Count Possible Decodings Of A Given Digit Sequence In Java Code

This article provides a detailed guide on counting the number of possible decodings for a given digit sequence in Java. You will learn about various approaches, from a straightforward recursive solution to optimized dynamic programming techniques.

Problem Statement

Imagine a coding system where letters A-Z are mapped to numbers 1-26, respectively (A=1, B=2, ..., Z=26). Given a sequence of digits, the challenge is to determine the total number of ways this sequence can be decoded into a string of letters. Digits can be grouped in two ways: as a single digit (1-9) or as a two-digit number (10-26). A '0' cannot be decoded on its own.

For example, the digit sequence "12" can be decoded in two ways:

1. "AB" (1 -> A, 2 -> B)
2. "L" (12 -> L)

This problem often arises in areas requiring interpretation of coded numerical data, where different groupings of digits lead to distinct meanings.

Example

Consider the digit sequence "226". Using the decoding rules, here are the possible interpretations:

• "BBF" (2 -> B, 2 -> B, 6 -> F)
• "VF" (22 -> V, 6 -> F)
• "BZ" (2 -> B, 26 -> Z)

For the input "226", the expected output is 3 possible decodings.

Background & Knowledge Prerequisites

To understand the solutions presented, familiarity with the following concepts is beneficial:

• Java Basics: Understanding of strings, loops, conditional statements, and method calls.
• Recursion: The concept of a function calling itself to solve smaller subproblems.
• Dynamic Programming (DP): An optimization technique used to avoid redundant computations in recursive problems by storing the results of subproblems. This includes:
• Memoization: Top-down DP, where results are stored as they are computed.
• Tabulation: Bottom-up DP, where results are computed iteratively from base cases.

Use Cases or Case Studies

This problem, while seemingly abstract, mirrors scenarios in various domains:

• Secure Communication Protocols: Decoding fragmented messages where individual digit sequences or combined pairs map to specific symbols, commands, or data fields in a secure communication channel.
• Biological Sequence Interpretation: In abstract bioinformatics models, where sequences of numerical data might represent components that can be grouped in different ways to form larger functional units, similar decoding logic could apply.
• Financial Transaction Codes: Interpreting specialized numerical codes in financial systems where certain single digits represent basic actions, and specific two-digit combinations denote compound transactions or sub-types.
• Data Serialization and Deserialization: In systems where data is serialized into compact digit sequences, and different interpretations (single vs. double-digit groupings) are valid depending on context, counting possibilities helps in validating or parsing data.

Solution Approaches

We will explore three main approaches to solve this problem: a recursive (brute-force) method, and two optimized dynamic programming techniques (memoization and tabulation).

Approach 1: Recursive Approach

The recursive approach directly translates the problem definition into code. For each position in the digit sequence, we explore two possibilities: decoding the current digit alone or decoding the current digit and the next one together (if valid).

• One-line summary: Recursively explores all possible ways to group digits, leading to potential re-computation of subproblems.

// Recursive Decoding Counter
public class Main {

    // Main function to initiate the recursive decoding process
    public static int countDecodingsRecursive(String s) {
        // Start the recursive helper function from the beginning of the string (index 0)
        return countDecodingsRecursiveHelper(s, 0);
    }

    // Helper function that performs the actual recursion
    private static int countDecodingsRecursiveHelper(String s, int index) {
        // Step 1: Base Case 1 - If we've reached the end of the string, it means we found one valid decoding.
        if (index == s.length()) {
            return 1;
        }

        // Step 2: Base Case 2 - If the current digit is '0', it cannot be decoded on its own.
        // It also cannot be part of a valid two-digit number (e.g., "01" is not 1, "00" is not valid).
        if (s.charAt(index) == '0') {
            return 0;
        }

        // Step 3: Count decodings by taking the current digit alone
        // Recursively call for the next position (index + 1)
        int count = countDecodingsRecursiveHelper(s, index + 1);

        // Step 4: Count decodings by taking the current digit and the next digit together
        // Check if there are at least two digits remaining and if they form a valid number (10-26)
        if (index + 1 < s.length()) {
            // Extract the two-digit number
            int twoDigitNum = Integer.parseInt(s.substring(index, index + 2));
            // If the two-digit number is within the valid range [10, 26]
            if (twoDigitNum >= 10 && twoDigitNum <= 26) {
                // Add the decodings from taking two digits together
                count += countDecodingsRecursiveHelper(s, index + 2);
            }
        }

