Decoded String at Index

class Solution {
public:
    string solve(string inputStr, int position) {
        long totalLength = 0;
        int length = inputStr.length();

        // Calculate total length when the string is fully decoded
        for (int index = 0; index < length; ++index) {
            if (isdigit(inputStr[index]))
                totalLength *= inputStr[index] - '0';
            else
                totalLength++;
        }

        for (int index = length - 1; index >= 0; --index) {
            position %= totalLength;
            if (position == 0 && isalpha(inputStr[index]))
                return (string) "" + inputStr[index];

            if (isdigit(inputStr[index]))
                totalLength /= inputStr[index] - '0';
            else
                totalLength--;
        }
        return "";
    }
};

The provided C++ code defines a function solve within a class Solution that decodes a pseudo-encoded string to find the character at a specific index when the string is fully expanded.

• The function accepts two parameters—a string inputStr and an integer i representing the position in the decoded string to retrieve.
• The goal is to determine what character would be at the index position in the string inputStr after it's fully decoded.

The code operates in two main phases:

1. Calculate the fully expanded length of the pseudo-encoded string.

   • Loop through each character in the inputStr.
   • If the character is a digit, the decoded length is multiplied by the numeric value of the digit.
   • If the character is an alphabet, increment the total decoded length by one.

2. Identify the character located at position without fully expanding the string:

   • Start iterating from the end of the string.
   • Use modulo operation to manage the size of the string's effective length (totalLength % position) to find the corresponding character in the compressed format.
   • If a digit is encountered, decrease the previously calculated total length by dividing it by the numeric value of the digit.
   • If the effective position equals zero and the current character is alphabetic, return this character as the result.

The function achieves the objective efficiently by avoiding the actual expansion of the string, using a reverse iteration and modulo operation to account for repeated patterns determined by the digits in the input string. This method is particularly effective for strings with repeats represented by large numbers, ensuring a lower time complexity than if the string were expanded fully.

class Solution {
    public String decodeStringAtIndex(String inputStr, int position) {
        long totalLength = 0;
        int inputLength = inputStr.length();

        // Calculate total length of the expanded string
        for (int i = 0; i < inputLength; ++i) {
            char currentChar = inputStr.charAt(i);
            if (Character.isDigit(currentChar))
                totalLength *= currentChar - '0';
            else
                totalLength++;
        }

        for (int i = inputLength - 1; i >= 0; --i) {
            char currentChar = inputStr.charAt(i);
            position %= totalLength;
            if (position == 0 && Character.isLetter(currentChar))
                return Character.toString(currentChar);

            if (Character.isDigit(currentChar))
                totalLength /= currentChar - '0';
            else
                totalLength--;
        }

        throw null; // This should not be reached normally
    }
}

The provided Java solution aims to find a specified character from the decoded version of an encoded string without fully expanding it, which would be computationally expensive for large input strings. This method efficiently handles the decoding process by leveraging properties of the encoded pattern and mathematical calculations to avoid full string expansion.

The method, decodeStringAtIndex, operates in two major parts:

1. Compute the total length of the fully expanded string:

   • Iterate over each character of the input string.
   • If the character is a digit, multiply the total length by this number (thus simulating the expansion).
   • If it's a letter, increment the total length by one.

2. Find the character at the specified position from the end to the start:

   • Iterate backwards over the string.
   • Use modulo operation with the current total length to find the effective position of the character.
   • If a letter is found at the adjusted position, return it.
   • If a digit is encountered, reduce the total length by dividing it by this digit (this undoes the expansion effect caused by that digit).

This approach ensures that you avoid the enormous memory usage and potential inefficiency of constructing the fully expanded string when it is unnecessary. Remember, an exception or error condition isn't expected to occur with correct input, and the condition to throw null serves as a fail-safe check in case of unexpected scenarios.

class Solution(object):
    def extractCharAtIndex(self, encodedStr, index):
        decodedLength = 0
        # Compute decoded string length
        for char in encodedStr:
            if char.isdigit():
                decodedLength *= int(char)
            else:
                decodedLength += 1

        for character in reversed(encodedStr):
            index %= decodedLength
            if index == 0 and character.isalpha():
                return character

            if character.isdigit():
                decodedLength /= int(character)
            else:
                decodedLength -= 1

This solution is designed to find the character at a given index from an encoded string without decoding the entire string, which increases efficiency, particularly with long strings. The function extractCharAtIndex in the Python class Solution accomplishes this through a two-step process:

1. Calculate the length of the fully decoded string:

   • Iterate through each character in the encoded input string. If the character is a digit, multiply the current length by this number to account for the repetition in the encoded pattern. If the character is alphabetical, simply increment the decoded length by one.

2. Identify the character at the specified index:

   • Traverse the string in reverse. During this reverse iteration, adjust the given index using modulo with the current computed length (index %= decodedLength). This accounts for repetitive patterns in decoding.
   • If the reversed character is alphabetical and matches the current position (index == 0), return this character as it is the one located at the original index.
   • If the reversed character is a digit, divide the current decoded length by this number to backtrack through the repetition of segments. If it is alphabetical, just decrement the decoded length reflecting the move back through the string one character at a time.

By using this approach, only the necessary parts of the string are computed, providing a result without unnecessary processing of the entire string. This method is particularly useful when dealing with very large encoded strings and reduces the memory and time complexity significantly.