2683. Neighboring Bitwise XOR

MediumBit ManipulationArray
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Problem Description

In this problem, you are given an array derived, which is said to be created from another binary array original by applying a bitwise XOR operation. The array original is a binary array, meaning it only contains 0's and 1's. The elements of the array derived are formed as follows:

  • Each element at index i in derived is the result of original[i] XOR original[i + 1], except for the last element.
  • The last element in the derived array is the result of original[n - 1] XOR original[0], creating a circular calculation from the end of the array back to the start.

Your task is to determine if there exists any valid original binary array that could have been used to obtain the given derived array through the described process.

Intuition

To solve this problem, we utilize the property of XOR operation. The fundamental point to notice is that XOR of a number with itself is zero, and the XOR of zero with any number is the number itself. With these properties in mind, let's consider the provided derived array and think about what happens if we take XOR of all its elements.

  1. If we XOR all elements of the hypothetical original array in a circular manner as described, we would end up with zero. This is because each element would be XORed with itself at some point in the process (since original[i] XOR original[i] is always 0).

  2. Conversely, if we XOR all elements of the provided derived array, and the result is non-zero, this implies there is no such original array that could have produced derived, as this would violate the property stated in step 1.

  3. Hence, if the cumulative XOR of all the elements of derived is zero, a valid original array could exist as it indicates that each number has been XORed with itself.

  4. The provided solution uses Python's reduce function and xor operator from the operator module to apply the XOR operation cumulatively across all the elements of the derived array. It checks whether the final result of the cumulative XOR is equal to zero.

By following this logic, we are lead to a simple and effective approach to solve the problem using just one line of code.

Solution Approach

The solution approach is surprisingly straightforward due to the properties of the XOR operation. The algorithm doesn't require any additional data structures, complex patterns, or multiple iterations over the data; it relies purely on a single pass over the array to reduce it to one value. Here's an explanation of the code:

  • The reduce function in Python is a tool from the functools module that is used to apply a particular function cumulatively to the items of an iterable, from left to right, so as to reduce the iterable to a single value. In this scenario, it is being used to apply the xor operation to the elements of the derived array.

  • The xor operation is a bitwise operation that is found in the operator module. This operation takes two numbers and returns their bitwise XOR. The XOR of two bits is 1 if the bits are different, and 0 if they are the same.

  • The implementation reduce(xor, derived) continuously applies the XOR operation across all elements of the derived array. As Python processes each element, it calculates the cumulative XOR from the start of the array up to the current element. This process results in a single integer value, which represents the XOR of the entire array.

  • Once the cumulative XOR is computed, we compare it with 0. If it is equal to zero (reduce(xor, derived) == 0), it means a valid original array can exist based on the properties of XOR discussed earlier. Otherwise, if the cumulative XOR is not zero, no such valid original array exists that could produce the derived array.

  • The decision is made in a single line of code, thanks to the efficiency of the reduce function and the xor operator. It elegantly verifies the possibility of the existence of a valid original array without explicitly reconstructing it, which makes this solution both efficient and clever.

Here is the full implementation in Python:

1from functools import reduce
2from operator import xor
3
4class Solution:
5    def doesValidArrayExist(self, derived: List[int]) -> bool:
6        return reduce(xor, derived) == 0

This implementation utilizes functional programming concepts in Python and showcases how a combination of mathematics and efficient use of built-in functions can lead to optimal and elegant solutions.

Discover Your Strengths and Weaknesses: Take Our 2-Minute Quiz to Tailor Your Study Plan:

What's the output of running the following function using input [30, 20, 10, 100, 33, 12]?

