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# Count Possible Decodings of a given Digit Sequence

• Difficulty Level : Medium
• Last Updated : 31 Mar, 2021

Let 1 represent ‘A’, 2 represents ‘B’, etc. Given a digit sequence, count the number of possible decodings of the given digit sequence.

Examples:

```Input:  digits[] = "121"
Output: 3
// The possible decodings are "ABA", "AU", "LA"

Input: digits[] = "1234"
Output: 3
// The possible decodings are "ABCD", "LCD", "AWD"```

An empty digit sequence is considered to have one decoding. It may be assumed that the input contains valid digits from 0 to 9 and there are no leading 0’s, no extra trailing 0’s, and no two or more consecutive 0’s.

This problem is recursive and can be broken into sub-problems. We start from the end of the given digit sequence. We initialize the total count of decodings as 0. We recur for two subproblems.
1) If the last digit is non-zero, recur for the remaining (n-1) digits and add the result to the total count.
2) If the last two digits form a valid character (or smaller than 27), recur for remaining (n-2) digits and add the result to the total count.

Following is the implementation of the above approach.

## C++

 `// C++ implementation to count number of``// decodings that can be formed from a``// given digit sequence``#include ``#include ``using` `namespace` `std;` `// recuring function to find``// ways in how many ways a``// string can be decoded of length``// greater than 0 and starting with``// digit 1 and greater.``int` `countDecoding(``char``* digits, ``int` `n)``{``    ``// base cases``    ``if` `(n == 0 || n == 1)``        ``return` `1;``    ``if` `(digits == ``'0'``)``        ``return` `0;` `    ``// for base condition "01123" should return 0``    ``// Initialize count``    ``int` `count = 0;` `    ``// If the last digit is not 0,``    ``// then last digit must add``    ``// to the number of words``    ``if` `(digits[n - 1] > ``'0'``)``        ``count = countDecoding(digits, n - 1);` `    ``// If the last two digits form a number smaller``    ``// than or equal to 26, then consider``    ``// last two digits and recur``    ``if` `(digits[n - 2] == ``'1'``        ``|| (digits[n - 2] == ``'2'``        ``&& digits[n - 1] < ``'7'``))``        ``count += countDecoding(digits, n - 2);` `    ``return` `count;``}` `// Given a digit sequence of length n,``// returns count of possible decodings by``// replacing 1 with A, 2 woth B, ... 26 with Z``int` `countWays(``char``* digits, ``int` `n)``{``    ``if` `(n == 0 || (n == 1 && digits == ``'0'``))``        ``return` `0;``    ``return` `countDecoding(digits, n);``}` `// Driver code``int` `main()``{``    ``char` `digits[] = ``"1234"``;``    ``int` `n = ``strlen``(digits);``    ``cout << ``"Count is "` `<< countWays(digits, n);``    ``return` `0;``}``// Modified by Atanu Sen`

## Java

 `// A naive recursive Java implementation``// to count number of decodings that``// can be formed from a given digit sequence` `class` `GFG {` `    ``// recuring function to find``    ``// ways in how many ways a``    ``// string can be decoded of length``    ``// greater than 0 and starting with``    ``// digit 1 and greater.``    ``static` `int` `countDecoding(``char``[] digits, ``int` `n)``    ``{``        ``// base cases``        ``if` `(n == ``0` `|| n == ``1``)``            ``return` `1``;` `        ``// for base condition "01123" should return 0``        ``if` `(digits[``0``] == ``'0'``)``            ``return` `0``;` `        ``// Initialize count``        ``int` `count = ``0``;` `        ``// If the last digit is not 0, then``        ``// last digit must add to``        ``// the number of words``        ``if` `(digits[n - ``1``] > ``'0'``)``            ``count = countDecoding(digits, n - ``1``);` `        ``// If the last two digits form a number``        ``// smaller than or equal to 26,``        ``// then consider last two digits and recur``        ``if` `(digits[n - ``2``] == ``'1'``            ``|| (digits[n - ``2``] == ``'2'``                ``&& digits[n - ``1``] < ``'7'``))``            ``count += countDecoding(digits, n - ``2``);` `        ``return` `count;``    ``}` `    ``// Given a digit sequence of length n,``    ``// returns count of possible decodings by``    ``// replacing 1 with A, 2 woth B, ... 26 with Z``    ``static` `int` `countWays(``char``[] digits, ``int` `n)``    ``{``        ``if` `(n == ``0` `|| (n == ``1` `&& digits[``0``] == ``'0'``))``            ``return` `0``;``        ``return` `countDecoding(digits, n);``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``char` `digits[] = { ``'1'``, ``'2'``, ``'3'``, ``'4'` `};``        ``int` `n = digits.length;``        ``System.out.printf(``"Count is %d"``,``                          ``countWays(digits, n));``    ``}``}` `// This code is contributed by Smitha Dinesh Semwal.``// Modified by Atanu Sen`

