Leetcode 1720: Decode XORed Array

Input: You are given an array `encoded` of length `n - 1` and an integer `first` which is the first element of the original array `arr`.

Example: Input: encoded = [5, 8, 10], first = 2

Constraints:

• 2 <= n <= 10^4

• encoded.length == n - 1

• 0 <= encoded[i] <= 10^5

• 0 <= first <= 10^5

Output: Return the decoded array `arr`.

Example: Output: [2, 7, 15, 5]

Constraints:

• The decoded array will always exist and be unique.

Goal: The goal is to reverse the encoding process and reconstruct the original array `arr` using the given `encoded` array and the first element of `arr`.

Steps:

• 1. Start by initializing the array `arr` with the first element `first`.

• 2. Iterate through the `encoded` array, and for each value, compute the next element in the original array using XOR: `arr[i + 1] = arr[i] XOR encoded[i]`.

• 3. Continue this process until the entire original array is reconstructed.

Goal: The input constraints are designed to ensure that the solution can be computed efficiently for large arrays.

Steps:

• The `encoded` array is always of length `n - 1`.

• The value of `first` is non-negative and fits within the constraints.

Assumptions:

• It is guaranteed that a unique solution exists for every input.

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

• Explanation: The original array is `[1, 0, 2, 1]`. Using the formula `arr[i+1] = arr[i] XOR encoded[i]`, we can reconstruct the array as follows: `arr[0] = 1`, `arr[1] = arr[0] XOR encoded[0] = 1 XOR 1 = 0`, `arr[2] = arr[1] XOR encoded[1] = 0 XOR 2 = 2`, and `arr[3] = arr[2] XOR encoded[2] = 2 XOR 3 = 1`.

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

• Explanation: The original array is `[4, 2, 0, 7, 4]`. Reconstructing each element: `arr[0] = 4`, `arr[1] = arr[0] XOR encoded[0] = 4 XOR 6 = 2`, `arr[2] = arr[1] XOR encoded[1] = 2 XOR 2 = 0`, `arr[3] = arr[2] XOR encoded[2] = 0 XOR 7 = 7`, and `arr[4] = arr[3] XOR encoded[3] = 7 XOR 3 = 4`.

Approach: We can decode the original array by iterating through the `encoded` array and using XOR operations to reconstruct the original elements, starting with the first element.

Observations:

• The problem is based on the XOR operation, which has properties that allow us to reverse the encoding process.

• The XOR operation is its own inverse, meaning that we can recover the original array using the formula `arr[i + 1] = arr[i] XOR encoded[i]`.

Steps:

• 1. Initialize a result array `arr` with the first element `first`.

• 2. Iterate through the `encoded` array, and for each element, compute the next element in the original array using the XOR operation.

• 3. Append each computed value to the `arr` array.

• 4. Return the final array `arr` as the result.

Empty Inputs:

• There are no cases with empty inputs, as the length of the `encoded` array is always at least 1.

Large Inputs:

• The algorithm should efficiently handle cases where `n` is large (up to 10^4).

Special Values:

• If `first` is 0, the first element of the array will be 0, and the rest of the array will be decoded accordingly.

Constraints:

• The solution should work within the provided time and space limits for all valid inputs.

vector<int> decode(vector<int>& encoded, int first) {
    vector<int> res;
    res.push_back(first);
    for(int i = 0; i < encoded.size(); i++)
        res.push_back(res[i] ^ encoded[i]);
    return res;
}

1 : Function Declaration

vector<int> decode(vector<int>& encoded, int first) {

The function `decode` is declared, taking an encoded array `encoded` and an integer `first`, which is the first element of the original array.

2 : Variable Initialization

    vector<int> res;

Declaring an empty vector `res` which will store the reconstructed array.

3 : Push First Element

    res.push_back(first);

Adding the first element `first` to the `res` vector, which is the starting point of the decoded array.

4 : Loop Iteration

    for(int i = 0; i < encoded.size(); i++)

Iterating through the `encoded` array to apply the XOR operation and reconstruct the original array.

5 : XOR Operation

        res.push_back(res[i] ^ encoded[i]);

Using the XOR operation to decode the next element of the original array by XORing the current value in `res` with the corresponding element in `encoded`, then pushing it into the `res` vector.

6 : Return Statement

    return res;

Returning the reconstructed array `res` containing the decoded values.

Best Case: O(n), where n is the length of the `encoded` array.

Average Case: O(n), as we are iterating through the `encoded` array once.

Worst Case: O(n), since we need to process each element of the `encoded` array.

Description: The time complexity is linear in the size of the `encoded` array.

Best Case: O(n), since we must store the result array regardless of input size.

Worst Case: O(n), as we are storing the original array `arr` of size `n`.

Description: The space complexity is linear in the size of the `encoded` array, as we need to store the decoded result.

Leetcode 1720: Decode XORed Array

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