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  4. Valid PAN format
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Valid PAN format

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votes

  • oats247 + 0 comments

    Bash.

    pan='[A-Z]\{5\}[0-9]\{4\}[A-Z]'

sed \
  -e '1d' \
  -e "s/$pan/YES/" \
  -e '/YES/! c\NO'

  • melroy_tellis + 0 comments

    It would be great to have support for C++14.Code that runs perfectly fine under C++14 throws errors here.

  • bryank + 0 comments

    Javascript 1 liner, well, the logic is anyway.

    function processData(input) {
    input.trim().split("\n").slice(1).forEach(str => console.log(test(str)));
}

function test(str) {
    return /^[A-Z]{5}\d{4}[A-Z]/.test(str) ? "YES" : "NO";
}

  • zerotohero + 0 comments
    # python3
n=int(input())
arr=[input() for i in range(n)]
for i in arr:
    if i[:5].isupper() and i[-1].isupper() and i[5:9].isdigit() and i[:5].isalpha():
        print('YES')
    else:
        print('NO')

  • nikola_cekic1 + 0 comments

    function processData(input) {

    var inputs = input.split("\n");
var regex = /^[A-Z]{5}\d{4}[A-Z]$/;

for (var i = 1; i < inputs.length; i++){

    var match = regex.exec(inputs[i]);

    if (match)
        console.log("YES");
    else
        console.log("NO");
}

    }

  • Ihor_L + 0 comments

    My not one-liner solution in Python.

    import re

pattern = re.compile(r'^[A-Z]{5}\d{4}[A-Z]$')

n = int(input())
for _ in range(n):
    pan_number = input()
    result = 'YES' if pattern.match(pan_number) else 'NO'
    print(result)

  • masaki_kyosuke22 + 0 comments

    I didn't know that PAN is not just for Primary Account Number (which is like CCN) but for Permanent Account Number too (which is like SSN). Interesting what we can learn here (not essentialy in the computer science).

  • wuyulin_212 + 0 comments
    import re
for _ in range(int(input())):
    print('YES' if bool(re.match(r'^[A-Z]{5}[0-9]{4}[a-zA-Z]$',input())) else 'NO')

  • tpgreene + 0 comments
    #include<stdio.h>

int get_length(char* arr){
    int i=0;
    while(arr[i]!='\0'){i++;}
    return i-1;
}

int valid_input(char* arr){
    int ans=1;
    for(int i=0; i<10; i++){
        if(i<=4||i==9){if(!(arr[i] >= 'A' && arr[i] <= 'Z')) ans=0;}
        else{if(!(arr[i] >= '0' && arr[i] <= '9')) ans=0;}
    }
    return ans;
}

int main(){
    char str[12]={};
    int n, l;
    
    scanf("%d%c", &n, &str[0]);
    
    while(n--){
        fgets(str, 12, stdin);
        l=get_length(str);
        if(n==0) l++;
        if(l==10){
            if(valid_input(str))printf("YES\n");
            else{printf("NO\n");}
        }
        else{printf("NO\n");}
        for(int i=0; i<12; i++){str[i]='\0';}
    }
    
    return 0;
}

  • itserin44 + 0 comments

    C#

    using System;
using System.Collections.Generic;
using System.IO;
using System.Text.RegularExpressions;

class Solution {
    static void Main(String[] args) {
        int lines = Convert.ToInt32(Console.ReadLine());
        
        for (int i = 0; i < lines; i++) {
            string pan = Console.ReadLine();
            if (Regex.IsMatch(pan, @"^[A-Z]{5}\d{4}[A-Z]$"))
	        Console.WriteLine("YES");
            else Console.WriteLine("NO");
        }
    }
}
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