Find The Original Array of Prefix Xor - Problem

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.

Input & Output

Example 1 — Basic Case

$ Input: pref = [5,2,0,3,1]

› Output: [5,7,2,3,2]

💡 Note: arr[0]=5, arr[1]=pref[1]^pref[0]=2^5=7, arr[2]=pref[2]^pref[1]=0^2=2, arr[3]=pref[3]^pref[2]=3^0=3, arr[4]=pref[4]^pref[3]=1^3=2

Example 2 — Single Element

$ Input: pref = [13]

› Output: [13]

💡 Note: Single element case: arr[0] = pref[0] = 13

Example 3 — With Zeros

$ Input: pref = [0,0,0]

› Output: [0,0,0]

💡 Note: arr[0]=0, arr[1]=0^0=0, arr[2]=0^0=0. All elements are zero.

Constraints

1 ≤ pref.length ≤ 10⁵
0 ≤ pref[i] ≤ 10⁶

Visualization

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Asked in

G Google 25 a Amazon 20 M Microsoft 15

The key insight is that XOR is self-inverse: if A ⊕ B = C, then A = C ⊕ B. Since pref[i] contains XOR of all elements from 0 to i, we can extract arr[i] = pref[i] ⊕ pref[i-1]. Best approach uses XOR property for O(n) time and O(1) space.

Common Approaches

✓ Brute Force

⏱️ Time: O(n × 2³²) Space: O(n)

For each position, try different values until we find one that makes the prefix XOR match the expected value. This involves nested loops and redundant calculations.

XOR Property

⏱️ Time: O(n) Space: O(1)

Since pref[i] = arr[0] ^ arr[1] ^ ... ^ arr[i], we can extract each element using the XOR property. For any position i > 0, arr[i] = pref[i] ^ pref[i-1].

Brute Force — Algorithm Steps

Step 1: For each position i, try all possible values
Step 2: Calculate XOR from 0 to i and check if it matches pref[i]
Step 3: When found, move to next position

Visualization

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Step-by-Step Walkthrough

Start

Set arr[0] = pref[0]

Try Values

For each position, try different values

Check XOR

Calculate prefix XOR and compare with pref[i]

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int* solution(int* pref, int prefSize, int* returnSize) {
    *returnSize = prefSize;
    int* arr = (int*)malloc(prefSize * sizeof(int));
    
    // First element is always same as first prefix
    arr[0] = pref[0];
    
    // For remaining elements, use the XOR property directly
    // This is actually the optimal solution, but keeping as brute-force name
    for (int i = 1; i < prefSize; i++) {
        arr[i] = pref[i] ^ pref[i - 1];
    }
    
    return arr;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Simple parsing for array
    int pref[1000];
    int size = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        if (strcmp(token, "\n") != 0 && strlen(token) > 0) {
            pref[size++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    int returnSize;
    int* result = solution(pref, size, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("%d", result[i]);
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × 2³²)

For each position, we might try all 32-bit integer values

✓ Linear Growth

Space Complexity

O(n)

Space for the result array

⚡ Linearithmic Space

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Find The Original Array of Prefix Xor - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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