3307. Find the K-th Character in String Game II

3307. Find the K-th Character in String Game II

Difficulty: Hard

Topics: Math, Bit Manipulation, Recursion

Alice and Bob are playing a game. Initially, Alice has a string word = "a".

You are given a positive integer k. You are also given an integer array operations, where operations[i] represents the type of the ith operation.

Now Bob will ask Alice to perform all operations in sequence:

• If operations[i] == 0, append a copy of word to itself.
• If operations[i] == 1, generate a new string by changing each character in word to its next character in the English alphabet, and append it to the original word. For example, performing the operation on "c" generates "cd" and performing the operation on "zb" generates "zbac".

Return the value of the kth character in word after performing all the operations.

Note that the character 'z' can be changed to 'a' in the second type of operation.

Example 1:

• Input: k = 5, operations = [0,0,0]
• Output: "a"
• Explanation: Initially, word == "a". Alice performs the three operations as follows:
  • Appends "a" to "a", word becomes "aa".
  • Appends "aa" to "aa", word becomes "aaaa".
  • Appends "aaaa" to "aaaa", word becomes "aaaaaaaa".

Example 2:

• Input: k = 10, operations = [0,1,0,1]
• Output: "b"
• Explanation: nitially, word == "a". Alice performs the four operations as follows:
  • Appends "a" to "a", word becomes "aa".
  • Appends "bb" to "aa", word becomes "aabb".
  • Appends "aabb" to "aabb", word becomes "aabbaabb".
  • Appends "bbccbbcc" to "aabbaabb", word becomes "aabbaabbbbccbbcc".

Constraints:

• 1 <= k <= 1014
• 1 <= operations.length <= 100
• operations[i] is either 0 or 1.
• The input is generated such that word has at least k characters after all operations.

Hint:

1. Try to replay the operations kth character was part of.
2. The kth character is only affected if it is present in the first half of the string.

Solution:

We need to determine the k-th character in a string after performing a series of operations. The string starts as "a", and each operation either appends a copy of the current string to itself (operation 0) or appends a transformed version of the string where each character is incremented by 1 in the alphabet (operation 1, with 'z' wrapping around to 'a'). Given that k can be as large as 10¹⁴ and the number of operations can be up to 100, we need an efficient approach that avoids explicitly constructing the string.

Approach

1. Problem Analysis: The key observation is that each operation doubles the length of the string. For operation 0, the second half is an exact copy of the first half. For operation 1, the second half is each character of the first half incremented by 1. The challenge is to efficiently determine the k-th character without constructing the entire string, especially since the final string length can be 2¹⁰⁰, which is infeasible to handle directly.

2. Backward Simulation: Instead of processing operations forward to build the string, we simulate the operations backward. Starting from the final string length 2ⁿ (where n is the number of operations), we reduce the problem step-by-step:

   • For each operation from last to first, check if the k-th character is in the first half or the second half of the current string.
   • If it's in the first half, we simply proceed to the previous state (half the length).
   • If it's in the second half, we adjust k by subtracting half the length. If the operation was type 1, we note that the character was transformed (incremented by 1) and will need to reverse this transformation later.
   • To handle large lengths efficiently, we skip detailed processing when the half-length exceeds 2⁴⁸ (since 2⁴⁸ > 10¹⁴), ensuring k is always in the first half in such cases.

3. Base Case Handling: After processing all operations backward, the base string is "a". The total transformations (shifts) applied during the backward pass are summed up modulo 26. The final character is derived by applying these shifts to 'a'.

Let's implement this solution in PHP: 3307. Find the K-th Character in String Game II

<?php
/**
 * @param Integer $k
 * @param Integer[] $operations
 * @return String
 */
function kthCharacter($k, $operations) {
    ...
    ...
    ...
    /**
     * go to ./solution.php
     */
}

// Test cases
// Example 1
$k = 5;
$operations = [0, 0, 0];
echo kthCharacter($k, $operations) . "\n"; // Output: a

// Example 2
$k = 10;
$operations = [0, 1, 0, 1];
echo kthCharacter($k, $operations) . "\n"; // Output: b
?>

Explanation:

1. Initialization: We start with total_shift set to 0, which will accumulate the number of transformations (increments) applied to the base character 'a'. The exponent exp is initialized to the number of operations n, representing the initial string length 2ⁿ.

2. Backward Processing:

   • For each operation from last to first, we check if the current half-length (2^exp-1) is manageable (less than 2⁴⁸). If so, we compute the half-length.
   • If k is in the second half of the current string, we adjust k by subtracting the half-length. If the operation was type 1, we increment total_shift.
   • If the half-length is too large (≥ 2⁴⁸), we skip detailed processing since k will always be in the first half.
   • We decrement exp after each operation to move to the previous state (half the string length).

3. Final Character Calculation: After processing all operations, total_shift is taken modulo 26 to handle wrap-around. The result is obtained by shifting 'a' by total_shift positions in the alphabet.

This approach efficiently narrows down the position k through backward simulation, avoiding the need to handle the exponentially large string directly, and leverages modular arithmetic to handle character transformations. The complexity is O(n), where n is the number of operations, making it suitable for large k.

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