        // Step 5: Return the total count for the current subproblem
        return count;
    }

    public static void main(String[] args) {
        // Test cases for the recursive approach
        System.out.println("--- Recursive Approach ---");
        System.out.println("Input: \\"12\\" | Possible decodings: " + countDecodingsRecursive("12"));       // Expected: 2 (AB, L)
        System.out.println("Input: \\"226\\" | Possible decodings: " + countDecodingsRecursive("226"));     // Expected: 3 (BBF, VF, BZ)
        System.out.println("Input: \\"12345\\" | Possible decodings: " + countDecodingsRecursive("12345")); // Expected: 3 (ABCDE, LCDE, AWDE)
        System.out.println("Input: \\"10\\" | Possible decodings: " + countDecodingsRecursive("10"));       // Expected: 1 (J)
        System.out.println("Input: \\"0\\" | Possible decodings: " + countDecodingsRecursive("0"));         // Expected: 0
        System.out.println("Input: \\"1\\" | Possible decodings: " + countDecodingsRecursive("1"));         // Expected: 1
        System.out.println("Input: \\"27\\" | Possible decodings: " + countDecodingsRecursive("27"));       // Expected: 1 (B7)
    }
}

Run

• Sample Output:

  --- Recursive Approach ---
  Input: "12" | Possible decodings: 2
  Input: "226" | Possible decodings: 3
  Input: "12345" | Possible decodings: 3
  Input: "10" | Possible decodings: 1
  Input: "0" | Possible decodings: 0
  Input: "1" | Possible decodings: 1
  Input: "27" | Possible decodings: 1

• Stepwise Explanation:

1. Base Case (End of String): If the index reaches the end of the string, it means we have successfully decoded all digits from the starting point to the end. This counts as one valid decoding, so we return 1.
2. Base Case (Leading Zero): If the digit at the current index is '0', it cannot be decoded as a single digit (0 has no letter mapping) and also cannot start a valid two-digit number (e.g., "01" is invalid). Therefore, no decodings are possible from this point, and we return 0.
3. Decode Single Digit: Always try to decode the current digit s.charAt(index) alone. This leads to a recursive call for the next position index + 1. The result of this call is added to our count.
4. Decode Two Digits: If there are at least two digits remaining (index + 1 < s.length()), check if the two-digit number formed by s.charAt(index) and s.charAt(index + 1) is within the valid range [10, 26]. If it is, recursively call for the position index + 2 (skipping two digits) and add its result to count.
5. Return: The sum of counts from both possibilities (single-digit decoding and two-digit decoding) is returned.

Approach 2: Dynamic Programming (Memoization)

The recursive approach suffers from redundant computations for overlapping subproblems. Memoization optimizes this by storing the results of already computed subproblems in an array (or map) and returning the stored result if the subproblem is encountered again.

• One-line summary: Optimizes the recursive approach by storing the results of subproblems to avoid re-computation, using a top-down strategy.

// Memoized Decoding Counter
import java.util.Arrays; // Needed for Arrays.fill()

public class Main {

    // Main function to initiate the memoized dynamic programming decoding process
    public static int countDecodingsMemoized(String s) {
        // Create a memoization array to store results of subproblems.
        // Size is s.length() + 1 to include states for index s.length().
        int[] memo = new int[s.length() + 1];
        // Initialize memoization array with -1 to indicate uncomputed states
        Arrays.fill(memo, -1);
        // Start the memoized helper function from the beginning of the string (index 0)
        return countDecodingsMemoizedHelper(s, 0, memo);
    }

    // Helper function that performs the memoized recursion
    private static int countDecodingsMemoizedHelper(String s, int index, int[] memo) {
        // Step 1: Base Case 1 - If we've reached the end of the string, one valid decoding found.
        if (index == s.length()) {
            return 1;
        }

        // Step 2: Base Case 2 - If the current digit is '0', it cannot be decoded.
        if (s.charAt(index) == '0') {
            return 0;
        }

        // Step 3: Check if the result for this index has already been computed and stored
        if (memo[index] != -1) {
            return memo[index]; // Return the stored result
        }

        // Step 4: Count decodings by taking the current digit alone
        int count = countDecodingsMemoizedHelper(s, index + 1, memo);