1def fun(arr: List[int]) -> List[int]:
2    import heapq
3    heapq.heapify(arr)
4    res = []
5    for i in range(3):
6        res.append(heapq.heappop(arr))
7    return res
8
1public static int[] fun(int[] arr) {
2    int[] res = new int[3];
3    PriorityQueue<Integer> heap = new PriorityQueue<>();
4    for (int i = 0; i < arr.length; i++) {
5        heap.add(arr[i]);
6    }
7    for (int i = 0; i < 3; i++) {
8        res[i] = heap.poll();
9    }
10    return res;
11}
12
1class HeapItem {
2    constructor(item, priority = item) {
3        this.item = item;
4        this.priority = priority;
5    }
6}
7
8class MinHeap {
9    constructor() {
10        this.heap = [];
11    }
12
13    push(node) {
14        // insert the new node at the end of the heap array
15        this.heap.push(node);
16        // find the correct position for the new node
17        this.bubble_up();
18    }
19
20    bubble_up() {
21        let index = this.heap.length - 1;
22
23        while (index > 0) {
24            const element = this.heap[index];
25            const parentIndex = Math.floor((index - 1) / 2);
26            const parent = this.heap[parentIndex];
27
28            if (parent.priority <= element.priority) break;
29            // if the parent is bigger than the child then swap the parent and child
30            this.heap[index] = parent;
31            this.heap[parentIndex] = element;
32            index = parentIndex;
33        }
34    }
35
36    pop() {
37        const min = this.heap[0];
38        this.heap[0] = this.heap[this.size() - 1];
39        this.heap.pop();
40        this.bubble_down();
41        return min;
42    }
43
44    bubble_down() {
45        let index = 0;
46        let min = index;
47        const n = this.heap.length;
48
49        while (index < n) {
50            const left = 2 * index + 1;
51            const right = left + 1;
52
53            if (left < n && this.heap[left].priority < this.heap[min].priority) {
54                min = left;
55            }
56            if (right < n && this.heap[right].priority < this.heap[min].priority) {
57                min = right;
58            }
59            if (min === index) break;
60            [this.heap[min], this.heap[index]] = [this.heap[index], this.heap[min]];
61            index = min;
62        }
63    }
64
65    peek() {
66        return this.heap[0];
67    }
68
69    size() {
70        return this.heap.length;
71    }
72}
73
74function fun(arr) {
75    const heap = new MinHeap();
76    for (const x of arr) {
77        heap.push(new HeapItem(x));
78    }
79    const res = [];
80    for (let i = 0; i < 3; i++) {
81        res.push(heap.pop().item);
82    }
83    return res;
84}
85

Example Walkthrough

Let's consider a small example to illustrate the solution approach. Suppose we have a derived array given as:

1derived = [1, 0, 1]

We want to determine if there is an original binary array that XORs to the given derived array. To do this, we can use the XOR properties to our advantage:

  1. We start by XOR-ing all the elements of the derived array:

    • Step 1: XOR 1 and 0 which gives us 1.
    • Step 2: XOR the result from step 1 with the next element, which is 1, so we get 1 XOR 1 = 0.
  2. After XOR-ing all the elements of derived array, we end up with 0, which confirms that there might be a valid original binary array, as the cumulative XOR equals zero.

Following the steps of the proposed solution:

  • We apply reduce with xor from the operator module to all items of the derived array:
1from functools import reduce
2from operator import xor
3
4result = reduce(xor, [1, 0, 1]) # This will be 0
  • The result from the reduce function would give us 0 which complies with our intuition that a valid original binary array exists.
  1. Since the result is equal to 0, our function doesValidArrayExist([1, 0, 1]) returns True, indicating that there is a possibility that an original binary array could exist to arrive at this derived array.
1derived = [1, 0, 1]
2Solution().doesValidArrayExist(derived) # Returns True

This small example demonstrates the effectiveness of the XOR operation for solving this problem and how a cumulative XOR of derived being equal to zero serves as the condition to determine the existence of a corresponding original array.

