Check if integer has repeating digits. No string methods or arrays

19,800

Here is a short and sweet version :)

 private static boolean hasDistinctDigits(int number) {
     int numMask = 0;
     int numDigits = (int) Math.ceil(Math.log10(number+1));
     for (int digitIdx = 0; digitIdx < numDigits; digitIdx++) {
         int curDigit = (int)(number / Math.pow(10,digitIdx)) % 10;
         int digitMask = (int)Math.pow(2, curDigit);             
         if ((numMask & digitMask) > 0) return false;
         numMask = numMask | digitMask;
     }
     return true;
 }

It works in a pretty simply way. numMask is an integer used to store what digits have already been encountered (since a decimal system number has only 10 digits and an integer gives use 16-bits, we have enough bits to store each decimal digit as it occurs).

We loop over all digits in the number. For each digit index, we get the actual digit in curDigit. Let's say the current digit is 5. We then check it the 5th bit is raised in out numMask: if it is, then we have already encounter a 5 in the past, so we can immediately tell that the number does not have all distinct digits and return false; otherwise, we modify numMask and raise the 5th bit.

If we make it to the end, then no dupicate digits were encountered.

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Updated on June 15, 2022

Comments

  • Nobody
    Nobody almost 2 years

    I'm trying to see if an int has multiples of the same digit. Trying to do it without string methods or arrays. The main method I'm having trouble with is hasDistinctDigits(). It works when the repeating digits are at the end, but not when they come at the beginning or middle.

    public static void main(String[] args) {
        System.out.println(hasDistinctDigits(12234));
    }
    
    public static boolean hasDistinctDigits(int number) {
        boolean returner = true;
        int count = 1;
        int newNum = number;
        int digit = 0;
    
        while (count < numDigits(number)) {         
            while (count < numDigits(newNum)) {
                digit = newNum % 10;
                newNum/=10;
                if (digit == getDigit(newNum, count)) {
                    returner = false;
                }
                count++;                
            }
            count++;
        }
        return returner;
    }
    
    public static int numDigits(int number) {
        int count = 0;
        while (number != 0) {
            number /= 10;
            count++;
        }
        return count;
    }
    
    public static int getDigit(int number, int i) {
        int digit = 0;
        int count = 0;
        int originalNum = number;
    
        while (count <= i) {
            if (count == i) {
                digit = number % 10;
            }
            number /= 10;
            count++;
        }
        if (i > numDigits(originalNum)) {
            return -1;
        } else {
            return digit;
        }
    }
    

    }

    If this is run, you will see '2' repeats itself, but the method still evaluates to true when it should be false.

  • Anantha Raju C
    Anantha Raju C about 7 years
    please provide an explanation as to how your answer helps