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

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++

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// C++ implementation to count decodings
 
#include <bits/stdc++.h>
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] == '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;
}

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Java

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// Java implementation to count decodings
class GFG{
 
// A Dynamic programming based function
// to count decodings in digit sequence
static int countDecodingDP(String digits, int n)
{
     
    // For base condition "01123"
    // should return 0
    if (digits.charAt(0) == '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++)
    {
        int dig1 = 0, dig2, dig3 = 0;
         
        // Change boolean to int
        if(digits.charAt(i - 1) != '0')
        {
            dig1 = 1;
        }
        if(digits.charAt(i - 2) == '1')
        {
            dig2 = 1;
        }
        else
            dig2 = 0;
             
        if(digits.charAt(i - 2) == '2' &&
           digits.charAt(i - 1) < '7')
        {
            dig3 = 1;
        }
        count2 = dig1 * count1 +
                 dig2 + dig3 * count0;
         
        count0 = count1;
        count1 = count2;
    }
 
    // Return the required answer
    return count1;
}
 
// Driver Code
public static void main(String[] args)
{
    String digits = "1234";
    int n = digits.length();
 
    // Function call
    System.out.print(countDecodingDP(digits, n));
}
}
 
// This code is contributed by 29AjayKumar

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Python3

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# Python3 implementation
# to count decodings
 
# A Dynamic programming based
# function to count decodings
# in digit sequence
def countDecodingDP(digits, n):
 
    # For base condition "01123"
    # should return 0
    if (digits[0] == '0'):
        return 0;   
 
    count0 = 1; count1 = 1;
 
    # Using last two calculated values,
    # calculate for ith index
    for i in range(2, n + 1):
        dig1 = 0; dig3 = 0;
 
        # Change boolean to int
        if (digits[i-1] != '0'):
            dig1 = 1;
         
        if (digits[i - 2] == '1'):
            dig2 = 1;
        else:
            dig2 = 0;
 
        if (digits[i - 2] == '2' and
            digits[i-1] < '7'):
            dig3 = 1;
         
        count2 = dig1 * count1 +
                 dig2 + dig3 * count0;
 
        count0 = count1;
        count1 = count2;   
 
    # Return the required answer
    return count1;
 
# Driver Code
if __name__ == '__main__':
    digits = "1234";
    n = len(digits);
 
    # Function call
    print(countDecodingDP(digits, n));
 
# This code is contributed by gauravrajput1

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C#

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// C# implementation to count decodings
using System;
 
class GFG{
 
// A Dynamic programming based function
// to count decodings in digit sequence
static int countDecodingDP(String digits, int n)
{
     
    // For base condition "01123"
    // should return 0
    if (digits[0] == '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++)
    {
        int dig1 = 0, dig2, dig3 = 0;
         
        // Change bool to int
        if(digits[i - 1] != '0')
        {
            dig1 = 1;
        }
        if(digits[i - 2] == '1')
        {
            dig2 = 1;
        }
        else
            dig2 = 0;
             
        if(digits[i - 2] == '2' &&
           digits[i - 1] < '7')
        {
            dig3 = 1;
        }
        count2 = dig1 * count1 +
                 dig2 + dig3 * count0;
         
        count0 = count1;
        count1 = count2;
    }
 
    // Return the required answer
    return count1;
}
 
// Driver Code
public static void Main(String[] args)
{
    String digits = "1234";
    int n = digits.Length;
 
    // Function call
    Console.Write(countDecodingDP(digits, n));
}
}
 
// This code is contributed by Amit Katiyar

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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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