C Code to Count Possible Decoding Of A Given Digit Sequence

Counting possible decoding of a given digit sequence in C

Here we will discuss how to count all the possible decoding of a given digit sequence in C. Before counting the number of decodings first let’s see how the numbers are assumed to be coded.

1 = A, 2 = B, 3 = C, . . . . , 26 = Z.
The decoding programs are the most possible questions asked and are largely practiced in C  programming. The program counts the number of possible decodings of the entered digit by the user of the given sequence.
For example, if the digit sequence is “12” then there are two possible decodings for this – One of them is ‘AB’ when we decode 1 as ‘A’ and 2 as ‘B’. Now we can also decode this digit sequence “12” as ‘L’ when we decode digits 1 and 2 taken together as an integer 12.

Now let’s take another example for reference,

Sequence = 131

  • Possible decoding (1, 3, 1,)    = ACA
  • Possible decoding (13, 1)   = MA
    So, the total possible decodings of sequence 131 is 2.
Possible decoding of a digit sequence in C

Working:-

Step 1: Start

Step 2: User is required to insert a digit sequence as an input

Step 3: Set count = 0

Step 4: If the last number is not a zero, then return for the next remaining (n-1) numbers and add the results then to the total count.

Step 5: If the last two digits form a valid variable (or smaller than 27), return for the remaining (n-2) numbers and add the outcome to the total calculation.

 Step 6: Stop

Possible decoding of a digit sequence
Competitive Coding Techniques

C Program:

//C Program to Count possible decodings of a given digit sequence
#include<stdio.h>
#include<math.h> 

int cnt_decoding_digits(char *dig, int a)
{
    // Initializing an array to store results     
     int cnt[a+1]; 
     cnt[0] = 1;
     cnt[1] = 1;

     for (int k = 2; k <= a; k++) { cnt[k] = 0; 
     // If the last digit not equal to 0, then last digit must added to the number of words if (dig[k-1] > '0')
     cnt[k] = cnt[k-1];

     // In case second last digit is smaller than 2 and last digit is smaller than 7, then last two digits form a valid character
     if (dig[k-2] == '1' || (dig[k-2] == '2' && dig[k-1] < '7') )
          cnt[k] += cnt[k-2];
 }
 return cnt[a];
}

 int main()
 {
     char dig[15];
     printf("\n Enter the sequence : ");
     gets(dig);
     int a = strlen(dig);
     printf("\n Possible count of decoding of the sequence : %d\n", cnt_decoding_digits(dig, a));
   return 0;
}
Output:
Insert the digit sequence: 123456
Possible count of decoding of the sequence: 3