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K-th Symbol in Grammar

Medium

Asked by:

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Topics:

RecursionBit Manipulation

We build a table of n rows (1-indexed). We start by writing 0 in the 1^st 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 1^st row is 0, the 2^nd row is 01, and the 3^rd row is 0110.

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

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 <= 2^{n - 1}

Try coding it in LeetCode

K-th Symbol in Grammar

Medium

Asked by:

22 views

Topics:

RecursionBit Manipulation

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

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

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 <= 2^{n - 1}

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on n and k? What are the maximum values they can take?
Is n always a positive integer, and is k always a positive integer less than or equal to the length of the n-th row?
Could you provide an example for n=3 and k=3 to ensure I understand the pattern?
Are there any specific error conditions I should handle, like invalid inputs such as n=0 or k=0?
If n is large, how should I avoid generating the entire string for the n-th row, as that could lead to memory issues?

Brute Force Solution

Approach

We want to find a specific symbol in a sequence that grows according to a set of rules. The brute force way is to fully build out this sequence step-by-step, based on the provided rules, until we have a sequence large enough to find the symbol we need.

Here's how the algorithm would work step-by-step:

Start with the first sequence which is just a single symbol, either 0 or 1, depending on the rule.
Use the rules to transform the first sequence into the second sequence.
Then, use the rules again to transform the second sequence into the third sequence.
Keep repeating this transformation process to build each sequence from the previous one.
Do this until you have built the sequence for the row you are interested in.
Once you have the entire sequence for that row, just look at the position corresponding to the symbol you want to find, and that's your answer.

Code Implementation

def kth_symbol_in_grammar_brute_force(row_number, k_position):
    current_sequence = "0"

    # Iterate to build each row up to the desired row
    for _ in range(1, row_number):
        next_sequence = ""

        # Build the next row based on the current row.
        for symbol in current_sequence:

            # Apply grammar rules: 0 becomes 01, 1 becomes 10
            if symbol == '0':
                next_sequence += "01"
            else:
                next_sequence += "10"
        current_sequence = next_sequence

    # Return the k-th symbol (adjust index because it starts from 1)
    return int(current_sequence[k_position - 1])

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach constructs each row of the grammar from the previous row. To build row n, we first need to build row n-1, and so on, all the way down to row 1. Each row doubles in size compared to the previous row. Therefore, building row n requires creating a string of length 2^(n-1). The time complexity is dominated by the construction of the final string, which takes O(2^n) time as the length of the string we construct grows exponentially with n.

Space Complexity

O(2^(N-1)) – The brute force approach builds each sequence row by row. To construct the Nth row, it must store the entire sequence from the (N-1)th row. The length of each row doubles with each iteration. Therefore, the (N-1)th row has a length of 2^(N-1). Since the entire (N-1)th row sequence needs to be stored in memory to construct the Nth row sequence, the auxiliary space required is proportional to the length of the (N-1)th row, which is 2^(N-1). This dominates space usage.

Optimal Solution

Approach

This problem can be solved efficiently by recognizing a pattern instead of building the entire grammar. We use recursion and focus on figuring out how the 'k-th' symbol is generated from the previous row, based on the rules of the grammar.

Here's how the algorithm would work step-by-step:

Think about the pattern: each row is derived from the previous row's bits. A '0' becomes '01', and a '1' becomes '10'.
The key is to trace the 'k-th' position back to the first row. Figure out which half of the previous row it came from.
If 'k' is in the second half of the previous row, it means the first half was copied and then the second half was added, shifted to the right.
Adjust 'k' to represent its position relative to the first half of the previous row if it was in the second half.
Invert the symbol if 'k' was originally in the second half of the previous row, because 0s become 1s, and 1s become 0s.
Repeat this process by going up row by row until you reach the first row, which always contains a '0'.
The final '0' or '1' you arrive at represents the k-th symbol in the specified row.

Code Implementation

def kth_grammar(row_number, k_index):
    inversion_count = 0

    # Keep iterating up to the root of the implicit binary tree
    while row_number > 1:
        row_length = 2**(row_number - 2)

        # If k_index is in the second half of the string,
        if k_index > row_length:
            # Mark that we need to invert the bit.
            inversion_count += 1
            k_index -= row_length

        row_number -= 1

    # Apply accumulated inversions to the base '0' to get the result
    return inversion_count % 2

Big(O) Analysis

Time Complexity

O(n) – The algorithm uses recursion to trace the k-th symbol back to the first row. In each recursive call, we determine if k is in the second half of the previous row and adjust k accordingly, effectively halving the problem size with each step. The number of recursive calls is determined by the row number n, as we go up row by row until we reach the first row. Therefore, the time complexity is directly proportional to n, resulting in O(n).

Space Complexity

O(N) – The dominant factor for space complexity is the recursion depth. The provided explanation suggests tracing back from the k-th position to the first row through recursion. Each recursive call adds a new frame to the call stack. Since the recursion goes up row by row until reaching the first row, the maximum depth of the recursion equals N, where N is the row number. Therefore, the space complexity is O(N) due to the call stack.

