LeetCode #779 — MEDIUM

K-th Symbol in Grammar

Move from brute-force thinking to an efficient approach using math strategy.

The Problem

Problem Statement

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}

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Bit Manipulation

Example 1

1
1

Example 2

2
1

Example 3

2
2

Step 02

Core Insight

What unlocks the optimal approach

Try to represent the current (N, K) in terms of some (N-1, prevK). What is prevK ?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #779: K-th Symbol in Grammar
class Solution {
    public int kthGrammar(int n, int k) {
        if (n == 1) {
            return 0;
        }
        if (k <= (1 << (n - 2))) {
            return kthGrammar(n - 1, k);
        }
        return kthGrammar(n - 1, k - (1 << (n - 2))) ^ 1;
    }
}

// Accepted solution for LeetCode #779: K-th Symbol in Grammar
func kthGrammar(n int, k int) int {
	if n == 1 {
		return 0
	}
	if k <= (1 << (n - 2)) {
		return kthGrammar(n-1, k)
	}
	return kthGrammar(n-1, k-(1<<(n-2))) ^ 1
}

# Accepted solution for LeetCode #779: K-th Symbol in Grammar
class Solution:
    def kthGrammar(self, n: int, k: int) -> int:
        if n == 1:
            return 0
        if k <= (1 << (n - 2)):
            return self.kthGrammar(n - 1, k)
        return self.kthGrammar(n - 1, k - (1 << (n - 2))) ^ 1

// Accepted solution for LeetCode #779: K-th Symbol in Grammar
struct Solution;

impl Solution {
    fn kth_grammar(n: i32, k: i32) -> i32 {
        let n = n as usize - 1;
        let k = k as usize - 1;
        Self::kth(n, k)
    }
    fn kth(n: usize, k: usize) -> i32 {
        if n == 0 {
            0
        } else {
            Self::kth(n - 1, k / 2) ^ (k % 2) as i32
        }
    }
}

#[test]
fn test() {
    let n = 1;
    let k = 1;
    let res = 0;
    assert_eq!(Solution::kth_grammar(n, k), res);
    let n = 2;
    let k = 1;
    let res = 0;
    assert_eq!(Solution::kth_grammar(n, k), res);
    let n = 2;
    let k = 2;
    let res = 1;
    assert_eq!(Solution::kth_grammar(n, k), res);
    let n = 4;
    let k = 5;
    let res = 1;
    assert_eq!(Solution::kth_grammar(n, k), res);
    let n = 30;
    let k = 417_219_134;
    let res = 1;
    assert_eq!(Solution::kth_grammar(n, k), res);
}

// Accepted solution for LeetCode #779: K-th Symbol in Grammar
function kthGrammar(n: number, k: number): number {
    if (n == 1) {
        return 0;
    }
    if (k <= 1 << (n - 2)) {
        return kthGrammar(n - 1, k);
    }
    return kthGrammar(n - 1, k - (1 << (n - 2))) ^ 1;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.