No Duplicates!

Problem

Read 11 random one-digit numbers. Exactly one of the digits appears twice. Find and print the duplicate digit.

38. No duplicates!, Comet 64

Solution 1

Use an array of 10 elements, because the one-digit numbers start from 0 and end at 9, inclusive. The only array we have access to is a string of characters. Interpret a string as an array of characters. Let i be an input one-digit integer. If the character in str at index i is lower-case O (i.e. o), this is the first time we see i. Set the character in str at index i to x, meaning we have seen the digit i. However, in case the character in str at index i is x, then we have seen i already and this is the second time we are seeing i. Therefore i is a duplicate.

str = oooooooooo;         // Array of 10 elements.
loop:                     // Start of "loop".
    int = input;          // Read a one-digit integer i.
    char = str[int];      // Peek at character with index i.
    check char = x;       // Is character = x?
    jump if true: print;  // If so, go to "print".
    char = x;             // Otherwise, have not seen i yet.
    str[int] = char;      // Mark the number as seen.
    jump to: loop;        // Go to "loop".
print:                    // The "print" branch.
    output = int;         // Print the duplicate number.

Solution 2

All the one-digit integers are:

0, 1, 2, 3, ..., 8, 9

The sum of these digits is

0 + 1 + 2 + 3 + ... + 8 + 9 = 45

Let x be a repeating digit. The sum of the 11 random one-digit numbers is S := 45 + x. The value of the repeating digit is therefore x := S - 45.

loop:                     // Start of "loop".
    check input = null;   // Any more digits to read?
    jump if true: print;  // If not, go to "print".
    int = int + input;    // Update cumulative sum.
    jump to: loop;        // Go to "loop".
print:                    // The "print" branch.
    int = int - 45;       // The duplicate digit.
    output = int;         // Print the duplicate digit.

Solution 3

Similar to Solution 2. However, after we have updated the cumulative sum, we check whether there are any more integers to read. This allows us to remove a jump instruction and a label. The register int initially holds zero. There is no need to explicitly set int to zero at the start of the program. This solution is optimal, according to the game.

loop:                     // Start of "loop".
    int = int + input;    // Update cumulative sum.
    check input = null;   // Any more digits to read?
    jump if false: loop;  // If so, go to "loop".
int = int - 45;           // Otherwise, get duplicate digit.
output = int;             // Print the duplicate digit.