        // Step 5: Count decodings by taking the current digit and the next digit together
        if (index + 1 < s.length()) {
            int twoDigitNum = Integer.parseInt(s.substring(index, index + 2));
            if (twoDigitNum >= 10 && twoDigitNum <= 26) {
                count += countDecodingsMemoizedHelper(s, index + 2, memo);
            }
        }

        // Step 6: Store the computed result for the current index before returning
        memo[index] = count;
        return count;
    }

    public static void main(String[] args) {
        // Test cases for the memoized approach
        System.out.println("\\n--- Memoized Dynamic Programming Approach ---");
        System.out.println("Input: \\"12\\" | Possible decodings: " + countDecodingsMemoized("12"));
        System.out.println("Input: \\"226\\" | Possible decodings: " + countDecodingsMemoized("226"));
        System.out.println("Input: \\"12345\\" | Possible decodings: " + countDecodingsMemoized("12345"));
        System.out.println("Input: \\"10\\" | Possible decodings: " + countDecodingsMemoized("10"));
        System.out.println("Input: \\"0\\" | Possible decodings: " + countDecodingsMemoized("0"));
        System.out.println("Input: \\"1\\" | Possible decodings: " + countDecodingsMemoized("1"));
        System.out.println("Input: \\"27\\" | Possible decodings: " + countDecodingsMemoized("27"));
    }
}

Run

• Sample Output:

  --- Memoized Dynamic Programming Approach ---
  Input: "12" | Possible decodings: 2
  Input: "226" | Possible decodings: 3
  Input: "12345" | Possible decodings: 3
  Input: "10" | Possible decodings: 1
  Input: "0" | Possible decodings: 0
  Input: "1" | Possible decodings: 1
  Input: "27" | Possible decodings: 1

• Stepwise Explanation:

1. Memoization Array: An array memo of size s.length() + 1 is created and initialized with a special value (e.g., -1) to indicate that no result has been computed for that index yet.
2. Base Cases: The base cases (index == s.length() and s.charAt(index) == '0') remain the same as in the recursive approach.
3. Check Memo: Before performing any computations, the function checks memo[index]. If it's not the initialized special value (i.e., it holds a pre-computed result), that result is immediately returned, avoiding redundant work.
4. Compute and Store: If the result is not in memo, the computations for single-digit and two-digit decodings are performed just like in the recursive approach.
5. Store Result: Once count is calculated, it is stored in memo[index] before being returned, making it available for future calls to the same subproblem.

Approach 3: Dynamic Programming (Tabulation)

Tabulation is a bottom-up dynamic programming approach. It builds up solutions for smaller subproblems iteratively, starting from the base cases, and uses them to solve larger subproblems until the final solution is reached.

• One-line summary: Iteratively builds up the solution from base cases, filling a DP table in a bottom-up manner, ensuring each subproblem is computed once.

// Tabulated Decoding Counter
public class Main {

    // Main function to initiate the tabulated dynamic programming decoding process
    public static int countDecodingsTabulation(String s) {
        int n = s.length();
        // Step 1: Handle edge case for an empty string. An empty string has 1 way to decode (empty sequence).
        if (n == 0) return 1;

        // Step 2: Create a DP array. dp[i] will store the number of ways to decode the substring s[i...n-1].
        // Size is n + 1 to handle index n (representing an empty suffix).
        int[] dp = new int[n + 1];

        // Step 3: Initialize base cases for the DP table.
        // dp[n] represents the number of ways to decode an empty string (suffix starting at index n).
        // There is one way to decode an empty string: by doing nothing.
        dp[n] = 1;

        // dp[n-1] represents the number of ways to decode the last character.
        // If the last character is '0', it cannot be decoded (0 ways).
        // Otherwise (1-9), it can be decoded in one way (1 way).
        dp[n - 1] = (s.charAt(n - 1) == '0') ? 0 : 1;

        // Step 4: Fill the DP table from right to left (from n-2 down to 0)
        // This ensures that dp[i+1] and dp[i+2] are already computed when we calculate dp[i].
        for (int i = n - 2; i >= 0; i--) {
            // If the current digit is '0', it cannot be decoded. So, dp[i] is 0.
            if (s.charAt(i) == '0') {
                dp[i] = 0;
                continue; // Move to the next iteration
            }

