Problem Statement

We build a table of n rows (1-indexed). We start by writing 0 in the 1st row. Now in every subsequent row, we look at the previous row and replace each occurrence of 0 with 01, and each occurrence of 1 with 10.

• For example, for n = 3, the 1st row is 0, the 2nd row is 01, and the 3rd row is 0110.

Given two integer n and k, return the kth (1-indexed) symbol in the nth row of a table of n rows.

Examples

Example 1:

Input: n = 1, k = 1
Output: 0

Explanation: row 1: 0

Example 2:

Input: n = 2, k = 1
Output: 0

Explanation:
row 1: 0
row 2: 01
Here is row 2, at index 1 is 0, while at index 2 is 1.

Example 3:

Input: n = 2, k = 2
Output: 1

Explanation:
row 1: 0
row 2: 01
Here in row 2, at index 2 is 1.

Different Approaches

Intuition

The problem revolves around understanding how the grammar sequence is constructed:

1. Starting from 0, each subsequent row is formed by replacing 0 with 01 and 1 with 10.
2. As a result, the sequence grows exponentially, and we cannot generate the full sequence for large values of n efficiently. Instead, we can use a recursive approach to directly determine the k-th symbol without constructing the whole sequence.

Approach

1. Recursive Definition:
   • If n = 1, the sequence is simply 0. Therefore, the k-th symbol is 0.
   • For n > 1, the sequence at row n has a length of 2^(n-1). We can split this sequence into two halves:
     • The first half (length 2^(n-2)) is identical to the sequence of row n-1.
     • The second half (length 2^(n-2)) is the complement of the sequence of row n-1.
   • To find the k-th symbol in the n-th row:
     • If k is in the first half, the k-th symbol is the same as in the (n-1)-th row.
     • If k is in the second half, the k-th symbol is the complement of the symbol in the (k - mid) position of the (n-1)-th row.
   • This observation allows us to recursively determine the k-th symbol in O(n) time.

Code:

#include <iostream>
using namespace std;

// Function to determine the k-th symbol in the n-th row
int kthSymbolInGrammar(int n, int k) {
    // Base case: If we're at the first row, the only symbol is 0
    if (n == 1 && k == 1) return 0;

    // Find the length of the row
    int mid = (1 << (n - 1)) / 2; // 2^(n-1) / 2

    // If k is less than or equal to mid, the symbol is the same as in the first half
    if (k <= mid) {
        return kthSymbolInGrammar(n - 1, k);
    } else {
        // Otherwise, it's the complement of the corresponding symbol in the first half
        return !kthSymbolInGrammar(n - 1, k - mid);
    }
}

int main() {
    int n, k;
    cout << "Enter n: ";
    cin >> n;
    cout << "Enter k: ";
    cin >> k;

    cout << "The " << k << "-th symbol in the " << n << "-th row is: " 
         << kthSymbolInGrammar(n, k) << endl;

    return 0;
}