Edge Cases

Case	How to Handle
n = 0 (invalid row)	Return an error or define the result as 0 or 1 depending on the specified requirement, ensuring invalid input is accounted for.
k = 0 (invalid position)	Return an error or define the result as 0 or 1 depending on the specified requirement, ensuring invalid input is accounted for.
n = 1, k = 1 (base case)	Return 0 as the base case for the first row and first position, ensuring correct start of the grammar.
k > 2^(n-1) (k exceeds row length)	Return an error or a specific value (e.g., -1) indicating that the requested position is beyond the row's length, ensuring bounds are respected.
Maximum value of n (potential stack overflow with recursion)	Use an iterative approach or memoization to avoid excessive recursion depth that could lead to stack overflow for large n.
Large values of n and k, leading to integer overflow during calculations of lengths	Use appropriate data types (e.g., long) to prevent integer overflow when calculating row lengths or positions, ensuring accurate results.
n is negative or very large (outside realistic range)	Validate the input n and k within a reasonable range, and return an error if it's outside those limits, safeguarding against incorrect calculations.
Solution incorrectly calculates the parity / flips bits	Thoroughly test with a wide variety of n and k to confirm correct calculation logic, potentially using bit manipulation techniques.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on n and k? What are the maximum values they can take?
Is n always a positive integer, and is k always a positive integer less than or equal to the length of the n-th row?
Could you provide an example for n=3 and k=3 to ensure I understand the pattern?
Are there any specific error conditions I should handle, like invalid inputs such as n=0 or k=0?
If n is large, how should I avoid generating the entire string for the n-th row, as that could lead to memory issues?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start with the first sequence which is just a single symbol, either 0 or 1, depending on the rule.
Use the rules to transform the first sequence into the second sequence.
Then, use the rules again to transform the second sequence into the third sequence.
Keep repeating this transformation process to build each sequence from the previous one.
Do this until you have built the sequence for the row you are interested in.
Once you have the entire sequence for that row, just look at the position corresponding to the symbol you want to find, and that's your answer.

Code Implementation

def kth_symbol_in_grammar_brute_force(row_number, k_position):
    current_sequence = "0"

    # Iterate to build each row up to the desired row
    for _ in range(1, row_number):
        next_sequence = ""

        # Build the next row based on the current row.
        for symbol in current_sequence:

            # Apply grammar rules: 0 becomes 01, 1 becomes 10
            if symbol == '0':
                next_sequence += "01"
            else:
                next_sequence += "10"
        current_sequence = next_sequence

    # Return the k-th symbol (adjust index because it starts from 1)
    return int(current_sequence[k_position - 1])

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Think about the pattern: each row is derived from the previous row's bits. A '0' becomes '01', and a '1' becomes '10'.
The key is to trace the 'k-th' position back to the first row. Figure out which half of the previous row it came from.
If 'k' is in the second half of the previous row, it means the first half was copied and then the second half was added, shifted to the right.
Adjust 'k' to represent its position relative to the first half of the previous row if it was in the second half.
Invert the symbol if 'k' was originally in the second half of the previous row, because 0s become 1s, and 1s become 0s.
Repeat this process by going up row by row until you reach the first row, which always contains a '0'.
The final '0' or '1' you arrive at represents the k-th symbol in the specified row.

Code Implementation

def kth_grammar(row_number, k_index):
    inversion_count = 0

    # Keep iterating up to the root of the implicit binary tree
    while row_number > 1:
        row_length = 2**(row_number - 2)

        # If k_index is in the second half of the string,
        if k_index > row_length:
            # Mark that we need to invert the bit.
            inversion_count += 1
            k_index -= row_length

        row_number -= 1

    # Apply accumulated inversions to the base '0' to get the result
    return inversion_count % 2

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
n = 0 (invalid row)	Return an error or define the result as 0 or 1 depending on the specified requirement, ensuring invalid input is accounted for.
k = 0 (invalid position)	Return an error or define the result as 0 or 1 depending on the specified requirement, ensuring invalid input is accounted for.
n = 1, k = 1 (base case)	Return 0 as the base case for the first row and first position, ensuring correct start of the grammar.
k > 2^(n-1) (k exceeds row length)	Return an error or a specific value (e.g., -1) indicating that the requested position is beyond the row's length, ensuring bounds are respected.
Maximum value of n (potential stack overflow with recursion)	Use an iterative approach or memoization to avoid excessive recursion depth that could lead to stack overflow for large n.
Large values of n and k, leading to integer overflow during calculations of lengths	Use appropriate data types (e.g., long) to prevent integer overflow when calculating row lengths or positions, ensuring accurate results.
n is negative or very large (outside realistic range)	Validate the input n and k within a reasonable range, and return an error if it's outside those limits, safeguarding against incorrect calculations.
Solution incorrectly calculates the parity / flips bits	Thoroughly test with a wide variety of n and k to confirm correct calculation logic, potentially using bit manipulation techniques.