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2433. Find The Original Array of Prefix Xor

Description

You are given an integer array pref of size n. Find and return the array arr of size n that satisfies:

  • pref[i] = arr[0] ^ arr[1] ^ ... ^ arr[i].

Note that ^ denotes the bitwise-xor operation.

It can be proven that the answer is unique.

 

Example 1:

Input: pref = [5,2,0,3,1]
Output: [5,7,2,3,2]
Explanation: From the array [5,7,2,3,2] we have the following:
- pref[0] = 5.
- pref[1] = 5 ^ 7 = 2.
- pref[2] = 5 ^ 7 ^ 2 = 0.
- pref[3] = 5 ^ 7 ^ 2 ^ 3 = 3.
- pref[4] = 5 ^ 7 ^ 2 ^ 3 ^ 2 = 1.

Example 2:

Input: pref = [13]
Output: [13]
Explanation: We have pref[0] = arr[0] = 13.

 

Constraints:

  • 1 <= pref.length <= 105
  • 0 <= pref[i] <= 106

Solutions

Solution 1: Bit Manipulation

According to the problem statement, we have equation one:

$$ pref[i]=arr[0] \oplus arr[1] \oplus \cdots \oplus arr[i] $$

So, we also have equation two:

$$ pref[i-1]=arr[0] \oplus arr[1] \oplus \cdots \oplus arr[i-1] $$

We perform a bitwise XOR operation on equations one and two, and get:

$$ pref[i] \oplus pref[i-1]=arr[i] $$

That is, each item in the answer array is obtained by performing a bitwise XOR operation on the adjacent two items in the prefix XOR array.

The time complexity is $O(n)$, where $n$ is the length of the prefix XOR array. Ignoring the space consumption of the answer, the space complexity is $O(1)$.

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class Solution:
    def findArray(self, pref: List[int]) -> List[int]:
        return [a ^ b for a, b in pairwise([0] + pref)]

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class Solution {
    public int[] findArray(int[] pref) {
        int n = pref.length;
        int[] ans = new int[n];
        ans[0] = pref[0];
        for (int i = 1; i < n; ++i) {
            ans[i] = pref[i - 1] ^ pref[i];
        }
        return ans;
    }
}

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class Solution {
public:
    vector<int> findArray(vector<int>& pref) {
        int n = pref.size();
        vector<int> ans = {pref[0]};
        for (int i = 1; i < n; ++i) {
            ans.push_back(pref[i - 1] ^ pref[i]);
        }
        return ans;
    }
};

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func findArray(pref []int) []int {
    n := len(pref)
    ans := []int{pref[0]}
    for i := 1; i < n; i++ {
        ans = append(ans, pref[i-1]^pref[i])
    }
    return ans
}

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function findArray(pref: number[]): number[] {
    let ans = pref.slice();
    for (let i = 1; i < pref.length; i++) {
        ans[i] = pref[i - 1] ^ pref[i];
    }
    return ans;
}

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impl Solution {
    pub fn find_array(pref: Vec<i32>) -> Vec<i32> {
        let n = pref.len();
        let mut res = vec![0; n];
        res[0] = pref[0];
        for i in 1..n {
            res[i] = pref[i] ^ pref[i - 1];
        }
        res
    }
}

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/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* findArray(int* pref, int prefSize, int* returnSize) {
    int* res = (int*) malloc(sizeof(int) * prefSize);
    res[0] = pref[0];
    for (int i = 1; i < prefSize; i++) {
        res[i] = pref[i - 1] ^ pref[i];
    }
    *returnSize = prefSize;
    return res;
}

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