# Count Possible Decodings of a given Digit Sequence in O(N) time and Constant Auxiliary space

Given a digit sequence S, the task is to find the number of possible decodings of the given digit sequence where 1 represents ‘A’, 2 represents ‘B’ … and so on up to 26, where 26 represents ‘Z’.

Examples:

Input: S = “121”
Output: 3
The possible decodings are “ABA”, “AU”, “LA”

Input: S = “1234”
Output: 3
The possible decodings are “ABCD”, “LCD”, “AWD”

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: In order to solve this problem in O(N) time complexity, Dynamic Programming is used. And in order to reduce the auxiliary space complexity to O(1), we use the space optimized version of recurrence relation discussed in the Fibonacci Number Post.

Similar to the Fibonacci Numbers, the key observations of any current ‘ith‘ index can be calculated using its previous two indices. So the Recurrence Relation to calculate the ith index can be denoted as

```// Condition to check last
// digit can be included or not
if (digit[i-1] is not '0')
count[i] += count[i-1]

// Condition to check the last
// two digits contribution
if (digit[i-2] is 1 or
(digit[i-2] is 2 and
digit[i-1] is less than 7))
count[i] += count[i-2]
```

Below is the implementation of the above approach:

## C++

 `// C++ implementation to count decodings ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// A Dynamic Programming based function ` `// to count decodings in digit sequence ` `int` `countDecodingDP(string digits, ``int` `n) ` `{ ` `    ``// For base condition "01123" ` `    ``// should return 0 ` `    ``if` `(digits == ``'0'``) ` `        ``return` `0; ` ` `  `    ``int` `count0 = 1, count1 = 1, count2; ` ` `  `    ``// Using last two calculated values, ` `    ``// calculate for ith index ` `    ``for` `(``int` `i = 2; i <= n; i++) { ` `        ``count2 = ((``int``)(digits[i - 1] != ``'0'``) *  count1) +  ` `                  ``(``int``)((digits[i - 2] == ``'1'``) or  ` `                  ``(digits[i - 2] == ``'2'` `and ` `                   ``digits[i - 1] < ``'7'``)) * count0; ` `        ``count0 = count1; ` `        ``count1 = count2; ` `    ``} ` ` `  `    ``// Return the required answer ` `    ``return` `count1; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``string digits = ``"1234"``; ` `    ``int` `n = digits.size(); ` ` `  `    ``// Function call ` `    ``cout << countDecodingDP(digits, n); ` ` `  `    ``return` `0; ` `} `

Output:

```3
```

Time Complexity: O(N)
Auxiliary Space Complexity: O(1)

Related Article: Count Possible Decodings of a given Digit Sequence

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