Decode Ways

class Solution {
public:
    int decodeWays(string seq) {
        if (seq[0] == '0') return 0;

        size_t len = seq.size();
        int previous = 1, beforePrevious = 1;

        for (size_t idx = 1; idx < len; idx++) {
            int count = 0;
            if (seq[idx] != '0') {
                count = beforePrevious;
            }
            int pair = stoi(seq.substr(idx - 1, 2));
            if (pair >= 10 && pair <= 26) {
                count += previous;
            }

            previous = beforePrevious;
            beforePrevious = count;
        }
        return beforePrevious;
    }
};

The "Decode Ways" solution in C++ focuses on counting how many ways a given sequence of digits can be decoded, assuming each number corresponds to a letter (similar to how '1' to '26' would map to 'A' to 'Z'). The function decodeWays implements a dynamic programming approach to solve the problem efficiently.

Here's a breakdown of the logic:

• Start by handling cases where the sequence begins with '0', as no valid encoding is possible. Immediately return 0.
• Initialize two integer variables, previous and beforePrevious, both set to 1. These track the number of ways to decode subsequences up to the current and previous positions.
• Iterate through the string starting from the second character (since single-digit decoding is straightforward).
• For each character, determine how many new ways to decode are possible:
  • Check if the current character is not '0' because '0' does not correspond to any letter by itself but is valid in combinations like '10' or '20'.
  • Convert the current and previous characters to a double-digit number. If this number is between 10 and 26, it validly maps to a letter, hence increasing the count of decoding ways by adding the number of ways up to the previous character.
  • Update the variables previous and beforePrevious for the next iteration.

The function ultimately returns beforePrevious which holds the total number of ways the full sequence can be decoded.

This approach ensures that each potential decoding is counted precisely, handling cases where single or double-digit numbers can form valid mappings. Thus, the program efficiently calculates the number without generating all possible strings but by dynamically adjusting the count of possible decodings.

class Solution {
    public int decodeWays(String sequence) {
        if (sequence.charAt(0) == '0') {
            return 0;
        }

        int length = sequence.length();
        int prevPrev = 1;
        int prev = 1;
        for (int index = 1; index < length; index++) {
            int currentDecode = 0;
            if (sequence.charAt(index) != '0') {
                currentDecode = prev;
            }
            int pairNum = Integer.parseInt(sequence.substring(index - 1, index + 1));
            if (pairNum >= 10 && pairNum <= 26) {
                currentDecode += prevPrev;
            }

            prevPrev = prev;
            prev = currentDecode;
        }
        return prev;
    }
}

The Java method decodeWays in the provided code interprets a string of digits as encoding options, similar to how letters are encoded in telephone keypads. The method returns the number of ways to decode the given string based on certain constraints. Here's how the algorithm functions:

• First, the method checks if the string starts with '0'. If it does, there are no valid ways to decode it, so it returns 0.
• Two integer variables, prevPrev and prev, are initialized to 1. These are used to store the counts of decode ways from previous computations.
• A loop runs from the second character to the last character of the string. Within this loop:
  • currentDecode is initialized to 0 for each character.
  • If the current character is not '0', currentDecode is updated to the value of prev (which represents the number of ways to decode the substring ending at the previous character).
  • The algorithm then checks if the two-character substring ending at the current character (composed of the current character and its predecessor) represents a valid pair number between 10 and 26. If valid, currentDecode is incremented by prevPrev (ways to decode excluding the last two characters).
  • The values of prevPrev and prev are updated for the next iteration.

At the end of the loop, prev holds the total number of ways to decode the entire string, which is then returned. This solution effectively uses dynamic programming principles to build up the number of decodings iteratively, thereby achieving a time-efficient decoding process.

int decodeWays(char* str) {
    if (str[0] == '0') return 0;
    int length = strlen(str);
    int prev = 1, nextToPrev = 1;
    for (int index = 1; index < length; index++) {
        int sum = 0;
        if (str[index] != '0') {
            sum = nextToPrev;
        }
        char pairDigits[3] = {str[index - 1], str[index], '\0'};
        int pairNum = atoi(pairDigits);
        if (pairNum >= 10 && pairNum <= 26) {
            sum += prev;
        }
        prev = nextToPrev;
        nextToPrev = sum;
    }
    return nextToPrev;
}

The provided C code illustrates a function decodeWays designed to compute the number of ways a given numeric string can be decoded, adhering to rules similar to the encoding schemes where 'A' to 'Z' correspond to '1' to '26'.

