Count Possible Decodings of a given Digit Sequence
Last Updated :
22 Mar, 2025
Let 1 maps to 'A', 2 maps to 'B', ..., 26 to 'Z'. Given a digit sequence, count the number of possible decodings of the given digit sequence.
Consider the input string "123". There are three valid ways to decode it:
- "ABC": The grouping is (1, 2, 3) → 'A', 'B', 'C'
- "AW": The grouping is (1, 23) → 'A', 'W'
- "LC": The grouping is (12, 3) → 'L', 'C'
Note: Groupings that contain invalid codes (e.g., "0" by itself or numbers greater than "26") are not allowed.
For instance, the string "230" is invalid because "0" cannot stand alone, and "30" is greater than "26", so it cannot represent any letter. The task is to find the total number of valid ways to decode a given string.
Examples:
Input: digits = "121"
Output: 3
Explanation: The possible decodings are "ABA", "AU", "LA"
Input: digits = "1234"
Output: 3
Explanation: The possible decodings are "ABCD", "LCD", "AWD"
[Naive Approach] Using Recursion - O(2^n) Time and O(n) Space
For the recursive approach to count decoding ways, there will be two cases:
- If the current digit is not '0', the problem reduces to solving for the remaining digits starting from the next index.
- If the two digits form a valid number between 10 and 26, the problem reduces to solving for the digits starting two positions ahead.
The recurrence relation will look like this:
- decodeHelper(digits, index) = decodeHelper(digits, index + 1) + decodeHelper(digits, index + 2)
Base Case: if index >= digits.length()
- decodeHelper(digits, index) = 1
C++
// C++ program to count decoding ways of a digit string
// using recursion.
#include <bits/stdc++.h>
using namespace std;
// Helper function to recursively calculate decoding ways.
int decodeHelper(string &digits, int index)
{
int n = digits.length();
// Base case: If we reach the end of the string,
// return 1 as it signifies a valid decoding.
if (index >= n)
{
return 1;
}
int ways = 0;
// Single-digit decoding: check if current digit is not '0'.
if (digits[index] != '0')
{
ways = decodeHelper(digits, index + 1);
}
// Two-digit decoding: check if next two digits are valid.
if ((index + 1 < n) && ((digits[index] == '1' && digits[index + 1] <= '9') ||
(digits[index] == '2' && digits[index + 1] <= '6')))
{
ways += decodeHelper(digits, index + 2);
}
return ways;
}
// Function to count decoding ways for the
// entire string.
int countWays(string &digits)
{
return decodeHelper(digits, 0);
}
int main()
{
string digits = "121";
cout << countWays(digits) << endl;
return 0;
}
// Java program to count decoding ways of a digit string
// using recursion.
import java.util.*;
class GfG {
// Helper function to recursively calculate decoding
// ways.
static int decodeHelper(String digits, int index)
{
int n = digits.length();
// Base case: If we reach the end of the string,
// return 1 as it signifies a valid decoding.
if (index >= n) {
return 1;
}
int ways = 0;
// Single-digit decoding: check if current digit is
// not '0'.
if (digits.charAt(index) != '0') {
ways = decodeHelper(digits, index + 1);
}
// Two-digit decoding: check if next two digits are
// valid.
if ((index + 1 < n)
&& ((digits.charAt(index) == '1'
&& digits.charAt(index + 1) <= '9')
|| (digits.charAt(index) == '2'
&& digits.charAt(index + 1) <= '6'))) {
ways += decodeHelper(digits, index + 2);
}
return ways;
}
// Function to count decoding ways for the
// entire string.
static int countWays(String digits)
{
return decodeHelper(digits, 0);
}
public static void main(String[] args)
{
String digits = "121";
System.out.println(countWays(digits));
}
}
# Python program to count decoding ways of a digit string
# using recursion.
# Helper function to recursively calculate decoding ways
def decodeHelper(digits, index):
n = len(digits)
# Base case: If we reach the end of the string,
# return 1 as it signifies a valid decoding.
if index >= n:
return 1
ways = 0
# Single-digit decoding: check if current digit is not '0'.
if digits[index] != '0':
ways = decodeHelper(digits, index + 1)
# Two-digit decoding: check if next two digits are valid.
if (index + 1 < n and
((digits[index] == '1' and digits[index + 1] <= '9') or
(digits[index] == '2' and digits[index + 1] <= '6'))):
ways += decodeHelper(digits, index + 2)
return ways
# Function to count decoding ways for the entire string
def countWays(digits):
return decodeHelper(digits, 0)
if __name__ == "__main__":
digits = "121"
print(countWays(digits))
// C# program to count decoding ways of a digit string
// using recursion.
