Hello Programmers, In this post, you will learn how to solve HackerRank Valid PAN format Solution. This problem is a part of the Regex HackerRank Series.
One more thing to add, don’t straight away look for the solutions, first try to solve the problems by yourself. If you find any difficulty after trying several times, then look for the solutions. We are going to solve the Regex HackerRank Solutions using CPP, JAVA, PYTHON, JavaScript & PHP Programming Languages.
HackerRank Valid PAN format Solution
Problem
The equivalent of SSN in India is a PAN number, which is unique to each of its citizens. In any of the country’s official documents, the PAN number is listed as follows
<char><char><char><char><char><digit><digit><digit><digit><char>
Your task is to figure out if the PAN number is valid or not. A valid PAN number will have all its letters in uppercase and digits in the same order as listed above.
Input Format
First line contains N. N lines follow, each having a PAN number.
Constraints
- 1 <= N <= 10
- Each char is an uppercase letter, i.e., char ∈ [‘A‘, ‘Z‘].
- Each digit lies between 0 and 9, i.e., digit ∈ [0, 9].
- The length of the PAN number is always 10, i.e., length(PAN) = 10
- Every character in PAN is either char or digit satisfying the above constraints.
Output Format
For every PAN number listed, print YES if it is valid and NO if it isn’t.
Sample Input
3
ABCDS1234Y
ABAB12345Y
avCDS1234Y
Sample Output
YES
NO
NO
The first PAN number is valid since the first 5 characters are letters, the next 4 are digits and the last character is an alphabet. All letters in input is in uppercase.
The second PAN number is invalid as the fifth character is a digit as opposed to an letter.
The third PAN number contains lowercase characters. Hence invalid
Viewing Submissions
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HackerRank Valid PAN format Solution in Cpp
#include <iostream>
#include <string>
using namespace std;
int main()
{
int n;
string s;
cin >> n;
cin.ignore();
bool IsPAN;
for (int i = 0; i < n; ++i)
{
getline(cin,s);
IsPAN = true;
for (int i = 0; i < 5; ++i)
if (int(s[i]) < 65 || int(s[i]) > 90)
IsPAN = false;
for (int i = 5; i < 9; ++i)
if (int(s[i]) < 48 || int(s[i]) > 57)
IsPAN = false;
if (int(s[9]) < 65 || int(s[9]) > 90)
IsPAN = false;
if (IsPAN)
cout << "YES" << endl;
else
cout << "NO" << endl;
}
return 0;
}
HackerRank Valid PAN format Solution in Java
import java.util.Scanner;
import java.util.regex.Pattern;
import java.util.regex.Matcher;
public class Solution {
public static void main(String args[])
{
Scanner reader = new Scanner(System.in);
int nTestCases = Integer.parseInt(reader.nextLine());
Pattern pattern = Pattern.compile("([A-Z]|[a-z]){5}[0-9]{4}([A-Z]|[a-z])");
for( int i = 0 ; i < nTestCases ; i++)
{
String input = reader.nextLine();
Matcher matcher = pattern.matcher(input);
if( matcher.find())
{
boolean hasUppercase = !input.equals(input.toLowerCase());
boolean hasLowercase = !input.equals(input.toUpperCase());
if( hasLowercase && hasUppercase)
System.out.println("NO");
else
System.out.println("YES");
}
else
System.out.println("NO");
}
}
}HackerRank Valid PAN format Solution in Python
n = input()
for i in xrange(n):
PAN = raw_input()
alp = PAN[:5]+PAN[-1]
if(alp.isalpha() and alp.isupper() and PAN[5:-1].isdigit()):
print "YES"
else:
print "NO"HackerRank Valid PAN format Solution in JavaScript
process.stdin.resume();
process.stdin.setEncoding("ascii");
process.stdin.on("data", function (input) {
input = input.match(/^(.*)$/igm);
var n = parseInt(input[0]),
strs = input.slice(1,n+1);
for (i=0, j=strs.length; i<j; i+=1) {
if (strs[i].match(/[A-Z]{5}[0-9]{4}[A-Z]/g)) {
process.stdout.write('YES\n');
} else {
process.stdout.write('NO\n');
}
}
});HackerRank Valid PAN format Solution in PHP
<?php
$_fp = fopen("php://stdin", "r");
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
fscanf($_fp, "%d", $m);
for ($i = 0; $i < $m; $i++) {
$line = trim(fgets($_fp));
$search = '/^[A-Z]{5}[0-9]{4}[A-Z]$/';
print (preg_match($search, $line) ? 'YES' : 'NO') . PHP_EOL;
}Disclaimer: This problem (Valid PAN format) is generated by HackerRank but the solution is provided by Chase2learn. This tutorial is only for Educational and Learning purposes.