            // Way 1: Decode the current digit alone.
            // This contributes dp[i+1] ways (decodings of the rest of the string from i+1).
            // This is always possible if s.charAt(i) is not '0'.
            dp[i] = dp[i + 1];

            // Way 2: Decode the current digit and the next digit together.
            // Check if s[i]s[i+1] forms a valid two-digit number (10-26).
            int twoDigitNum = Integer.parseInt(s.substring(i, i + 2));
            if (twoDigitNum >= 10 && twoDigitNum <= 26) {
                // If valid, add dp[i+2] ways (decodings of the rest of the string from i+2).
                dp[i] += dp[i + 2];
            }
        }

        // Step 5: The result for the entire string s[0...n-1] is stored in dp[0].
        return dp[0];
    }

    public static void main(String[] args) {
        // Test cases for the tabulated approach
        System.out.println("\\n--- Tabulated Dynamic Programming Approach ---");
        System.out.println("Input: \\"12\\" | Possible decodings: " + countDecodingsTabulation("12"));
        System.out.println("Input: \\"226\\" | Possible decodings: " + countDecodingsTabulation("226"));
        System.out.println("Input: \\"12345\\" | Possible decodings: " + countDecodingsTabulation("12345"));
        System.out.println("Input: \\"10\\" | Possible decodings: " + countDecodingsTabulation("10"));
        System.out.println("Input: \\"0\\" | Possible decodings: " + countDecodingsTabulation("0"));
        System.out.println("Input: \\"1\\" | Possible decodings: " + countDecodingsTabulation("1"));
        System.out.println("Input: \\"27\\" | Possible decodings: " + countDecodingsTabulation("27"));
    }
}

Run

• Sample Output:

  --- Tabulated Dynamic Programming Approach ---
  Input: "12" | Possible decodings: 2
  Input: "226" | Possible decodings: 3
  Input: "12345" | Possible decodings: 3
  Input: "10" | Possible decodings: 1
  Input: "0" | Possible decodings: 0
  Input: "1" | Possible decodings: 1
  Input: "27" | Possible decodings: 1

• Stepwise Explanation:

1. DP Array Initialization: An array dp of size n + 1 (where n is string length) is created. dp[i] will store the number of ways to decode the substring starting from index i to the end.
2. Base Cases:

• dp[n] = 1: An empty suffix (starting after the last character) has one way to be decoded (the empty decoding itself). This is crucial for the recurrence relation.
• dp[n - 1]: For the last character, if it's '0', dp[n-1] is 0. Otherwise (1-9), dp[n-1] is 1.

1. Iterative Filling: The loop runs from i = n - 2 down to 0. This bottom-up approach ensures that dp[i+1] and dp[i+2] are already computed when dp[i] is calculated.
2. Handling '0': If s.charAt(i) is '0', no valid decoding can start at this position, so dp[i] is 0.
3. Single Digit Decoding: If s.charAt(i) is not '0', it can be decoded alone. This contributes dp[i+1] ways (the number of ways to decode the rest of the string starting from i+1).
4. Two Digit Decoding: If s.charAt(i) and s.charAt(i+1) form a valid two-digit number (10-26), they can be decoded together. This contributes dp[i+2] ways (the number of ways to decode the rest of the string starting from i+2).
5. Final Result: dp[0] contains the total number of ways to decode the entire string.

Conclusion

Counting the possible decodings of a digit sequence is a classic dynamic programming problem. While a direct recursive approach can solve it, it is inefficient due to overlapping subproblems. Both memoization (top-down DP) and tabulation (bottom-up DP) provide efficient solutions by storing and reusing computed results. The tabulation method often offers a cleaner iterative solution with explicit loop control, making it preferred for many DP problems. Understanding these approaches helps in tackling a wide range of combinatorial problems in programming.

Summary

• Problem: Map digit sequences (1-9, 10-26) to letters (A-Z) and count all unique decoding combinations.
• Recursive Approach: Directly explores all possibilities, simple to understand but inefficient for longer sequences due to redundant calculations.
• Memoization (Top-down DP): Optimizes recursion by caching results of subproblems, preventing repeated computations.
• Tabulation (Bottom-up DP): Builds the solution iteratively from smallest subproblems to the full problem, typically more efficient regarding stack space compared to recursion.
• Key Constraints: '0' cannot be decoded alone, and two-digit numbers must be between 10 and 26 (inclusive).