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# 1238. Circular Permutation in Binary Representation

Given 2 integers `n` and `start`. Your task is return **any** permutation `p` of `(0,1,2.....,2^n -1)` such that :

* `p[0] = start`
* `p[i]` and `p[i+1]` differ by only one bit in their binary representation.
* `p[0]` and `p[2^n -1]` must also differ by only one bit in their binary representation.

**Example 1:**

```
Input: n = 2, start = 3
Output: [3,2,0,1]
Explanation: The binary representation of the permutation is (11,10,00,01). 
All the adjacent element differ by one bit. Another valid permutation is [3,1,0,2]
```

**Example 2:**

```
Input: n = 3, start = 2
Output: [2,6,7,5,4,0,1,3]
Explanation: The binary representation of the permutation is (010,110,111,101,100,000,001,011).
```

**Constraints:**

* `1 <= n <= 16`
* `0 <= start < 2 ^ n`

```cpp
vector<int> oneBitDiffPermutation(int n) {
    vector<int> res;
    res.reserve(1 << n);
    res.push_back(0);
    for (int i = 0; i < n; ++i) {
        for (int j = res.size() - 1; j >= 0; --j) {
            res.push_back(res[j] ^ (1 << i));
        }
    }
    return res;
}
vector<int> circularPermutation(int n, int start) { // time: O(2^n); space: O(2^n)
    vector<int> res = oneBitDiffPermutation(n);
    for (int num : res) cout << num << " ";
    cout << endl;
    auto start_pos = find(res.begin(), res.end(), start);
    reverse(res.begin(), start_pos);
    reverse(start_pos, res.end());
    reverse(res.begin(), res.end());
    return res;
}
```

```cpp
// Bit manipulation for gray code
vector<int> circularPermutation(int n, int start) { // time: O(2^n); space: O(2^n)
    vector<int> res;
    res.reserve(1 << n);
    for (int i = 0; i < (1 << n); ++i) {
        res.push_back(start ^ i ^ (i >> 1));
    }
    return res;
}
```