
Problem Statement
In this task, we need to create a unique table composed of n
rows, starting with the number 0
in the first row. As we proceed to fill the rows, each occurrence of 0
in a row is replaced with 01
and each occurrence of 1
is replaced with 10
in the subsequent row. The challenge is not only about constructing such a patterned table but also identifying the kth
character in the nth
row of this table.
For example, when n = 3
, the sequence in the rows would appear as:
- 1st row:
0
- 2nd row:
01
(transformation of first row) - 3rd row:
0110
(transformation of second row)
Testing comprehension of this sequence transformation and efficient retrieval is essential when given two integers n
(which indicates the row number) and k
(which indicates the position in that row).
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
Example 3
Input:
n = 2, k = 2
Output:
1
Explanation:
row 1: 0 row 2: 01
Constraints
1 <= n <= 30
1 <= k <= 2n - 1
Approach and Intuition
When approaching this problem, the key observations from the description and examples can guide us towards an effective solution:
Sequence Generation:
- The problem essentially transforms sequences through very specific rules:
0
becomes01
and1
becomes10
. - Recognizing that the sequence's length doubles with each row (starting at 1 for
n = 1
, then 2, 4, 8, etc.), the sequence's size for any rown
can be determined to be2^n - 1
.
- The problem essentially transforms sequences through very specific rules:
Pattern Realization:
- A crucial point here is realizing that the pattern displays a fractal-like recursion. Hence, each transformation is a function of the previous state, and the entire row can be seen as a function of transformations starting from a single
0
.
- A crucial point here is realizing that the pattern displays a fractal-like recursion. Hence, each transformation is a function of the previous state, and the entire row can be seen as a function of transformations starting from a single
Efficient Retrieval:
- Calculating every row up to
n
explicitly might be computationally expensive but can be avoided using recursive insights. - Recognizing whether the
kth
bit originated from a0
or a1
in the previous generation can significantly cut down the calculation time. Ifk
falls in the first half of the array, it resulted from a0
. Otherwise, it came from a1
. This allows us to determine the bit at positionk
recursively without constructing the entire sequence.
- Calculating every row up to
Edge and Constraint Handling:
- With
n
maxing out at 30, the maximum row length could become quite large (2^30 - 1
), making a direct computation of large rows inefficient. The constraints (1 <= n <= 30
and1 <= k <= 2^n - 1
) clarify the upper limits we need to manage.
- With
Through these observations, it is evident that understanding the recursive pattern and using it to jump directly to the kth
position in the nth
row without explicit full sequence generation is feasible and optimal.
Solutions
- C++
- Java
- JavaScript
- Python
class Solution {
public:
int kthSymbol(int N, int K) {
int popCount = __builtin_popcount(K - 1);
return popCount & 1 ? 1 : 0;
}
};
The given C++ solution defines a function named kthSymbol
which determines the K-th symbol in an N-th row of a specific grammar sequence. The essence of this solution leverages built-in bit manipulation functions provided by C++ to achieve its goal effectively. Here's how it operates:
- The function
kthSymbol
receives two parameters:N
andK
, whereN
represents the row number andK
indicates the position in that row. - The code computes the population count (number of 1's in the binary representation) of
K-1
using the function__builtin_popcount
. - It then checks if this population count is odd or even using a bitwise AND operation with 1 (
popCount & 1
). - If the result is odd (
popCount & 1 ? 1 : 0
), it returns 1; otherwise, it returns 0.
This method utilizes the properties of binary representation and population count to dissect the recursive pattern presented by this specific grammar sequence efficiently. Keep in mind the direct use of bit manipulation functions for optimized computation.
class Solution {
public int findKthSymbol(int level, int position) {
int bits = Integer.bitCount(position - 1);
return bits % 2 == 0 ? 0 : 1;
}
}
The problem involves finding the K-th symbol in a specific grammar sequence at a given level and position. The solution utilizes the Java programming language.
To determine the K-th symbol:
- The method
findKthSymbol
receives two parameters:level
, which specifies the depth of the sequence, andposition
, which identifies the specific position in that level for which the symbol is required. - Calculate the Hamming weight (number of 1s) in the binary representation of
position - 1
using the Java methodInteger.bitCount(position - 1)
. - Determine the symbol by checking if the count of 1s is even or odd. If even, the method returns 0; if odd, it returns 1.
This approach effectively uses bitwise calculation to find the required symbol, utilizing properties of binary numbers to determine the pattern of the sequence at any depth and position. The use of bitCount
provides a compact and efficient way to compute the solution.
let grammarValue = function(level, position) {
const bitCount = (position - 1).toString(2).split('1').length - 1;
return bitCount % 2 === 0 ? 0 : 1;
};
The problem "K-th Symbol in Grammar" involves determining the nth symbol in a binary grammar sequence at a given level, where the sequence grows exponentially. The provided JavaScript solution leverages bitwise operations and properties of binary numbers to efficiently solve the problem.
The function grammarValue
accepts two parameters:
level
: The specific level in the grammar sequence where the symbol is located.position
: The position of the symbol in the sequence at the specified level.
Here’s a breakdown of how the solution works:
- Compute the zero-based index by subtracting one from
position
. - Convert this index to a binary string to easily count the number of '1's. This is crucial because the k-th grammar sequence can be visualized as a binary tree where flips in values occur with each '1' in the binary sequence.
- Count the occurrences of '1's in the binary representation. This is done by converting the number to a binary string, splitting the string by '1', and subtracting one to get the total count of '1's.
- Determine the final value at the position by checking if the count of '1's is even or odd. If even, the function returns 0; if odd, it returns 1.
This solution efficiently uses binary properties to decide the value at any given position in the sequence without generating the entire sequence, effectively reducing time and space complexity.
class Solution:
def findKthBit(self, level: int, position: int) -> int:
bitCount = bin(position - 1).count('1')
return 0 if bitCount % 2 == 0 else 1
The Python3 program titled "K-th Symbol in Grammar" presents a solution for determining the K-th bit in a specific grammar sequence rule. In this sequence, the way bits are generated involves flipping and inverting elements from the previous level.
- The
findKthBit
function in theSolution
class accepts two parameters:level
andposition
. - It calculates the bit's value by determining the number of 1s in the binary representation of
position-1
. - This count determines the parity of the number of 1s. If the count of 1s is even, the function returns 0, otherwise, it returns 1.
Ensure the level and position values adhere to your specific case constraints when implementing or testing this function in different scenarios to guarantee appropriate functionality. This algorithm optimally determines the desired position bit, leveraging binary operations for efficiency.
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