Wsrtjtyk

My question is how do I check if a digit is repeated in an integer without using arrays?

For example: 123145... 1 is repeated twice. so the output should be 15 (1+2+3+4+5)

My current code is:

# include "stdio.h"

int main () {

  int input = 0;

  int sum = 0;

  int input = 0;

  int sum = 0;



  int digit;



  printf("Please enter a number: ");

  scanf("%d" ,&input);



  while(input > 0) {

    digit = input % 10;



    if(d0 < 1) {

        sum += digit;

        d0 = 1;

    }



    input /= 10;

  }



  printf("Sum of different digits is: %dn", sum);

  return 0;

}

This is a really dumb way to solve this problem, but it uses neither arrays nor bit vectors:

#include <stdbool.h>

bool numberHasDigit(unsigned n, unsigned digit) {

  while (n) {

    if (n % 10 == digit) return true;

    n /= 10;

  }

  return false;

}

unsigned sumOfUniqueDigits(unsigned n) {

  unsigned sum = 0;

  for (unsigned digit = 1; digit <= 9; ++digit) {

    if (numberHasDigit(n, digit)) sum += digit;

  }

  return sum;

}

It's dumb because using an array (or bit vector) of flags is much faster, particularly for big numbers, and the code is just as simple:

unsigned sumOfUniqueDigits(unsigned n) {

  bool seen[10] = {false};

  unsigned sum = 0;

  while (n) {

    unsigned digit = n % 10;

    if (!seen[digit]) {

      sum += digit;

      seen[digit] = true;

    }

    n = n / 10;

  }

  return sum;

}

I have avoided for loops and arrays and function calls.

Basically, we run a while loop that looks at the 9 interesting digits (1,2,3,4,5,6,7,8,9).

While looking at each digit, we run another while loop to see if the current digit is in the number by using your modulus and division looping. If the digit is in the number, then we add it to the sum.

We only look at each digit one time, so duplicates are ignored.

# include "stdio.h"

int main()

{

    int input = 0;

    int sum = 0;





    printf("Please enter a number: ");

    scanf("%d", &input);



    //, now, subtract those values that do not exist in your number

    int digit = 1;  // start at 1, since 0 does not add to the sum

    while (digit < 10) {

        // check to see if the digit is in the number

        int testInput = input;



        while (testInput > 0) {

            if (testInput % 10 == digit) {

                sum += digit;

                break; // Don't test for this digit any more, so go back to the outer loop.

            }

            testInput /= 10;

        }



        digit++; // increment the digit

    }





    printf("Sum of different digits is: %dn", sum);

    return 0;

}

A variation on the @rici good answer (and written as a one function call as @John Murray) with reduced operations - on average about 2x faster.

Rather than iterate 1 to 9, it looks for a repeat of the least significant digit in the rest of the number.

unsigned SumUniqueDigits(unsigned n) {

  unsigned sum = 0;

  while (n) {

    unsigned lsdigit = n % 10;

    n /= 10;

    unsigned mssdigits = n;

    while (mssdigits) {

      if (mssdigits % 10 == lsdigit) {

        lsdigit = 0;

        break;

      }

      mssdigits /= 10;

    }

    sum += lsdigit;

  }

  return sum;

}

Alternative even faster: On re-examination - failed test code. Removed.