1720. Decode XORed Array
Problem Description
In the given problem, there is an array named arr
that contains n
non-negative integers which we can't see because it is hidden. This arr
array has been transformed into another array named encoded
with a length of n - 1
. This transformation uses the bitwise XOR operation: each element of encoded
is the result of XORing consecutive elements in arr
, i.e., encoded[i] = arr[i] XOR arr[i + 1]
. The XOR operation is a bitwise operation that outputs 1 only if the input bits are different, and 0 otherwise.
The challenge is to reconstruct the original arr
array given the encoded
array and the first element of arr
(referred to as first
or arr[0]
). The description assures that there is a unique solution for arr
.
Intuition
The XOR operation plays a crucial role here because it has a unique property: it's reversible if we know one of the operands. Meaning, if we have a XOR b = c
, we can find a
by doing c XOR b
, and similarly, find b
by doing c XOR a
. This is because performing XOR twice with the same number cancels out the operation (e.g., a XOR b XOR b
is equal to a
).
This property makes it possible to recover the arr
array starting from the provided first
element (arr[0]
). The intuition behind the solution is simply to iterate through the encoded
array and apply the XOR operation between the last known value of arr
and the current element in encoded
to find the next value of arr
. In essence, given arr[i]
and encoded[i]
, we can solve for arr[i + 1]
by calculating arr[i] XOR encoded[i]
.
By starting with arr[0]
as first
and iteratively applying the XOR operation with the encoded
values, we can decode the entire arr
array. The solution process unfolds one element at a time, until all values in arr
are revealed.
Solution Approach
The implementation utilizes a simple for
loop and the append
method on lists in Python. For those unfamiliar with Python, appending to a list adds a new element to the end of the list. The algorithm works as follows:
- Initially, the known first element of
arr
(given asfirst
) is appended to an empty list namedans
. - We then iterate over each element
e
in theencoded
array. - In each iteration, we XOR the last element of
ans
withe
. In terms of the algorithm, ifans[-1]
is the last element of the listans
, thene
is XORed withans[-1]
, and the result is the next element inarr
, which is then appended toans
. - This method leverages the reversible property of XOR mentioned earlier:
encoded[i] = arr[i] XOR arr[i + 1]
impliesarr[i + 1] = encoded[i] XOR arr[i]
. Sincearr[i]
is the last known element (initiallyfirst
), we can decodearr[i + 1]
using the elements ofencoded
. - The loop continues until we have reconstructed the entire
arr
array.
The Solution
class and its decode
method, provided in the reference solution, are examples of the use of Python's object-oriented programming paradigm. The decode
method encapsulates the aforementioned algorithm. There are no specific data structures besides the list used for the result, and no complex patterns -- it's a straight application of XOR to decode each subsequent number in the sequence.
Here's an explicit breakdown of the steps in the decode
function:
- Start by creating a result list
ans
with the first elementfirst
. - Iterate over the
encoded
array using afor
loop. - In each iteration, XOR the last element of
ans
with the current element inencoded
and append the result toans
. - Continue until all elements in
encoded
have been used. - The list
ans
is now the decodedarr
array, which we return from the function.
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Let's illustrate the solution approach with a small example. Suppose we have the following encoded
array and first
element:
encoded = [6, 1, 4]
first = 5
(which is actuallyarr[0]
)
We know that encoded[i]
is derived from arr[i] XOR arr[i + 1]
. So the original array arr
starts with first
and has n
elements, with n = len(encoded) + 1
.
Now we will use the given solution approach to decode the original array arr
.
Step 1: Initialize the result list with first
ans = [5]
Step 2: XOR first
with the first element of encoded
- Calculate
5 XOR 6
which equals3
- Append
3
toans
ans
becomes[5, 3]
Step 3: XOR the last element of ans
(now 3
) with the next element of encoded
- Calculate
3 XOR 1
which equals2
- Append
2
toans
ans
becomes[5, 3, 2]
Step 4: XOR the last element of ans
(now 2
) with the next element of encoded
- Calculate
2 XOR 4
which equals6
- Append
6
toans
ans
becomes[5, 3, 2, 6]
At this point, we've performed an XOR operation with all elements of the encoded
array, and our result list now contains all elements of the original array arr
.
