Decode XORed Array - Practice Coding Problems

Decode XORed Array - Problem

Decode XORed Array is a fascinating bit manipulation problem that demonstrates the reversible nature of XOR operations.

You're given a mysterious encoded array that was created by XOR-ing adjacent elements of an original array. For example, if the original array was [1,0,2,1], the encoded array would be [1,2,3] where each element represents arr[i] XOR arr[i+1].

Your mission: Given the encoded array and the first element of the original array, reconstruct the complete original array.

Key insight: Since XOR is its own inverse (a XOR b XOR b = a), you can use the encoded values to step through and rebuild the original array one element at a time.

Input & Output

example_1.py — Basic Case

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

› Output: [1,0,2,1]

💡 Note: Starting with first=1, we calculate: arr[1] = 1⊕1 = 0, arr[2] = 2⊕0 = 2, arr[3] = 3⊕2 = 1. The XOR verification works: [1⊕0, 0⊕2, 2⊕1] = [1,2,3]

example_2.py — Single Element

$ Input: encoded = [6], first = 1

› Output: [1,7]

💡 Note: With only one encoded value, we get arr[1] = 6⊕1 = 7. Verification: 1⊕7 = 6 ✓

example_3.py — Zeros and Large Numbers

$ Input: encoded = [0,2,1], first = 4

› Output: [4,4,6,7]

💡 Note: Starting with 4: arr[1] = 0⊕4 = 4, arr[2] = 2⊕4 = 6, arr[3] = 1⊕6 = 7. Shows that XOR with 0 preserves the value.

Constraints

1 ≤ encoded.length ≤ 10³
0 ≤ encoded[i] ≤ 10⁵
0 ≤ first ≤ 10⁵
The answer is guaranteed to exist and be unique

Visualization

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Understanding the Visualization

Start with the known first element

This is our anchor point - we know arr[0] = first

Use XOR's self-inverse property

Since encoded[0] = arr[0] ⊕ arr[1], we get arr[1] = encoded[0] ⊕ arr[0]

Continue the chain reaction

Each decoded element helps decode the next: arr[i+1] = encoded[i] ⊕ arr[i]

Complete reconstruction

Process all encoded values to rebuild the entire original array

Key Takeaway

🎯 Key Insight: XOR is its own inverse operation, making it perfect for encoding/decoding schemes. Once you know one element and the XOR relationships, you can reconstruct the entire sequence in linear time!

Asked in

G Google 25 a Amazon 18 ⊞ Microsoft 12 f Meta 8

The optimal solution uses XOR's self-inverse property: if encoded[i] = arr[i] ⊕ arr[i+1], then arr[i+1] = encoded[i] ⊕ arr[i]. Starting with the given first element, we can directly compute each subsequent element in O(n) time with O(1) extra space.

Common Approaches

Approach	Time	Space	Notes
✓ Direct XOR Reconstruction (Optimal)	O(n)	O(1)	Use XOR's self-inverse property to directly compute each element
Brute Force (Try All Combinations)	O(k^n)	O(n)	Generate all possible arrays and check which one produces the given encoded array

Direct XOR Reconstruction (Optimal) — Algorithm Steps

Initialize result array with the given first element
For each position i from 1 to n-1:
Calculate arr[i] = encoded[i-1] XOR arr[i-1]
Continue until all elements are computed
Return the reconstructed array

Visualization

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Step-by-Step Walkthrough

Start with first element

We know arr[0] = first

Apply XOR formula

arr[1] = encoded[0] XOR arr[0]

Continue the chain

Each subsequent element uses the previous result

Complete reconstruction

All elements computed in single pass

Code -

solution.c — C

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

int* decode(int* encoded, int encodedSize, int first, int* returnSize) {
    int n = encodedSize + 1;
    int* result = (int*)malloc(n * sizeof(int));
    result[0] = first;
    *returnSize = n;
    
    for (int i = 0; i < encodedSize; i++) {
        // arr[i+1] = encoded[i] XOR arr[i]
        result[i + 1] = encoded[i] ^ result[i];
    }
    
    return result;
}

int main() {
    int encoded[1000];
    int encodedSize = 0;
    int num;
    
    char line[10000];
    fgets(line, sizeof(line), stdin);
    char* token = strtok(line, " \n");
    while (token != NULL) {
        encoded[encodedSize++] = atoi(token);
        token = strtok(NULL, " \n");
    }
    
    int first;
    scanf("%d", &first);
    
    int returnSize;
    int* result = decode(encoded, encodedSize, first, &returnSize);
    
    for (int i = 0; i < returnSize; i++) {
        printf("%d%s", result[i], i == returnSize-1 ? "\n" : " ");
    }
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the encoded array to compute each element

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra space (not counting output array)

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Direct XOR Reconstruction (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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