1486. XOR Operation in an Array

EasyBit ManipulationMath

Problem Description

You need to create an array and find the XOR of all its elements.

Given two integers n and start, you construct an array nums with n elements where each element follows a specific pattern:

nums[0] = start + 2 * 0 = start
nums[1] = start + 2 * 1 = start + 2
nums[2] = start + 2 * 2 = start + 4
And so on...

In general, nums[i] = start + 2 * i where i goes from 0 to n-1.

Your task is to calculate the bitwise XOR of all elements in this array and return the result.

For example, if n = 5 and start = 0:

The array would be [0, 2, 4, 6, 8]
The XOR operation would be 0 ^ 2 ^ 4 ^ 6 ^ 8 = 8

The solution uses Python's reduce function with the XOR operator to efficiently compute the result without explicitly creating the array. It generates each element start + 2 * i for i in range [0, n) and applies XOR operation cumulatively across all elements.

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Intuition

The key observation is that we don't actually need to create and store the entire array in memory. Since we only need the XOR result of all elements, we can generate each element on-the-fly and compute the XOR as we go.

The XOR operation has a useful property: it's associative, meaning (a ^ b) ^ c = a ^ (b ^ c). This allows us to process elements one by one without storing them all first.

Starting from the problem requirements:

Each element follows the pattern start + 2 * i
We need the XOR of all n elements

The straightforward approach would be:

Generate each number using the formula start + 2 * i for i from 0 to n-1
Apply XOR operation across all these numbers

Since XOR is a bitwise operation that processes numbers bit by bit, and we're dealing with a simple arithmetic sequence (with common difference of 2), we can compute the result by iterating through the sequence once.

The solution leverages Python's reduce function which takes a binary operation (XOR in this case) and applies it cumulatively to all elements in a sequence. Combined with a generator expression (start + 2 * i) for i in range(n), we efficiently compute the XOR without extra space for storing the array.

This approach has O(n) time complexity for generating and XOR-ing n numbers, and O(1) space complexity since we're not storing the array.

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Solution Approach

The solution implements a direct simulation approach to calculate the XOR of all elements in the virtual array.

Implementation Details:

The solution uses Python's reduce function from the functools module combined with the XOR operator to compute the result:

return reduce(xor, ((start + 2 * i) for i in range(n)))

Let's break down how this works:

Generator Expression: (start + 2 * i) for i in range(n)
- This creates a generator that produces values on-demand
- For each i from 0 to n-1, it yields the value start + 2 * i
- No actual array is created in memory
XOR Operation: The xor function (imported from operator module) performs bitwise XOR between two numbers
Reduce Function: reduce(xor, generator)
- Takes the first two elements from the generator and applies XOR
- Takes the result and XORs it with the next element
- Continues until all elements are processed
- Returns the final XOR result

Step-by-step execution example with n = 4 and start = 3:

Generate sequence: 3, 5, 7, 9 (from 3+2*0, 3+2*1, 3+2*2, 3+2*3)
Apply reduce with XOR:
- Step 1: 3 ^ 5 = 6
- Step 2: 6 ^ 7 = 1
- Step 3: 1 ^ 9 = 8
Return 8

This simulation approach directly computes what's asked without any optimization tricks, making it straightforward to understand and implement. The time complexity is O(n) as we process each of the n elements once, and space complexity is O(1) since we only maintain the running XOR result.

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Example Walkthrough

Let's walk through the solution with n = 3 and start = 4.

Step 1: Understand what array we're building

nums[0] = 4 + 2 * 0 = 4
nums[1] = 4 + 2 * 1 = 6
nums[2] = 4 + 2 * 2 = 8
So our virtual array is [4, 6, 8]

Step 2: Apply the reduce function with XOR

The solution uses reduce(xor, ((start + 2 * i) for i in range(n))).

The generator produces values in sequence:

When i = 0: yields 4 + 2 * 0 = 4
When i = 1: yields 4 + 2 * 1 = 6
When i = 2: yields 4 + 2 * 2 = 8

The reduce function processes these values:

First operation: 4 ^ 6
- Binary: 100 ^ 110 = 010 (which is 2 in decimal)
Second operation: 2 ^ 8
- Binary: 0010 ^ 1000 = 1010 (which is 10 in decimal)

Step 3: Return the result The final XOR result is 10.

Let's verify manually: 4 ^ 6 ^ 8

In binary: 100 ^ 110 ^ 1000
100 ^ 110 = 010
010 ^ 1000 = 1010
Converting back to decimal: 1010 = 10

The solution efficiently computes this result without creating the actual array in memory, processing each element as it's generated.