## Python3

 `# Recursive implementation of numDecodings``def` `numDecodings(s: ``str``) ``-``> ``int``:``    ``if` `len``(s) ``=``=` `0``    ``or` `(``len``(s) ``=``=` `1``        ``and` `s[``0``] ``=``=` `'0'``):``        ``return` `0``    ``return` `numDecodingsHelper(s, ``len``(s))`  `def` `numDecodingsHelper(s: ``str``, n: ``int``) ``-``> ``int``:``    ``if` `n ``=``=` `0` `or` `n ``=``=` `1``:``        ``return` `1``    ``count ``=` `0``    ``if` `s[n``-``1``] > ``"0"``:``        ``count ``=` `numDecodingsHelper(s, n``-``1``)``    ``if` `(s[n ``-` `2``] ``=``=` `'1'``        ``or` `(s[n ``-` `2``] ``=``=` `'2'``            ``and` `s[n ``-` `1``] < ``'7'``)):``        ``count ``+``=` `numDecodingsHelper(s, n ``-` `2``)``    ``return` `count`  `# Driver code``digits ``=` `"1234"``print``(``"Count is "``, numDecodings(digits))``# This code is contributed by Frank Hu`

## C#

 `// A naive recursive C# implementation``// to count number of decodings that``// can be formed from a given digit sequence``using` `System;` `class` `GFG {` `    ``// recuring function to find``    ``// ways in how many ways a``    ``// string can be decoded of length``    ``// greater than 0 and starting with``    ``// digit 1 and greater.``    ``static` `int` `countDecoding(``char``[] digits, ``int` `n)``    ``{` `        ``// base cases``        ``if` `(n == 0 || n == 1)``            ``return` `1;` `        ``// Initialize count``        ``int` `count = 0;` `        ``// If the last digit is not 0, then``        ``// last digit must add to``        ``// the number of words``        ``if` `(digits[n - 1] > ``'0'``)``            ``count = countDecoding(digits, n - 1);` `        ``// If the last two digits form a number``        ``// smaller than or equal to 26, then``        ``// consider last two digits and recur``        ``if` `(digits[n - 2] == ``'1'``            ``|| (digits[n - 2] == ``'2'``                ``&& digits[n - 1] < ``'7'``))``            ``count += countDecoding(digits, n - 2);` `        ``return` `count;``    ``}` `    ``// Given a digit sequence of length n,``    ``// returns count of possible decodings by``    ``// replacing 1 with A, 2 woth B, ... 26 with Z``    ``static` `int` `countWays(``char``[] digits, ``int` `n)``    ``{``        ``if` `(n == 0 || (n == 1 && digits == ``'0'``))``            ``return` `0;``        ``return` `countDecoding(digits, n);``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``char``[] digits = { ``'1'``, ``'2'``, ``'3'``, ``'4'` `};``        ``int` `n = digits.Length;``        ``Console.Write(``"Count is "``);``        ``Console.Write(countWays(digits, n));``    ``}``}` `// This code is contributed by nitin mittal.`

## PHP

 ` ``'0'``)``        ``\$count` `= countDecoding(``\$digits``, ``\$n` `- 1);` `    ``// If the last two digits form a number``    ``// smaller than or equal to 26, then``    ``// consider last two digits and recur``    ``if` `(``\$digits``[``\$n` `- 2] == ``'1'` `||``       ``(``\$digits``[``\$n` `- 2] == ``'2'` `&&``        ``\$digits``[``\$n` `- 1] < ``'7'``) )``        ``\$count` `+= countDecoding(``\$digits``, ``\$n` `- 2);` `    ``return` `\$count``;``}` `// Given a digit sequence of length n,``// returns count of possible decodings by``// replacing 1 with A, 2 woth B, ... 26 with Z``function` `countWays(&``\$digits``, ``\$n``){``  ``if``(``\$n``==0 || (``\$n` `== 1 && ``\$digits`` == ``'0'``))``     ``return` `0;``  ``return` `countDecoding(``\$digits``, ``\$n``);``}` `// Driver Code``\$digits` `= ``"1234"``;``\$n` `= ``strlen``(``\$digits``);``echo` `"Count is "` `. countWays(``\$digits``, ``\$n``);` `// This code is contributed by ita_c``?>`

## Javascript

 ``

Output:

`Count is 3`

The time complexity of above the code is exponential. If we take a closer look at the above program, we can observe that the recursive solution is similar to Fibonacci Numbers. Therefore, we can optimize the above solution to work in O(n) time using Dynamic Programming

Following is the implementation for the same.

