Suppose we have an array enc. There is an array perm that is a permutation of the first n(odd) positive integers. This list will be encoded into array enc of length n-1, such that enc[i] = perm[i] XOR perm[i+1]. We have to find the original array perm.
So, if the input is like enc = [2,5,6,3], then the output will be [7, 5, 0, 6, 5], here [7 XOR 5 XOR 0 XOR 6 XOR 5] = [2, 5, 6, 3]
To solve this, we will follow these steps −
- n := size of enc
- result := an array of size (n+1) and fill with 0
- x := 0
- for i in range 1 to n+1, do
- x := x XOR i
- result[0] := x
- for i in range 1 to n, increase by 2, do
- result[0] := result[0] XOR enc[i]
- for i in range 1 to n, do
- result[i] := result[i-1] XOR enc[i-1]
- return result
Example
Let us see the following implementation to get better understanding −
def solve(enc): n = len(enc) result = [0] * (n+1) x = 0 for i in range(1, n+2): x ^= i result[0] = x for i in range(1, n+1, 2): result[0] ^= enc[i] for i in range(1, n+1): result[i] = result[i-1] ^ enc[i-1] return result enc = [2,5,6,3] print(solve(enc))
Input
[2,5,6,3]
Output
[7, 5, 0, 6, 5]