Solution Implementation

1from functools import reduce
2from operator import xor
3
4class Solution:
5    def xorOperation(self, n: int, start: int) -> int:
6        """
7        Calculate the XOR of n elements in an array where each element 
8        is computed as start + 2 * i for i from 0 to n-1.
9      
10        Args:
11            n: The number of elements in the array
12            start: The starting value for generating array elements
13          
14        Returns:
15            The XOR result of all elements in the generated array
16        """
17        # Generate array elements using the formula: start + 2 * i
18        # where i ranges from 0 to n-1
19        # Then apply XOR operation across all elements using reduce
20        return reduce(xor, (start + 2 * i for i in range(n)))
21

1class Solution {
2    /**
3     * Performs XOR operation on an array where nums[i] = start + 2 * i.
4     * 
5     * @param n     The number of elements in the array
6     * @param start The starting value for generating array elements
7     * @return      The XOR of all elements in the generated array
8     */
9    public int xorOperation(int n, int start) {
10        // Initialize the result variable to store XOR result
11        int result = 0;
12      
13        // Iterate through n elements
14        for (int i = 0; i < n; i++) {
15            // Calculate current element: start + 2 * i
16            int currentElement = start + 2 * i;
17          
18            // XOR the current element with the accumulated result
19            result ^= currentElement;
20        }
21      
22        // Return the final XOR result
23        return result;
24    }
25}
26

1class Solution {
2public:
3    int xorOperation(int n, int start) {
4        // Initialize the result variable to store XOR of all elements
5        int result = 0;
6      
7        // Iterate through n elements
8        for (int i = 0; i < n; ++i) {
9            // Calculate the current element: start + 2*i
10            // XOR it with the accumulated result
11            result ^= (start + 2 * i);
12        }
13      
14        // Return the final XOR result
15        return result;
16    }
17};
18

1/**
2 * Performs XOR operation on an array of n elements
3 * where each element is calculated as start + 2 * i
4 * @param n - The number of elements to generate and XOR
5 * @param start - The starting value for the sequence
6 * @returns The XOR result of all generated elements
7 */
8function xorOperation(n: number, start: number): number {
9    // Initialize the result variable to store cumulative XOR
10    let xorResult: number = 0;
11  
12    // Iterate through n elements
13    for (let i: number = 0; i < n; i++) {
14        // Generate current element: start + 2 * i
15        // XOR it with the accumulated result
16        xorResult ^= start + 2 * i;
17    }
18  
19    // Return the final XOR result
20    return xorResult;
21}
22

Time and Space Complexity

The time complexity is O(n), where n is the input parameter representing the number of elements to XOR. The algorithm iterates through a generator expression that produces n values (from i = 0 to i = n-1), and for each iteration, it performs a constant-time calculation start + 2 * i followed by an XOR operation. The reduce function applies the XOR operation sequentially across all n generated values, resulting in linear time complexity.

The space complexity is O(1). The code uses a generator expression ((start + 2 * i) for i in range(n)) instead of a list comprehension, which means values are computed on-the-fly rather than stored in memory. The reduce function only maintains a single accumulator variable to store the running XOR result, and no additional data structures proportional to the input size are created. Therefore, the space usage remains constant regardless of the value of n.

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Common Pitfalls

1. Order of Operations Confusion

A common mistake is misunderstanding the operator precedence when computing start + 2 * i. Some might incorrectly parenthesize it as (start + 2) * i.

Incorrect Implementation:

# Wrong: This computes (start + 2) * i instead of start + (2 * i)
return reduce(xor, ((start + 2) * i for i in range(n)))

Correct Implementation:

# Correct: Multiplication has higher precedence than addition
return reduce(xor, (start + 2 * i for i in range(n)))
# Or be explicit with parentheses for clarity
return reduce(xor, (start + (2 * i) for i in range(n)))

2. Edge Case Handling with n = 1

When using reduce without providing an initial value, it requires at least one element in the iterable. While this works fine for n >= 1, developers might forget to handle or test the edge case where n = 1.

Potential Issue:

# This works but might not be obvious that n=1 is handled correctly
return reduce(xor, (start + 2 * i for i in range(n)))

More Explicit Solution:

# Handle single element case explicitly for clarity
if n == 1:
    return start
return reduce(xor, (start + 2 * i for i in range(n)))

3. Using List Comprehension Instead of Generator

Creating an actual list when a generator would suffice wastes memory, especially for large values of n.

Memory-Inefficient:

# Creates entire list in memory
return reduce(xor, [start + 2 * i for i in range(n)])

Memory-Efficient:

# Uses generator expression - computes values on-demand
return reduce(xor, (start + 2 * i for i in range(n)))

4. Manual Loop Implementation Errors

Some developers might try to implement the XOR manually but forget to initialize the result properly or use the wrong initial value.

Common Mistake:

# Wrong: Initializing with 1 instead of 0
result = 1  # Should be 0 for XOR identity
for i in range(n):
    result ^= start + 2 * i
return result

Correct Manual Implementation:

# Correct: XOR identity element is 0
result = 0
for i in range(n):
    result ^= start + 2 * i
return result

5. Forgetting XOR Properties

Not recognizing that XOR is associative and commutative, leading to unnecessarily complex solutions or worrying about the order of operations.

Overcomplicated:

# Unnecessarily sorting or ordering elements
elements = [start + 2 * i for i in range(n)]
elements.sort()  # Unnecessary - XOR is commutative
return reduce(xor, elements)

Simple and Correct:

# Order doesn't matter for XOR
return reduce(xor, (start + 2 * i for i in range(n)))

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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

[30, 20, 10]

[100, 33, 30]

[10, 12, 20]

I don't know