Leetcode 1734: Decode XORed Permutation

Input: The input consists of a single integer array `encoded` of length `n - 1`, where `encoded[i] = perm[i] XOR perm[i + 1]`. The permutation `perm` is a permutation of integers from `1` to `n` and `n` is odd.

Example: Input: encoded = [4, 3, 7]

Constraints:

• 3 <= n < 10^5

• n is odd

• encoded.length == n - 1

Output: Return the decoded permutation array `perm` of size `n`.

Example: Output: [2, 4, 1, 5, 3]

Constraints:

• The output is a list of integers representing the decoded permutation of the integers from 1 to n.

Goal: The goal is to decode the encoded array and reconstruct the original permutation by finding a way to use XOR operations efficiently.

Steps:

• 1. Compute the XOR of all integers from 1 to n.

• 2. XOR every second element of the `encoded` array with the computed XOR to obtain the first element of the permutation.

• 3. Use the XOR relationship between consecutive elements of the permutation to find the entire permutation array.

• 4. Return the reconstructed permutation array.

Goal: The given array `encoded` satisfies the following constraints:

Steps:

• The length of `encoded` is `n - 1` where n is an odd integer.

• The elements in `encoded` are derived using XOR operations from consecutive elements of the permutation array.

Assumptions:

• The encoded array always provides a valid and unique solution for the permutation.

• The permutation consists of the integers from 1 to n, and each integer appears exactly once.

• Input: Input: encoded = [4, 3, 7]

• Explanation: The XOR of all integers from 1 to 5 is 7. XORing every second element of `encoded` with 7 gives the first element of the permutation, which is 2. Then we can use the XOR of consecutive elements to find the rest of the permutation.

• Input: Input: encoded = [3, 1]

• Explanation: If perm = [1, 2, 3], then `encoded = [1 XOR 2, 2 XOR 3] = [3, 1]`. Therefore, the decoded permutation is [1, 2, 3].

Approach: To solve this problem, we leverage the XOR properties to decode the given array and reconstruct the permutation.

Observations:

• The XOR operation has properties that can be used to decode the encoded array efficiently.

• We can compute the XOR of all integers from 1 to n to use in decoding the array.

Steps:

• 1. Compute the XOR of all integers from 1 to n.

• 2. Use this result to XOR every second element of the `encoded` array, starting from the first, to get the first element of the permutation.

• 3. Using the XOR relationship, compute the rest of the permutation array.

• 4. Return the decoded permutation.

Empty Inputs:

• The array `encoded` cannot be empty because n >= 3.

Large Inputs:

• For large values of n, the solution should efficiently handle up to n < 10^5 elements.

Special Values:

• Since n is always odd, the total number of elements in `encoded` will always be even.

Constraints:

• The XOR operation is commutative and associative, which allows for efficient decoding.

vector<int> decode(vector<int>& enc) {
    int n = enc.size() + 1;
    vector<int> ans(n, 0);
    
    int x = 0;
    for(int i = 1; i < n + 1; i++)
        x ^= i;
    
    ans[0] = x;
    for(int i = 1; i < enc.size(); i += 2)
        ans[0] ^= enc[i];
    
    for(int i = 1; i < n; i++)
        ans[i] = ans[i - 1] ^ enc[i - 1];
    
    return ans;
}

1 : Function Declaration

vector<int> decode(vector<int>& enc) {

The function `decode` is defined, which takes a reference to a vector of integers `enc` (encoded values) and returns a vector of integers (decoded values).

2 : Variable Initialization

    int n = enc.size() + 1;

The size of the original decoded array `n` is determined by adding 1 to the size of the encoded array `enc`.

3 : Array Initialization

    vector<int> ans(n, 0);

A new vector `ans` of size `n` is initialized with all elements set to 0, which will hold the decoded values.

4 : Variable Initialization

    int x = 0;

The variable `x` is initialized to 0. This will be used for XOR operations to decode the values.

5 : Loop

    for(int i = 1; i < n + 1; i++)

A loop is initiated from 1 to `n`, used to XOR the values for decoding.

6 : Bitwise Operation

        x ^= i;

The variable `x` is XOR'd with the current value of `i`, effectively encoding the range of numbers.

7 : Assignment

    ans[0] = x;

The first element of the `ans` array is assigned the value of `x` after the XOR operations in the previous loop.

8 : Loop

    for(int i = 1; i < enc.size(); i += 2)

A loop iterates over the encoded array `enc`, starting from index 1 and incrementing by 2, ensuring only the elements at odd indices are processed.

9 : Bitwise Operation

        ans[0] ^= enc[i];

The first element of the `ans` array is XOR'd with the value from the encoded array `enc` at the current index `i`.

10 : Loop

    for(int i = 1; i < n; i++)

A second loop starts from index 1 to `n - 1`, used to calculate the remaining values in the `ans` array.

11 : Bitwise Operation

        ans[i] = ans[i - 1] ^ enc[i - 1];

Each element of the `ans` array is calculated by XOR'ing the previous element in `ans` with the corresponding element in the encoded array `enc`.

12 : Return Statement

    return ans;

The function returns the decoded `ans` array, which now contains the original sequence of numbers.

Best Case: O(n), where n is the number of elements in the permutation.

Average Case: O(n), as we need to process each element in the array once.

Worst Case: O(n), since the algorithm processes all elements in linear time.

Description: The time complexity is linear, O(n), since we need to process each element of the encoded array and compute the XOR operations.

Best Case: O(n), as we need space to store the decoded permutation array.

Worst Case: O(n), where n is the number of elements in the permutation.

Description: The space complexity is O(n) because we need to store the decoded permutation array, which has size n.

Leetcode 1734: Decode XORed Permutation

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