Grey Code

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An n-bit gray code sequence is a sequence of 2n integers where:

  • Every integer is in the inclusive range [0, 2n - 1],
  • The first integer is 0,
  • An integer appears no more than once in the sequence,
  • The binary representation of every pair of adjacent integers differs by exactly one bit, and
  • The binary representation of the first and last integers differs by exactly one bit.

Given an integer n, return any valid n-bit gray code sequence.

Example 1:

Input: n = 2

Output: [0,1,3,2]

Explanation:

The binary representation of [0,1,3,2] is [00,01,11,10].

  • 00 and 01 differ by one bit
  • 01 and 11 differ by one bit
  • 11 and 10 differ by one bit
  • 10 and 00 differ by one bit

[0,2,3,1] is also a valid gray code sequence, whose binary representation is [00,10,11,01].

  • 00 and 10 differ by one bit
  • 10 and 11 differ by one bit
  • 11 and 01 differ by one bit
  • 01 and 00 differ by one bit

Example 2:

Input: n = 1

Output: [0,1]

Constraints:

  • 1 <= n <= 16

Solution

In this question, we are dealing with binary representation of integers, we will be using bit manipulation. To learn more, checkout Bitmask Intro.

The two important operations we will use are the shift operator (<<) and the XOR (exclusive or) operator (^).

  • 1 << i: will shift 1 by i bits to the left, the new binary number will have i zeros after a 1.
  • code ^ (1 << i): flips the ith bit (from the right) in code's binary representation. By the property of XOR that 0^1 = 1 and 1^1 = 0.

Now, we can use these two operations to find the candidates for the next code. And since we are only looking for one solution, we can use the aggregation template to return earlier. We will apply the backtracking2 template and fill in the logic.

  • is_leaf: start_index == 2**n, we have used all integers
  • get_edges: new_code = code ^ (1 << i) where 0 <= i < n, this new_code has one flipped bit on the i-th bit, it can potentially be the next code
  • is_valid: new_code is valid when we have not used it (not in visited)

Implementation

1def grayCode(self, n: int) -> List[int]:
2    length = 1 << n   # same as 2**n
3    visited = [False] * length
4
5    def dfs(start_index, code):
6        if start_index == length:
7            return True
8        for i in range(n):
9            new_code = code ^ (1 << i)
10            print(new_code)
11            if not visited[new_code]:
12                path.append(new_code)
13                visited[new_code] = True
14                if dfs(start_index+1, new_code): return True
15                visited[new_code] = False
16                path.pop()
17        return False
18        
19    path = [0]
20    visited[0] = True
21    dfs(1, 0)
22    return path

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What is the worst case running time for finding an element in a binary tree (not necessarily binary search tree) of size n?

Solution Implementation

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Fast Track Your Learning with Our Quick Skills Quiz:

What's the output of running the following function using input [30, 20, 10, 100, 33, 12]?

1def fun(arr: List[int]) -> List[int]:
2    import heapq
3    heapq.heapify(arr)
4    res = []
5    for i in range(3):
6        res.append(heapq.heappop(arr))
7    return res
8
1public static int[] fun(int[] arr) {
2    int[] res = new int[3];
3    PriorityQueue<Integer> heap = new PriorityQueue<>();
4    for (int i = 0; i < arr.length; i++) {
5        heap.add(arr[i]);
6    }
7    for (int i = 0; i < 3; i++) {
8        res[i] = heap.poll();
9    }
10    return res;
11}
12
1class HeapItem {
2    constructor(item, priority = item) {
3        this.item = item;
4        this.priority = priority;
5    }
6}
7
8class MinHeap {
9    constructor() {
10        this.heap = [];
11    }
12
13    push(node) {
14        // insert the new node at the end of the heap array
15        this.heap.push(node);
16        // find the correct position for the new node
17        this.bubble_up();
18    }
19
20    bubble_up() {
21        let index = this.heap.length - 1;
22
23        while (index > 0) {
24            const element = this.heap[index];
25            const parentIndex = Math.floor((index - 1) / 2);
26            const parent = this.heap[parentIndex];
27
28            if (parent.priority <= element.priority) break;
29            // if the parent is bigger than the child then swap the parent and child
30            this.heap[index] = parent;
31            this.heap[parentIndex] = element;
32            index = parentIndex;
33        }
34    }
35
36    pop() {
37        const min = this.heap[0];
38        this.heap[0] = this.heap[this.size() - 1];
39        this.heap.pop();
40        this.bubble_down();
41        return min;
42    }
43
44    bubble_down() {
45        let index = 0;
46        let min = index;
47        const n = this.heap.length;
48
49        while (index < n) {
50            const left = 2 * index + 1;
51            const right = left + 1;
52
53            if (left < n && this.heap[left].priority < this.heap[min].priority) {
54                min = left;
55            }
56            if (right < n && this.heap[right].priority < this.heap[min].priority) {
57                min = right;
58            }
59            if (min === index) break;
60            [this.heap[min], this.heap[index]] = [this.heap[index], this.heap[min]];
61            index = min;
62        }
63    }
64
65    peek() {
66        return this.heap[0];
67    }
68
69    size() {
70        return this.heap.length;
71    }
72}
73
74function fun(arr) {
75    const heap = new MinHeap();
76    for (const x of arr) {
77        heap.push(new HeapItem(x));
78    }
79    const res = [];
80    for (let i = 0; i < 3; i++) {
81        res.push(heap.pop().item);
82    }
83    return res;
84}
85

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