using System;
class GfG {
// Helper function to recursively calculate decoding
// ways.
static int DecodeHelper(string digits, int index)
{
int n = digits.Length;
// Base case: If we reach the end of the string,
// return 1 as it signifies a valid decoding.
if (index >= n) {
return 1;
}
int ways = 0;
// Single-digit decoding: check if current digit is
// not '0'.
if (digits[index] != '0') {
ways = DecodeHelper(digits, index + 1);
}
// Two-digit decoding: check if next two digits are
// valid.
if ((index + 1 < n)
&& ((digits[index] == '1'
&& digits[index + 1] <= '9')
|| (digits[index] == '2'
&& digits[index + 1] <= '6'))) {
ways += DecodeHelper(digits, index + 2);
}
return ways;
}
// Function to count decoding ways for the entire
// string.
static int countWays(string digits)
{
return DecodeHelper(digits, 0);
}
static void Main()
{
string digits = "121";
Console.WriteLine(countWays(digits));
}
}
// JavaScript program to count decoding ways of a digit
// string using recursion.
function decodeHelper(digits, index)
{
const n = digits.length;
// Base case: If we reach the end of the string,
// return 1 as it signifies a valid decoding.
if (index >= n) {
return 1;
}
let ways = 0;
// Single-digit decoding: check if current digit is not
// '0'.
if (digits[index] !== "0") {
ways = decodeHelper(digits, index + 1);
}
// Two-digit decoding: check if next two digits are
// valid.
if (index + 1 < n
&& ((digits[index] === "1"
&& digits[index + 1] <= "9")
|| (digits[index] === "2"
&& digits[index + 1] <= "6"))) {
ways += decodeHelper(digits, index + 2);
}
return ways;
}
// Function to count decoding ways for the entire string
function countWays(digits)
{
return decodeHelper(digits, 0);
}
const digits = "121";
console.log(countWays(digits));
[Better Approach 1] Using Top-Down DP (Memoization) - O(n) Time and O(n) Space
1. Optimal Substructure: The solution to the problem can be broken down into smaller subproblems:
Mathematically, the recurrence relations are:
if digits[index] != '0':
decodeHelper(digits, index) = decodeHelper(digits, index + 1)
if digits[index] and digits[index + 1] form a valid number :
decodeHelper(digits, index) += decodeHelper(digits, index + 2)
2. Overlapping Subproblems: The same subproblems are recalculated multiple times, like computing the number of decodings from index i repeatedly. This overlap can be avoided using memoization to store already computed results.
memo[index] = decodeHelper(digits, index + 1) if digits[index] != '0'
memo[index] += decodeHelper(digits, index + 2) if digits[index] and digits[index + 1]
C++
// C++ program to count decoding ways of a digit string
// using Memoization
#include <bits/stdc++.h>
using namespace std;
// Helper function to recursively calculate decoding ways
// with memoization.
int decodeHelper(string &digits, int index, vector<int> &memo)
{
int n = digits.length();
// Base case: If we reach the end of the string,
// return 1 as it signifies a valid decoding.
if (index >= n)
{
return 1;
}
// If the value is already calculated for this
// index, return it.
if (memo[index] != -1)
{
return memo[index];
}
int ways = 0;
// Single-digit decoding: check if current digit is not '0'.
if (digits[index] != '0')
{
ways = decodeHelper(digits, index + 1, memo);
}
// Two-digit decoding: check if next two digits are valid.
if ((index + 1 < n) && ((digits[index] == '1' && digits[index + 1] <= '9') ||
(digits[index] == '2' && digits[index + 1] <= '6')))
{
ways += decodeHelper(digits, index + 2, memo);
}
// Memoize the result for the current index.