Final Result:
The decoded original array arr
is [5, 3, 2, 6]
.
Using the given solution approach, we are able to reconstruct arr
from encoded
and first
. This illustrates how the reversible property of the XOR operation can be utilized to solve this type of decoding problem iteratively.
Solution Implementation
1from typing import List
2
3class Solution:
4 def decode(self, encoded: List[int], first: int) -> List[int]:
5 # Initialize the result list with the first element
6 decoded_list = [first]
7
8 # Iterate over the encoded list and decode each element
9 for encoded_element in encoded:
10 # The next number is found by XORing the last number in the decoded list
11 # with the current encoded element
12 decoded_list.append(decoded_list[-1] ^ encoded_element)
13
14 # Return the fully decoded list
15 return decoded_list
16
1class Solution {
2
3 /**
4 * Decodes an encoded array with the given first element value.
5 *
6 * @param encoded The array of integers to be decoded.
7 * @param first The first element of the decoded array.
8 * @return The decoded array of integers.
9 */
10 public int[] decode(int[] encoded, int first) {
11 // The length of the decoded array is one more than the length of the encoded array.
12 int n = encoded.length;
13 int[] decodedArray = new int[n + 1];
14
15 // Setting the first element of the decoded array.
16 decodedArray[0] = first;
17
18 // Iterating through the encoded array to decode it.
19 for (int i = 0; i < n; ++i) {
20 // The current element is obtained by XORing the previous element of the decoded
21 // array with the current element of the encoded array.
22 decodedArray[i + 1] = decodedArray[i] ^ encoded[i];
23 }
24
25 // Returning the decoded array.
26 return decodedArray;
27 }
28}
29
1#include <vector> // Include the vector header to use std::vector
2
3class Solution {
4public:
5 // Decodes an encoded vector using the first element
6 // @param encoded: the encoded vector of integers
7 // @param first: the first element to start decoding
8 // @return the decoded vector of integers
9 std::vector<int> decode(std::vector<int>& encoded, int first) {
10 std::vector<int> decoded; // Create an empty vector to store the decoded numbers
11 decoded.push_back(first); // Add the first element to the decoded vector
12
13 // Decode the rest of the encoded vector
14 for (int i = 0; i < encoded.size(); ++i) {
15 // The next number is found by XORing the current number with the encoded number
16 decoded.push_back(decoded[i] ^ encoded[i]);
17 }
18
19 return decoded; // Return the fully decoded vector
20 }
21};
22
1// Import the Array type from TypeScript for type annotations
2import { Array } from "typescript";
3
4// Decode an encoded array using the first element
5// @param encoded - the encoded array of numbers
6// @param first - the first element to start decoding
7// @return the decoded array of numbers
8function decode(encoded: Array<number>, first: number): Array<number> {
9 let decoded: Array<number> = []; // Create an empty array to store the decoded numbers
10 decoded.push(first); // Add the first element to the decoded array
11
12 // Decode the rest of the encoded array
13 for (let i = 0; i < encoded.length; i++) {
14 // The next number is found by XORing the current number with the encoded number
15 decoded.push(decoded[i] ^ encoded[i]);
16 }
17
18 return decoded; // Return the fully decoded array
19}
20
Time and Space Complexity
Time Complexity
The given Python function decode
consists of a single loop that iterates through the encoded
list, which has n
elements, where n
is the length of the encoded
list. Within the loop, there is a constant time operation which performs an XOR operation (^
) and appends the result to the ans
list. Therefore, since each operation in the loop takes O(1)
time and the loop runs for n
iterations, the overall time complexity is O(n)
.
Space Complexity
The space complexity of the function decode
is determined by the ans
list which the function populates and returns. Since the ans
list will contain exactly n + 1
elements after processing an encoded
list of length n
, the space complexity is O(n)
. The space required grows linearly with the input size, making the space complexity linear as well.
Learn more about how to find time and space complexity quickly using problem constraints.
Problem: Given a list of tasks and a list of requirements, compute a sequence of tasks that can be performed, such that we complete every task once while satisfying all the requirements.
Which of the following method should we use to solve this problem?
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