Decode XORed Array

Decode XORed Array - Problem

There is a hidden integer array arr that consists of n non-negative integers.

It was encoded into another integer array encoded of length n - 1, such that encoded[i] = arr[i] XOR arr[i + 1]. For example, if arr = [1,0,2,1], then encoded = [1,2,3].

You are given the encoded array. You are also given an integer first, that is the first element of arr, i.e. arr[0].

Return the original array arr. It can be proved that the answer exists and is unique.

Input & Output

Example 1 — Basic Case

$ Input: encoded = [1,2,3], first = 1

› Output: [1,0,2,1]

💡 Note: Starting with arr[0] = 1, we decode: arr[1] = 1⊕1 = 0, arr[2] = 0⊕2 = 2, arr[3] = 2⊕3 = 1

Example 2 — Different Values

$ Input: encoded = [6,2,7,3], first = 4

› Output: [4,2,0,7,4]

💡 Note: arr[0] = 4, arr[1] = 4⊕6 = 2, arr[2] = 2⊕2 = 0, arr[3] = 0⊕7 = 7, arr[4] = 7⊕3 = 4

Example 3 — Minimum Size

$ Input: encoded = [5], first = 3

› Output: [3,6]

💡 Note: With only one encoded value: arr[0] = 3, arr[1] = 3⊕5 = 6

Constraints

2 ≤ encoded.length ≤ 10⁴
0 ≤ encoded[i] ≤ 10⁵
0 ≤ first ≤ 10⁵

Visualization

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Asked in

a Amazon 15 M Microsoft 12

The key insight is using XOR's reversible property: if A ⊕ B = C, then A ⊕ C = B. Best approach uses this to decode each element: arr[i+1] = arr[i] ⊕ encoded[i]. Time: O(n), Space: O(n)

Common Approaches

✓ Brute Force

⏱️ Time: O(k^n) Space: O(n)

Generate all possible arrays of length n+1, then for each array check if XORing consecutive elements produces the encoded array. This is extremely inefficient.

XOR Properties

⏱️ Time: O(n) Space: O(n)

Since XOR is reversible (A ⊕ B = C means A ⊕ C = B), we can decode each element using the previous element and the corresponding encoded value.

Brute Force — Algorithm Steps

Generate all possible arrays
For each array, compute XOR of consecutive elements
Check if result matches encoded array

Visualization

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Step-by-Step Walkthrough

Generate Candidates

Try all possible arrays starting with first

Validate

Check if XOR of consecutive elements matches encoded

Result

Return first valid array found

Code -

solution.c — C

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

int* solution(int* encoded, int encodedSize, int first, int* returnSize) {
    *returnSize = encodedSize + 1;
    int* result = (int*)malloc(*returnSize * sizeof(int));
    result[0] = first;
    
    // Use XOR property directly - brute force is impractical
    for (int i = 0; i < encodedSize; i++) {
        result[i + 1] = result[i] ^ encoded[i];
    }
    
    return result;
}

int main() {
    int encoded[1000];
    int encodedSize = 0;
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse JSON-like array format
    char* start = strchr(line, '[');
    char* end = strchr(line, ']');
    if (start && end) {
        start++; // Skip [
        *end = '\0'; // Terminate at ]
        
        char* token = strtok(start, ",");
        while (token != NULL) {
            encoded[encodedSize++] = atoi(token);
            token = strtok(NULL, ",");
        }
    }
    
    int first;
    scanf("%d", &first);
    
    int returnSize;
    int* result = solution(encoded, encodedSize, first, &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(k^n)

k^n possible arrays where k is range of values and n is array length

✓ Linear Growth

Space Complexity

O(n)

Space for candidate array and result

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

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Code -

Time & Space Complexity

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