memo[index] = ways;
return ways;
}
// Function to count decoding ways for the entire string.
int countWays(string &digits)
{
int n = digits.length();
// Create a memoization vector initialized to -1.
vector<int> memo(n, -1);
return decodeHelper(digits, 0, memo);
}
int main()
{
string digits = "121";
cout << countWays(digits) << endl;
return 0;
}
// Java program to count decoding ways of a digit string
// using recursion with memoization.
import java.util.*;
class GfG {
// Helper function to recursively calculate decoding
// ways with memoization.
static int decodeHelper(String digits, int index,
int[] memo)
{
int n = digits.length();
// Base case: If we reach the end of the string,
// return 1 as it signifies a valid decoding.
if (index >= n) {
return 1;
}
// If the value is already calculated for this
// index, return it.
if (memo[index] != -1) {
return memo[index];
}
int ways = 0;
// Single-digit decoding: check if current digit is
// not '0'.
if (digits.charAt(index) != '0') {
ways = decodeHelper(digits, index + 1, memo);
}
// Two-digit decoding: check if next two digits are
// valid.
if ((index + 1 < n)
&& ((digits.charAt(index) == '1'
&& digits.charAt(index + 1) <= '9')
|| (digits.charAt(index) == '2'
&& digits.charAt(index + 1) <= '6'))) {
ways += decodeHelper(digits, index + 2, memo);
}
// Memoize the result for the current index.
memo[index] = ways;
return ways;
}
// Function to count decoding ways for the entire
// string.
static int countWays(String digits)
{
int n = digits.length();
// Create a memoization array initialized to -1.
int[] memo = new int[n];
Arrays.fill(memo, -1);
return decodeHelper(digits, 0, memo);
}
public static void main(String[] args)
{
String digits = "121";
System.out.println(countWays(digits));
}
}
# Python program to count decoding ways of a digit string
# using recursion with memoization.
# Helper function to recursively calculate decoding ways
def decodeHelper(digits, index, memo):
n = len(digits)
# Base case: If we reach the end of the string,
# return 1 as it signifies a valid decoding.
if index >= n:
return 1
# If the result for this index is already
# calculated, return it.
if memo[index] != -1:
return memo[index]
ways = 0
# Single-digit decoding: check if current digit is not '0'.
if digits[index] != '0':
ways = decodeHelper(digits, index + 1, memo)
# Two-digit decoding: check if next two digits are valid.
if (index + 1 < n and
((digits[index] == '1' and digits[index + 1] <= '9') or
(digits[index] == '2' and digits[index + 1] <= '6'))):
ways += decodeHelper(digits, index + 2, memo)
# Memoize the result for the current index.
memo[index] = ways
return ways
# Function to count decoding ways for the entire string
def countWays(digits):
n = len(digits)
# Create a memoization list initialized to -1.
memo = [-1] * n
return decodeHelper(digits, 0, memo)
if __name__ == "__main__":
digits = "121"
print(countWays(digits))
// C# program to count decoding ways of a digit string
// using recursion with memoization.
using System;
class GfG {
// Helper function to recursively calculate
// decoding ways.
static int DecodeHelper(string digits, int index,
int[] memo)
{
int n = digits.Length;
// Base case: If we reach the end of the string,
// return 1 as it signifies a valid decoding.
if (index >= n) {
return 1;
}
// If result for this index is already computed,
// return it.
if (memo[index] != -1) {
return memo[index];
}
int ways = 0;
// Single-digit decoding: check if current
// digit is not '0'.
if (digits[index] != '0') {
ways = DecodeHelper(digits, index + 1, memo);
}
// Two-digit decoding: check if next two
// digits are valid.
if ((index + 1 < n)
&& ((digits[index] == '1'
&& digits[index + 1] <= '9')
|| (digits[index] == '2'
&& digits[index + 1] <= '6'))) {
ways += DecodeHelper(digits, index + 2, memo);
}
// Store the result in memo and return it.
memo[index] = ways;
return ways;
}
// Function to count decoding ways for the entire
// string.
static int countWays(string digits)
{
int n = digits.Length;
// Memoization array initialized to -1.