• The function first checks if the initial character of the string is '0'. Since there's no corresponding letter for '0', the function immediately returns 0.
• It determines the length of the string to manage the iteration over the characters.
• Two integer variables, prev and nextToPrev, are initialized to 1. These track the count of decoding possibilities up to the current and previous characters respectively.
• Inside a for loop, which iterates beginning from the second character of the string, it initializes an integer sum to 0. This variable accumulates the number of ways the substring up to the current index can be decoded.
• If the current character isn't '0', sum is updated with the value of nextToPrev because the current character could represent a valid letter alone.
• The code then constructs a two-character array pairDigits containing the current and previous characters, terminated by a null character. This array is converted to an integer pairNum using atoi.
• If pairNum falls between 10 and 26, both inclusive, it's a valid two-digit encoding, so sum is incremented by the value of prev.
• To proceed to the next iteration, prev is updated to nextToPrev and nextToPrev is updated to sum.

Finally, after the loop completes, nextToPrev is returned. This value represents the total number of ways the entire string can be decoded. This function efficiently uses dynamic programming principles, particularly memoization, to ensure each substring is only processed once, achieving optimal performance.

var decodeWays = function (encoded) {
    if (encoded.charAt(0) === "0") {
        return 0;
    }
    let len = encoded.length;
    let prevPrev = 1;
    let prev = 1;
    for (let index = 1; index < len; index++) {
        let temp = 0;
        if (encoded.charAt(index) !== "0") {
            temp = prev;
        }
        let doubleDigit = parseInt(encoded.slice(index - 1, index + 1));
        if (doubleDigit >= 10 && doubleDigit <= 26) {
            temp += prevPrev;
        }
        prevPrev = prev;
        prev = temp;
    }
    return prev;
};

The JavaScript function decodeWays is designed to determine the number of ways a given numeric string can be decoded, considering each number could represent a letter according to a predetermined mapping (similar to how 'A' might be 1, 'B' is 2, etc.). Here's a breakdown of how this function operates:

• First, check if the first character of the input string is "0". If so, return 0, as no valid encoding starts with zero.
• Initialize variables prevPrev and prev to 1. These will keep track of the number of ways to decode the string up to the characters at the current index minus two and minus one, respectively.
• Loop through the encoded string starting from the second character. For each character:
  • Set a temporary variable temp to zero.
  • If the character is not "0", update temp to the value of prev. This indicates that current digit can be used as a standalone decode unit, thus inheriting the decode possibilities of the previous sequence.
  • Check the two-digit number formed with the current and the previous character. If it's between 10 and 26, it can represent a letter. Increase temp by the value of prevPrev, adding the number of ways to decode considering this two-character combination.
• Update prevPrev with the value of prev, and prev with the value of temp for the next iteration of the loop.
• After completing the loop, prev will contain the total number of ways to decode the entire string.

This approach uses dynamic programming to efficiently solve the problem by breaking it down into smaller sub-problems, each depending only on the last couple of preceding solutions. Thus, the algorithm runs with a complexity of O(n), where n is the length of the input string.

class Solution:
    def decodeWays(self, data: str) -> int:
        if data[0] == "0":
            return 0

        prev = 1
        prev_prev = 1
        for index in range(1, len(data)):
            count = 0
            if data[index] != "0":
                count = prev
            pair = int(data[index - 1 : index + 1])
            if 10 <= pair <= 26:
                count += prev_prev
            prev_prev = prev
            prev = count

        return prev

The provided Python code solves the problem of counting the number of ways a numeric string can be decoded into characters. Assume each number corresponds to a letter (e.g., '1' is 'A', '2' is 'B', ..., '26' is 'Z'). This script uses dynamic programming to efficiently compute the solution by iterating through the string and calculating the number of ways it can be decoded up to each position.

Follow these main points in the code:

• Initialize two variables, prev and prev_prev, to keep track of the ways to decode the string up to the current and the previous position.
• Iterate through each character of the string starting from the second character.
  • Set count to zero. If the current character is not '0', set count to the value of prev since the current character alone can be decoded.
  • Check if the current and the previous characters together form a number between 10 and 26. If so, add the value of prev_prev to count because this pair can be a valid character.
  • Update prev_prev to the value of prev, and then update prev to the value of count.
• Return the value of prev, which represents the total number of ways to decode the entire string.

This method is efficient with a time complexity of O(n), where n is the length of the string, since it processes each character exactly once.