Solution Implementation

1from functools import reduce
2from typing import List
3from operator import xor
4
5class Solution:
6    def doesValidArrayExist(self, derived: List[int]) -> bool:
7        # The function checks if there exists a valid array
8        # An array is considered valid if the cumulative XOR of all elements is 0
9        # 'reduce' applies the 'xor' function cumulatively to the items of 'derived', from left to right
10        # Hence, the entire list is reduced to a single value
11        # If this final reduced value is 0, it means all pairs are matched (i.e., a valid array exists)
12      
13        return reduce(xor, derived) == 0
14
1class Solution {
2
3    // A method that checks if there's a valid array whose derived xor-sum is zero
4    public boolean doesValidArrayExist(int[] derivedArray) {
5        int xorSum = 0; // Initialize xorSum to 0 to use it as the initial value
6
7        // Iterate over each element in the derived array
8        for (int element : derivedArray) {
9            // Perform XOR operation between the xorSum and the current element
10            xorSum ^= element;
11        }
12      
13        // Return true if the xorSum is 0, which means a valid array exists
14        // Otherwise, return false
15        return xorSum == 0;
16    }
17}
18
1#include <vector> // Include vector header for using vectors
2
3// The Solution class definition
4class Solution {
5public:
6    // Function checks if there exists a valid array such that all of its elements XORed equals zero
7    bool doesValidArrayExist(std::vector<int>& derivedArray) {
8        // Initialize a variable to store the cumulative XOR of array elements
9        int cumulativeXOR = 0;
10
11        // Iterate through each element in the derivedArray
12        for (int element : derivedArray) {
13            // Perform XOR operation and store the result back in cumulativeXOR
14            cumulativeXOR ^= element;
15        }
16
17        // If the final result of cumulativeXOR is zero, a valid array exists (return true)
18        // Otherwise, no such array exists (return false)
19        return cumulativeXOR == 0;
20    }
21};
22
1/**
2 * Determines if a "valid" array exists; an array is considered valid
3 * if the XOR of all its elements is zero.
4 * 
5 * @param {number[]} numbers - The array of numbers to be checked.
6 * @returns {boolean} - True if the array is valid, False otherwise.
7 */
8function doesValidArrayExist(numbers: number[]): boolean {
9    // Initialize sum as zero to perform XOR operation.
10    let xorSum = 0;
11  
12    // Iterate over each number in the array.
13    for (const number of numbers) {
14        // Perform XOR operation with current number and update the xorSum.
15        xorSum ^= number;
16    }
17  
18    // Check if xorSum is zero (all pairs XOR to zero); return true if so.
19    return xorSum === 0;
20}
21

Time and Space Complexity

The provided Python function doesValidArrayExist uses the reduce function with the xor operator from the functools and operator modules respectively to determine if the XOR of all the numbers in a list is equal to 0. The XOR operation is applied pairwise to the elements of the list until a single result remains.

Time Complexity

The reduce function applies the xor operation to the list derived with a time complexity of O(n), where n is the number of elements in the list. This is because each element in the list must be accessed exactly once to perform the XOR operation with the accumulated result.

The overall time complexity is O(n).

Space Complexity

Since the reduce function applies the xor operation in-place and accumulates the result without using any additional data structures that grow with input size, the space complexity is constant.

The overall space complexity is O(1).

Learn more about how to find time and space complexity quickly using problem constraints.


Fast Track Your Learning with Our Quick Skills Quiz:

What's the output of running the following function using the following tree as input?

1def serialize(root):
2    res = []
3    def dfs(root):
4        if not root:
5            res.append('x')
6            return
7        res.append(root.val)
8        dfs(root.left)
9        dfs(root.right)
10    dfs(root)
11    return ' '.join(res)
12
1import java.util.StringJoiner;
2
3public static String serialize(Node root) {
4    StringJoiner res = new StringJoiner(" ");
5    serializeDFS(root, res);
6    return res.toString();
7}
8
9private static void serializeDFS(Node root, StringJoiner result) {
10    if (root == null) {
11        result.add("x");
12        return;
13    }
14    result.add(Integer.toString(root.val));
15    serializeDFS(root.left, result);
16    serializeDFS(root.right, result);
17}
18
1function serialize(root) {
2    let res = [];
3    serialize_dfs(root, res);
4    return res.join(" ");
5}
6
7function serialize_dfs(root, res) {
8    if (!root) {
9        res.push("x");
10        return;
11    }
12    res.push(root.val);
13    serialize_dfs(root.left, res);
14    serialize_dfs(root.right, res);
15}
16

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