int[] memo = new int[n];
Array.Fill(memo, -1);
return DecodeHelper(digits, 0, memo);
}
static void Main()
{
string digits = "121";
Console.WriteLine(countWays(digits));
}
}
// JavaScript program to count decoding ways of a digit
// string using recursion with memoization.
function decodeHelper(digits, index, memo)
{
const n = digits.length;
// Base case: If we reach the end of the string,
// return 1 as it signifies a valid decoding.
if (index >= n) {
return 1;
}
// If result for this index is already computed, return
// it.
if (memo[index] !== -1) {
return memo[index];
}
let ways = 0;
// Single-digit decoding: check if current digit is not
// '0'.
if (digits[index] !== "0") {
ways = decodeHelper(digits, index + 1, memo);
}
// Two-digit decoding: check if next two digits are
// valid.
if (index + 1 < n
&& ((digits[index] === "1"
&& digits[index + 1] <= "9")
|| (digits[index] === "2"
&& digits[index + 1] <= "6"))) {
ways += decodeHelper(digits, index + 2, memo);
}
// Store the result in memo and return it.
memo[index] = ways;
return ways;
}
// Function to count decoding ways for the entire string.
function countWays(digits)
{
const n = digits.length;
// Memoization array initialized to -1.
const memo = new Array(n).fill(-1);
return decodeHelper(digits, 0, memo);
}
const digits = "121";
console.log(countWays(digits));
[Better Approach 2] Using Bottom-Up DP (Tabulation) - O(n) Time and O(n) Space
The approach here is similar to the recursive method, but instead of breaking down the problem recursively, we solve it iteratively in a bottom-up manner using dynamic programming. We will create a 1D array dp of size (n + 1), where dp[i], represents the number of ways to decode the substring starting from index i of the string digits[].
Dynamic Programming Relation:
If the current digit is not '0', it can be decoded as a single digit, so we update the dp[i] as:
If the next two digits form a valid number between 10 and 26, then we also consider decoding the current and next digit together:
Base Case: dp[n] = 1, where an empty string has one valid decoding.
C++
// C++ program to count decoding ways of a digit string
// using Tabulation
#include <bits/stdc++.h>
using namespace std;
// Function to count decoding ways for the entire string.
int countWays(string &digits)
{
int n = digits.length();
// Create a dp array initialized to 0, with size n + 1.
vector<int> dp(n + 1, 0);
// Base case: An empty string has one valid decoding.
dp[n] = 1;
// Iterate from the end of the string to the beginning.
for (int i = n - 1; i >= 0; i--)
{
// Single-digit decoding: check if current
// digit is not '0'.
if (digits[i] != '0')
{
dp[i] = dp[i + 1];
}
// Two-digit decoding: check if next two digits are valid.
if ((i + 1 < n) &&
((digits[i] == '1' && digits[i + 1] <= '9') || (digits[i] == '2' && digits[i + 1] <= '6')))
{
dp[i] += dp[i + 2];
}
}
return dp[0];
}
int main()
{
string digits = "121";
cout << countWays(digits) << endl;
return 0;
}
// Java program to count decoding ways of a digit string
// using Tabulation
import java.util.ArrayList;
class GfG {
// Function to count decoding ways for the entire
// string.
public static int countWays(String digits)
{
int n = digits.length();
// Create a dp list initialized to 0, with size n
// + 1.
ArrayList<Integer> dp = new ArrayList<>();
for (int i = 0; i <= n; i++) {
dp.add(0);
}
// Base case: An empty string has one valid
// decoding.
dp.set(n, 1);
// Iterate from the end of the string to the
// beginning.
for (int i = n - 1; i >= 0; i--) {
// Single-digit decoding: check if current
// digit is not '0'.
if (digits.charAt(i) != '0') {
dp.set(i, dp.get(i + 1));
}
// Two-digit decoding: check if next two digits
// are valid.
