grid47 Exploring patterns and algorithms
Aug 21, 2024
5 min read
Solution to LeetCode 779: K-th Symbol in Grammar Problem
We build a sequence of rows starting with the string ‘0’ in the first row. For each subsequent row, every occurrence of ‘0’ from the previous row is replaced with ‘01’, and every occurrence of ‘1’ is replaced with ‘10’. Given two integers n and k, return the k-th (1-indexed) symbol in the n-th row.
Problem
Approach
Steps
Complexity
Input: The input consists of two integers n and k, where n represents the row number and k represents the position of the symbol in that row.
Example: Input: n = 3, k = 4
Constraints:
• 1 <= n <= 30
• 1 <= k <= 2^n - 1
Output: The output is the k-th (1-indexed) symbol in the n-th row of the sequence.
Example: Output: 1
Constraints:
• The output must be either 0 or 1, depending on the symbol at position k in row n.
Goal: The goal is to determine the k-th symbol in the n-th row without explicitly constructing the entire row.
Steps:
• If n is 1, return 0 (base case).
• If k is even, recursively find the symbol in the previous row at the index k / 2, and invert it.
• If k is odd, recursively find the symbol in the previous row at the index (k + 1) / 2, and return it as is.
Goal: The solution must handle large values of n and k efficiently. We cannot construct the entire row for large n due to exponential growth in size.
Steps:
• n is guaranteed to be at most 30, meaning we can handle 2^n symbols.
• k must be a valid index for the n-th row, i.e., 1 <= k <= 2^n - 1.
Assumptions:
• Both n and k are valid inputs, with k within the range for the n-th row.
• Input: Example 1: Input: n = 1, k = 1
• Explanation: The first row is '0', so the 1st symbol is '0'.
• Input: Example 2: Input: n = 3, k = 4
• Explanation: Row 3 is '0110', and the 4th symbol is '1'.
• Input: Example 3: Input: n = 2, k = 2
• Explanation: Row 2 is '01', and the 2nd symbol is '1'.
Approach: The approach involves leveraging the recursive pattern of the sequence generation. Instead of generating entire rows, we recursively find the k-th symbol in the previous row and determine its value.
Observations:
• The sequence generation follows a recursive pattern, where each row depends on the previous one.
• Instead of constructing the entire row, we can solve this problem by recursively narrowing down the position of the desired symbol.
Steps:
• Start from the n-th row and the k-th position.
• If n is 1, return 0 (base case).
• Otherwise, recursively compute the symbol by using the previous row's position based on whether k is odd or even.
Empty Inputs:
• There will always be valid values for n and k, as per the constraints.
Large Inputs:
• The solution must be efficient enough to handle n = 30, where the row size is 2^30.
Special Values:
• When k is 1 or 2^n - 1, the solution must handle these boundary cases.
Constraints:
• n is guaranteed to be at most 30, so handling up to 2^30 symbols is feasible.
intkthGrammar(int n, int k) {
if (n ==1) return0;
if (k%2==0) return kthGrammar(n -1, k /2) ==0?1:0;
elsereturn kthGrammar(n -1, (k +1) /2) ==0?0:1;
}
1 : Function Definition
intkthGrammar(int n, int k) {
Defines the function `kthGrammar` which takes two arguments: `n` (the row number) and `k` (the position in the row). It returns the value at the k-th position of the n-th row in a binary grammar sequence.
2 : Base Case
if (n ==1) return0;
The base case: when n equals 1, the first row of the grammar sequence is always [0]. Therefore, it returns 0.
3 : Recursive Case
if (k%2==0) return kthGrammar(n -1, k /2) ==0?1:0;
If k is even, the function calls itself recursively with n-1 and k/2, then returns the complement (1 if 0, 0 if 1) of the result.
4 : Recursive Case
elsereturnkthGrammar(n -1, (k +1) /2) ==0?0:1;
If k is odd, the function calls itself recursively with n-1 and (k+1)/2, and returns the same value (0 if 0, 1 if 1).
Best Case: O(log n), as we reduce the problem size by half at each step.
Average Case: O(log n), since the recursive depth is logarithmic.
Worst Case: O(log n), as the maximum depth of recursion is log n.
Description: Time complexity is logarithmic due to the halving nature of the recursion.
Best Case: O(1), when no recursion depth is needed.
Worst Case: O(log n), as the recursion depth can be at most log n.
Description: Space complexity is logarithmic due to recursion stack.
