory, with each number alone on a line. Each telephone number consists of a string composed of decimal digits, uppercase letters (excluding Q and Z) and hyphens. Exactly seven of the characters in the string will be digits or letters.
Output — Generate a line of output for each telephone number that appears more than once in any form. The line should give the telephone number in standard form, followed by a space, followed by the number of times the telephone number appears in the directory. Arrange the output lines by telephone number in ascending lexicographical order. If there are no duplicates in the input print the line:
No duplicates.
import java.util.Map;
import java.util.Scanner;
import java.util.TreeMap;
public class Main{
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int total = scan.nextInt();
scan.nextLine(); // But why?
String[] numbers = new String[total];
Map<String, Integer> dict = new TreeMap<String, Integer>();
for(int i = 0; i < total; i++) {
numbers[i] = scan.nextLine();
numbers[i] = convert(numbers[i]);
if(!dict.containsKey(numbers[i])) {
dict.put(numbers[i], 1);
} else {
dict.put(numbers[i], dict.get(numbers[i]) + 1);
}
}
scan.close();
boolean hasDuplication = true;
for(String number: dict.keySet()) {
if(dict.get(number) > 1) {
hasDuplication = false;
System.out.println(number + " " + dict.get(number));
}
}
if(hasDuplication) {
System.out.println("No duplicates.");
}
}
public static String convert(String raw) {
raw = raw.replaceAll("-", "");
raw = raw.toLowerCase();
String number = "";
for(int i = 0; i < raw.length(); i++) {
number += parse(raw.charAt(i));
}
number = number.substring(0, 3) + "-" + number.substring(3);
return number;
}
public static char parse(char digit) {
if(digit >= 'a' && digit < 'q') {
digit = (char) ((digit-'a') / 3 + '2');
} else if(digit > 'q' && digit < 'z') {
digit = (char) ((digit-'q') / 3 + '7');
}
return digit;
}
}