if ((i + 1 < n)
&& ((digits.charAt(i) == '1'
&& digits.charAt(i + 1) <= '9')
|| (digits.charAt(i) == '2'
&& digits.charAt(i + 1) <= '6'))) {
dp.set(i, dp.get(i) + dp.get(i + 2));
}
}
return dp.get(0);
}
public static void main(String[] args)
{
String digits = "121";
System.out.println(countWays(digits));
}
}
# Python program to count decoding ways of a digit string
# using Tabulation
# Function to count decoding ways for the entire string.
def countWays(digits):
n = len(digits)
# Create a dp list initialized to 0, with size n + 1.
dp = [0] * (n + 1)
# Base case: An empty string has one valid decoding.
dp[n] = 1
# Iterate from the end of the string to the beginning.
for i in range(n - 1, -1, -1):
# Single-digit decoding: check if current
# digit is not '0'.
if digits[i] != '0':
dp[i] = dp[i + 1]
# Two-digit decoding: check if next two digits are valid.
if (i + 1 < n and
((digits[i] == '1' and digits[i + 1] <= '9') or
(digits[i] == '2' and digits[i + 1] <= '6'))):
dp[i] += dp[i + 2]
return dp[0]
if __name__ == "__main__":
digits = "121"
print(countWays(digits))
// C# program to count decoding ways of a digit string
// using Tabulation
using System;
using System.Collections.Generic;
class GfG {
// Function to count decoding ways for the entire
// string.
static int countWays(string digits)
{
int n = digits.Length;
// Create a dp list initialized to 0, with size n
// + 1.
List<int> dp = new List<int>(new int[n + 1]);
// Base case: An empty string has one valid
// decoding.
dp[n] = 1;
// Iterate from the end of the string to the
// beginning.
for (int i = n - 1; i >= 0; i--) {
// Single-digit decoding: check if current
// digit is not '0'.
if (digits[i] != '0') {
dp[i] = dp[i + 1];
}
// Two-digit decoding: check if next two digits
// are valid.
if ((i + 1 < n)
&& ((digits[i] == '1'
&& digits[i + 1] <= '9')
|| (digits[i] == '2'
&& digits[i + 1] <= '6'))) {
dp[i] += dp[i + 2];
}
}
return dp[0];
}
static void Main()
{
string digits = "121";
Console.WriteLine(countWays(digits));
}
}
// Javascript program to count decoding ways of
// a digit string using Tabulation
// Function to count decoding ways for the entire string.
function countWays(digits)
{
const n = digits.length;
// Create a dp array initialized to 0, with size n + 1.
const dp = Array(n + 1).fill(0);
// Base case: An empty string has one valid decoding.
dp[n] = 1;
// Iterate from the end of the string to the beginning.
for (let i = n - 1; i >= 0; i--) {
// Single-digit decoding: check if current
// digit is not '0'.
if (digits[i] !== "0") {
dp[i] = dp[i + 1];
}
// Two-digit decoding: check if next two digits are
// valid.
if (i + 1 < n
&& ((digits[i] === "1" && digits[i + 1] <= "9")
|| (digits[i] === "2"
&& digits[i + 1] <= "6"))) {
dp[i] += dp[i + 2];
}
}
return dp[0];
}
const digits = "121";
console.log(countWays(digits));
[Expected Approach] Using Space Optimised DP - O(n) Time and O(1) Space
In previous approach of dynamic programming we have derive the relation between states as given below:
- dp[i] = dp[i + 1]
- dp[i] = dp[i]+ dp[i + 2]
If we observe that for calculating current dp[i] state we only need previous two values dp[i+1] and dp[i]+dp[i+2]. There is no need to store all the previous states.
C++
#include <bits/stdc++.h>
using namespace std;
// Function to count decoding ways for the
// entire string with optimized space.
int countWays(string digits)
{
int n = digits.length();
// If the string is empty or starts with '0',
// there are no valid decodings.
if (n == 0 || digits[0] == '0')
{
return 0;
}
// two variables to store the previous two results.
int prev1 = 1, prev2 = 0;
for (int i = 1; i <= n; ++i)
{
int current = 0;
// Check for valid single-digit decoding
if (digits[i - 1] != '0')
{
current += prev1;
}
// Check for valid two-digit decoding
// (previous digit and current digit form a
// valid number between 10 and 26)
if (i > 1)
{
int twoDigit = (digits[i - 2] - '0') * 10 + (digits[i - 1] - '0');
if (twoDigit >= 10 && twoDigit <= 26)
{
current += prev2;
}
}
// Update prev1 and prev2 for the next iteration.
