CODEWITHSUNDEEP
java
didil
didil

Reputation: 715

How to generate 2-digits distant PIN codes?

I want to generate 4-digits PIN codes that are at least 2-digits distant from each other.

For example:

Permutation is easy to check but how to check the distance?

Upvotes: 0

Views: 89

Answers (3)

user3437460
user3437460

Reputation: 17454

If have have no issues with permutation with your given problem description, it seems you only want to check if the digit of each position is having a difference of at least 2.

There are so many ways to do it in Java. One "primitive" way I personally like is using basic math operations:

public boolean isDistantPins (int pinA, int pinB, int requiredSpacedDigits, int distance){
    int spacedDigits = 0, aDigit = 0, bDigit = 0;
    while(pinA > 0 ){
        aDigit = pinA % 10;    //get last digit from the right
        bDigit = pinB % 10;    //get last digit from the right

        if(Math.abs(aDigit - bDigit) >= distance)    //find difference
            spacedDigits ++;

        pinA = pinA / 10;      //remove last digit from the right
        pinB = pinB / 10;      //remove last digit from the right
    }
    return spacedDigits == requiredSpacedDigits;
}

To invoke it:

System.out.println(isDistantPins(5235, 4068, 1, 2));
System.out.println(isDistantPins(5235, 5339, 1, 2));
System.out.println(isDistantPins(5235, 2553, 1, 2));
System.out.println(isDistantPins(5235, 5236, 1, 2));
System.out.println(isDistantPins(5235, 5235, 1, 2));

Output:

true
true
true
false
false

Note that this does not check the permutation. Hence you can run this after your permutation check (which you have no issues) so the codes for checking permutation is not shown here.

Upvotes: 1

anurag saxena
anurag saxena

Reputation: 287

Calculate the XOR of two numbers. Count the number of set bits.

Upvotes: 0

Alex Mandelias
Alex Mandelias

Reputation: 517

The logic is the following:

  • Check if pin1 is permutation
  • Check every digit of pin1 against the digit in the same position of pin2 using /, %.
  • Count the amount of same digits.
  • Return true/false based on the number of them.

Assuming you have a way to check for permutations, here's a complete solution:

class Solution {
    public static void main(String[] args) {
        /*
         * 5235 and 4068 is OK
         * 5235 and 5339 is OK
         * 5235 and 2553 is OK
         * 5235 and 5236 is NOT OK
         * 5325 and 5235 is NOT OK (permutation)
         */

        // threshold is the maximum number of digits that can be the same
        // while the pin1, pin2 are distant enough; in this case 2
        int threshold = 2;
        int pinLength = 4;

        System.out.println(areDistant(5235, 4068, threshold, pinLength));
        System.out.println(areDistant(5235, 5339, threshold, pinLength));
        System.out.println(areDistant(5235, 2553, threshold, pinLength));
        System.out.println(areDistant(5235, 5236, threshold, pinLength));
        System.out.println(areDistant(5325, 5235, threshold, pinLength));
    }

    public static boolean areDistant(int pin1, int pin2, int threshold, int pinLength) {
        if (isPermutation(pin1, pin2))
            return false;

        int sameDigits = 0;
        int ithDigit1, ithDigit2;
        for (int i=0; i<pinLength; i++) {
            ithDigit1 = (int) (pin1 / Math.pow(10, i)) % 10;
            ithDigit2 = (int) (pin2 / Math.pow(10, i)) % 10;
            // System.out.println(ithDigit1);
            // System.out.println(ithDigit2);
            if ( ithDigit1 == ithDigit2)
                sameDigits += 1;
        }
        return sameDigits <= threshold;
    }

    private static  boolean isPermutation(int pin1, int pin2) {
        return false;
        // fill the code here
    }
}

Upvotes: 1

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