## C++

 `// A Dynamic Programming based C++``// implementation to count decodings``#include ``#include ``using` `namespace` `std;` `// A Dynamic Programming based function``// to count decodings``int` `countDecodingDP(``char` `*digits, ``int` `n)``{``    ``// A table to store results of subproblems``    ``int` `count[n+1];``    ``count = 1;``    ``count = 1;``    ``//for base condition "01123" should return 0``    ``if``(digits==``'0'``) ``         ``return` `0;``    ``for` `(``int` `i = 2; i <= n; i++)``    ``{``        ``count[i] = 0;` `        ``// If the last digit is not 0,``        ``// then last digit must add to the number of words``        ``if` `(digits[i-1] > ``'0'``)``            ``count[i] = count[i-1];` `        ``// If second last digit is smaller``        ``// than 2 and last digit is smaller than 7,``        ``// then last two digits form a valid character``        ``if` `(digits[i-2] == ``'1'` `||``              ``(digits[i-2] == ``'2'` `&& digits[i-1] < ``'7'``) )``            ``count[i] += count[i-2];``    ``}``    ``return` `count[n];``}` `// Driver program to test above function``int` `main()``{``    ``char` `digits[] = ``"1234"``;``    ``int` `n = ``strlen``(digits);``    ``cout << ``"Count is "` `<< countDecodingDP(digits, n);``    ``return` `0;``}``// Modified by Atanu Sen`

## Java

 `// A Dynamic Programming based Java``// implementation to count decodings``import` `java.io.*;` `class` `GFG``{``    ` `// A Dynamic Programming based``// function to count decodings``static` `int` `countDecodingDP(``char` `digits[],``                           ``int` `n)``{``    ``// A table to store results of subproblems``    ``int` `count[] = ``new` `int``[n + ``1``];``    ``count[``0``] = ``1``;``    ``count[``1``] = ``1``;``    ``if``(digits[``0``]==``'0'``)   ``//for base condition "01123" should return 0``          ``return` `0``;``    ``for` `(``int` `i = ``2``; i <= n; i++)``    ``{``        ``count[i] = ``0``;` `        ``// If the last digit is not 0,``        ``// then last digit must add to``        ``// the number of words``        ``if` `(digits[i - ``1``] > ``'0'``)``            ``count[i] = count[i - ``1``];` `        ``// If second last digit is smaller``        ``// than 2 and last digit is smaller``        ``// than 7, then last two digits``        ``// form a valid character``        ``if` `(digits[i - ``2``] == ``'1'` `||``           ``(digits[i - ``2``] == ``'2'` `&&``            ``digits[i - ``1``] < ``'7'``))``            ``count[i] += count[i - ``2``];``    ``}``    ``return` `count[n];``}` `// Driver Code``public` `static` `void` `main (String[] args)``{``    ``char` `digits[] = {``'1'``,``'2'``,``'3'``,``'4'``};``    ``int` `n = digits.length;``    ``System.out.println(``"Count is "` `+``               ``countDecodingDP(digits, n));``}``}` `// This code is contributed by anuj_67``// Modified by Atanu Sen`