prev2 = prev1;
prev1 = current;
}
return prev1;
}
int main()
{
string digits = "121";
cout << countWays(digits) << endl;
return 0;
}
// Function to count decoding ways for the entire string
// with optimized space.
class GfG {
static int countWays(String digits)
{
int n = digits.length();
// If the string is empty or starts with '0',
// there are no valid decodings.
if (n == 0 || digits.charAt(0) == '0') {
return 0;
}
// two variables to store the previous two results.
int prev1 = 1, prev2 = 0;
for (int i = 1; i <= n; ++i) {
int current = 0;
// Check for valid single-digit decoding
if (digits.charAt(i - 1) != '0') {
current += prev1;
}
// Check for valid two-digit decoding
// (previous digit and current digit form
// a valid number between 10 and 26)
if (i > 1) {
int twoDigit
= (digits.charAt(i - 2) - '0') * 10
+ (digits.charAt(i - 1) - '0');
if (twoDigit >= 10 && twoDigit <= 26) {
current += prev2;
}
}
// Update prev1 and prev2 for the next
// iteration.
prev2 = prev1;
prev1 = current;
}
return prev1;
}
public static void main(String[] args)
{
String digits = "121";
System.out.println(countWays(digits));
}
}
def countWays(digits):
n = len(digits)
# If the string is empty or starts with '0',
# there are no valid decodings.
if n == 0 or digits[0] == '0':
return 0
# two variables to store the previous two results.
prev1, prev2 = 1, 0
for i in range(1, n + 1):
current = 0
# Check for valid single-digit decoding
if digits[i - 1] != '0':
current += prev1
# Check for valid two-digit decoding
# (previous digit and current digit form a valid number between 10 and 26)
if i > 1:
two_digit = (int(digits[i - 2]) * 10 + int(digits[i - 1]))
if 10 <= two_digit <= 26:
current += prev2
# Update prev1 and prev2 for the next iteration.
prev2 = prev1
prev1 = current
return prev1
if __name__ == '__main__':
digits = '121'
print(countWays(digits))
using System;
class GfG {
static int countWays(string digits)
{
int n = digits.Length;
// If the string is empty or starts with '0',
// there are no valid decodings.
if (n == 0 || digits[0] == '0') {
return 0;
}
// two variables to store the previous two results.
int prev1 = 1, prev2 = 0;
for (int i = 1; i <= n; ++i) {
int current = 0;
// Check for valid single-digit decoding
if (digits[i - 1] != '0') {
current += prev1;
}
// Check for valid two-digit decoding
// (previous digit and current digit form a
// valid number between 10 and 26)
if (i > 1) {
int twoDigit = (digits[i - 2] - '0') * 10
+ (digits[i - 1] - '0');
if (twoDigit >= 10 && twoDigit <= 26) {
current += prev2;
}
}
// Update prev1 and prev2 for the next
// iteration.
prev2 = prev1;
prev1 = current;
}
return prev1;
}
public static void Main()
{
string digits = "121";
Console.WriteLine(countWays(digits));
}
}
function countWays(digits)
{
const n = digits.length;
// If the string is empty or starts with '0',
// there are no valid decodings.
if (n === 0 || digits[0] === "0") {
return 0;
}
// two variables to store the previous two results.
let prev1 = 1, prev2 = 0;
for (let i = 1; i <= n; ++i) {
let current = 0;
// Check for valid single-digit decoding
if (digits[i - 1] !== "0") {
current += prev1;
}
// Check for valid two-digit decoding
// (previous digit and current digit form a valid
// number between 10 and 26)
if (i > 1) {
const twoDigit = (parseInt(digits[i - 2]) * 10
+ parseInt(digits[i - 1]));
if (twoDigit >= 10 && twoDigit <= 26) {
current += prev2;
}
}
// Update prev1 and prev2 for the next iteration.
prev2 = prev1;
prev1 = current;
}
return prev1;
}
const digits = "121";
console.log(countWays(digits));
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