LeetCode 2433. Find The Original Array of Prefix Xor Solution in Java, C++, Python & More | Explanation + Code

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 … \oplus arr[i]

So, we also have equation two:

pref[i-1]=arr[0] \oplus arr[1] \oplus … \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).

PythonJavaC++GoTypeScriptRustC

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

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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; }(code-box)

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 } }(code-box)

/** * 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; }(code-box)