README_EN.md

1238. Circular Permutation in Binary Representation

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Description

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

Solutions

Python3

class Solution:
    def circularPermutation(self, n: int, start: int) -> List[int]:
        g = [i ^ (i >> 1) for i in range(1 << n)]
        j = g.index(start)
        return g[j:] + g[:j]

class Solution:
    def circularPermutation(self, n: int, start: int) -> List[int]:
        return [i ^ (i >> 1) ^ start for i in range(1 << n)]

Java

class Solution {
    public List<Integer> circularPermutation(int n, int start) {
        int[] g = new int[1 << n];
        int j = 0;
        for (int i = 0; i < 1 << n; ++i) {
            g[i] = i ^ (i >> 1);
            if (g[i] == start) {
                j = i;
            }
        }
        List<Integer> ans = new ArrayList<>();
        for (int i = j; i < j + (1 << n); ++i) {
            ans.add(g[i % (1 << n)]);
        }
        return ans;
    }
}

class Solution {
    public List<Integer> circularPermutation(int n, int start) {
        List<Integer> ans = new ArrayList<>();
        for (int i = 0; i < 1 << n; ++i) {
            ans.add(i ^ (i >> 1) ^ start);
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> circularPermutation(int n, int start) {
        int g[1 << n];
        int j = 0;
        for (int i = 0; i < 1 << n; ++i) {
            g[i] = i ^ (i >> 1);
            if (g[i] == start) {
                j = i;
            }
        }
        vector<int> ans;
        for (int i = j; i < j + (1 << n); ++i) {
            ans.push_back(g[i % (1 << n)]);
        }
        return ans;
    }
};

class Solution {
public:
    vector<int> circularPermutation(int n, int start) {
        vector<int> ans(1 << n);
        for (int i = 0; i < 1 << n; ++i) {
            ans[i] = i ^ (i >> 1) ^ start;
        }
        return ans;
    }
};

Go

func circularPermutation(n int, start int) []int {
	g := make([]int, 1<<n)
	j := 0
	for i := range g {
		g[i] = i ^ (i >> 1)
		if g[i] == start {
			j = i
		}
	}
	return append(g[j:], g[:j]...)
}

func circularPermutation(n int, start int) (ans []int) {
	for i := 0; i < 1<<n; i++ {
		ans = append(ans, i^(i>>1)^start)
	}
	return
}

TypeScript

function circularPermutation(n: number, start: number): number[] {
    const ans: number[] = [];
    for (let i = 0; i < 1 << n; ++i) {
        ans.push(i ^ (i >> 1) ^ start);
    }
    return ans;
}

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