Welcome to Subscribe On Youtube

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

Description

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:

Initially, 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.

Solutions

Solution 1: Recurrence

Since the length of the string doubles after each operation, if we perform $i$ operations, the length of the string will be $2^i$.

We can simulate this process to find the first string length $n$ that is greater than or equal to $k$.

Next, we backtrack and discuss the following cases:

  • If $k \gt n / 2$, it means $k$ is in the second half. If $\textit{operations}[i - 1] = 1$, it means the character at position $k$ is obtained by adding $1$ to the character in the first half. We add $1$ to it. Then we update $k$ to $k - n / 2$.
  • If $k \le n / 2$, it means $k$ is in the first half and is not affected by $\textit{operations}[i - 1]$.
  • Next, we update $n$ to $n / 2$ and continue backtracking until $n = 1$.

Finally, we take the resulting number modulo $26$ and add the ASCII code of 'a' to get the answer.

The time complexity is $O(\log k)$, and the space complexity is $O(1)$.

  • class Solution {
    public char kthCharacter(long k, int[] operations) {
        long n = 1;
        int i = 0;
        while (n < k) {
            n *= 2;
            ++i;
        }
        int d = 0;
        while (n > 1) {
            if (k > n / 2) {
                k -= n / 2;
                d += operations[i - 1];
            }
            n /= 2;
            --i;
        }
        return (char) ('a' + (d % 26));
    }
}


  • class Solution {
public:
    char kthCharacter(long long k, vector<int>& operations) {
        long long n = 1;
        int i = 0;
        while (n < k) {
            n *= 2;
            ++i;
        }
        int d = 0;
        while (n > 1) {
            if (k > n / 2) {
                k -= n / 2;
                d += operations[i - 1];
            }
            n /= 2;
            --i;
        }
        return 'a' + (d % 26);
    }
};


  • class Solution:
    def kthCharacter(self, k: int, operations: List[int]) -> str:
        n, i = 1, 0
        while n < k:
            n *= 2
            i += 1
        d = 0
        while n > 1:
            if k > n // 2:
                k -= n // 2
                d += operations[i - 1]
            n //= 2
            i -= 1
        return chr(d % 26 + ord("a"))


  • func kthCharacter(k int64, operations []int) byte {
	n := int64(1)
	i := 0
	for n < k {
		n *= 2
		i++
	}
	d := 0
	for n > 1 {
		if k > n/2 {
			k -= n / 2
			d += operations[i-1]
		}
		n /= 2
		i--
	}
	return byte('a' + (d % 26))
}


  • function kthCharacter(k: number, operations: number[]): string {
    let n = 1;
    let i = 0;
    while (n < k) {
        n *= 2;
        i++;
    }
    let d = 0;
    while (n > 1) {
        if (k > n / 2) {
            k -= n / 2;
            d += operations[i - 1];
        }
        n /= 2;
        i--;
    }
    return String.fromCharCode('a'.charCodeAt(0) + (d % 26));
}

All Problems

All Solutions