Gray Code
Logical Breakdown
- [ ] Subproblem Identification:
- [ ] Optimal Substructure:
- [ ] Constraint Handling: (e.g., Modulo \(10^9 + 7\))
- [ ] Optimization: (Matrix Exponentiation if 'A' is huge)
Mathematical Rigor
State Definition
Let \(dp[i]\) be the state for the \(i\)-th subproblem.
Recurrence Relation
Visualization
Complexity Analysis
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Iterative DP | \(O(N)\) | \(O(1)\) |
Code Reference
package com.dsa.recursion;
import java.util.*;
/ * Problem: Gray Code * Group: 07. Recursion */ / * ============================================================================================ * PROBLEM: Gray Code * ============================================================================================ * * --------------------- * 1. Problem Description * --------------------- * The Gray Code is a binary numeral system where two successive values differ in only one bit. * Given a non-negative integer A representing the number of bits, print the sequence of Gray Codes. * The sequence must begin with 0. * * --------------------- * 2. Logical Breakdown & Mathematical Formulation * --------------------- * Let G(n) be the sequence of Gray Codes for n bits. * * Recursive Construction (Reflection Method): * To construct G(n) from G(n-1): * 1. Start with the sequence G(n-1). * 2. Create a reversed copy of G(n-1). * 3. Prefix '0' to the original sequence (values remain unchanged). * 4. Prefix '1' to the reversed sequence (adds 2^(n-1) to each value). * 5. Concatenate the two resulting sequences. * * Recurrence Relation: * G(n) = G(n-1) concatenated with [ (x + 2^(n-1)) for x in reverse(G(n-1)) ] * * Base Case: * G(1) = [0, 1] * * Mathematical Verification: * Since G(n-1) has the Gray Code property, the first half is valid. * Since we reflect (reverse) the second half, the transition point (middle) differs by exactly * the MSB (2^(n-1)). * All adjacent elements in the second half also preserve the 1-bit difference property. * * --------------------- * 3. Constraints & Complexity Analysis * --------------------- * Constraints: * 1 <= A <= 16 * * Time Complexity: * O(2^A) - We generate exactly 2^A elements. Each step involves iterating over the previous list. * * Space Complexity: * O(2^A) - To store the output sequence. * With A=16, 2^16 = 65,536 integers, which fits comfortably in memory. * * --------------------- * 4. Input / Output Format * --------------------- * Input: * - An integer A. * * Output: * - A list/array of integers representing the Gray Code sequence. * * --------------------- * 5. Examples * --------------------- * Input 1: A = 1 * Output 1: [0, 1] * * Input 2: A = 2 * Sequence Generation: * 1. Take G(1): [0, 1] * 2. Reverse G(1): [1, 0] * 3. Add 2^(2-1) = 2 to reversed: [3, 2] * 4. Concatenate: [0, 1, 3, 2] * * Explanation (Binary): * 00 (0) * 01 (1) -> differs by last bit * 11 (3) -> differs by first bit * 10 (2) -> differs by last bit * * ============================================================================================ */ public class GrayCode { public int solve(int A) { // TODO: Implement solution return 0; }
public ArrayList<Integer> grayCode(int i) {
return new ArrayList<>();
}
}