## Python3

 `# A Dynamic Programming based Python3``# implementation to count decodings` `# A Dynamic Programming based function``# to count decodings``def` `countDecodingDP(digits, n):` `    ``count ``=` `[``0``] ``*` `(n ``+` `1``); ``# A table to store``                           ``# results of subproblems``    ``count[``0``] ``=` `1``;``    ``count[``1``] ``=` `1``;` `    ``for` `i ``in` `range``(``2``, n ``+` `1``):` `        ``count[i] ``=` `0``;` `        ``# If the last digit is not 0, then last``        ``# digit must add to the number of words``        ``if` `(digits[i ``-` `1``] > ``'0'``):``            ``count[i] ``=` `count[i ``-` `1``];` `        ``# If second last digit is smaller than 2``        ``# and last digit is smaller than 7, then``        ``# last two digits form a valid character``        ``if` `(digits[i ``-` `2``] ``=``=` `'1'` `or``           ``(digits[i ``-` `2``] ``=``=` `'2'` `and``            ``digits[i ``-` `1``] < ``'7'``) ):``            ``count[i] ``+``=` `count[i ``-` `2``];` `    ``return` `count[n];` `# Driver Code``digits ``=` `"1234"``;``n ``=` `len``(digits);``print``(``"Count is"` `,``       ``countDecodingDP(digits, n));` `# This code is contributed by mits`

## C#

 `// A Dynamic Programming based C#``// implementation to count decodings``using` `System;` `class` `GFG``{``    ` `// A Dynamic Programming based``// function to count decodings``static` `int` `countDecodingDP(``char``[] digits,``                           ``int` `n)``{``    ``// A table to store results of subproblems``    ``int``[] count = ``new` `int``[n + 1];``    ``count = 1;``    ``count = 1;` `    ``for` `(``int` `i = 2; i <= n; i++)``    ``{``        ``count[i] = 0;` `        ``// If the last digit is not 0,``        ``// then last digit must add to``        ``// the number of words``        ``if` `(digits[i - 1] > ``'0'``)``            ``count[i] = count[i - 1];` `        ``// If second last digit is smaller``        ``// than 2 and last digit is smaller``        ``// than 7, then last two digits``        ``// form a valid character``        ``if` `(digits[i - 2] == ``'1'` `||``           ``(digits[i - 2] == ``'2'` `&&``            ``digits[i - 1] < ``'7'``))``            ``count[i] += count[i - 2];``    ``}``    ``return` `count[n];``}` `// Driver Code``public` `static` `void` `Main()``{``    ``char``[] digits = {``'1'``,``'2'``,``'3'``,``'4'``};``    ``int` `n = digits.Length;``    ``Console.WriteLine(``"Count is "` `+``            ``countDecodingDP(digits, n));``}``}` `// This code is contributed``// by Akanksha Rai``// Modified by Atanu Sen`

## PHP

 ` ``'0'``)``            ``\$count``[``\$i``] = ``\$count``[``\$i``-1];` `        ``// If second last digit is smaller than 2 and last digit is``        ``// smaller than 7, then last two digits form a valid character``        ``if` `(``\$digits``[``\$i``-2] == ``'1'` `|| (``\$digits``[``\$i``-2] == ``'2'` `&& ``\$digits``[``\$i``-1] < ``'7'``) )``            ``\$count``[``\$i``] += ``\$count``[``\$i``-2];``    ``}``    ``return` `\$count``[``\$n``];``}` `// Driver program to test above function``    ``\$digits` `= ``"1234"``;``    ``\$n` `= ``strlen``(``\$digits``);``    ``echo`  `"Count is "` `, countDecodingDP(``\$digits``, ``\$n``);` `#This code is contributed by ajit.``?>`

## Javascript

 ``

Output:

`Count is 3`

Time Complexity of the above solution is O(n) and it requires O(n) auxiliary space. We can reduce auxiliary space to O(1) by using the space-optimized version discussed in the Fibonacci Number Post