1486. XOR Operation in an Array

EasyBit ManipulationMath
Leetcode Link

Problem Description

You are tasked with creating an array named nums, where each element follows a specific pattern based on two given integers: n and start. The array should be constructed such that the element at index i is calculated by the formula start + 2 * i. In this formula, i is the index in the array, starting from 0 (hence it's 0-indexed), and n is the total number of elements that should exist within the array nums. Once the array is constructed, your goal is to calculate the result of applying the bitwise XOR operation to all the elements within the array.

Bitwise XOR is an operation that takes two bits and returns 1 if the bits are different and 0 if they are the same. When doing this operation between numbers, it's applied bit by bit from their binary representations. The XOR of a single number with itself is always 0, and the XOR of a number with 0 is always the number itself.

The challenge is to write a function that will carry out this process and return the single integer that results from the cumulative XOR of all the elements in nums.


To solve this problem, we can iterate over the range from 0 to n - 1 and compute the value of start + 2 * i for each i. After computing each element based on the given pattern, we perform a bitwise XOR operation sequentially on these elements.

We start with an initial value of 0 for the answer (ans). Then, for each i, we calculate the ith number in the sequence with start + 2 * i and use the XOR operator (^) to combine it with our running total in ans. This is equivalent to saying that ans becomes ans XOR (start + 2 * i) for each iteration. Since XOR is associative, the order in which we combine the numbers doesn't matter, meaning we can sequentially update ans as we go through the loop.

After completing the loop, ans will contain the cumulative XOR of all the array's elements, which is what we return as our final solution.

The solution is straightforward and relies on the properties of the XOR operation to combine each new element into our running total. This cumulative approach is useful because we don't need to maintain the array nums in memory; we only need to remember the current XOR value through each iteration, which is a more memory-efficient solution.

Learn more about Math patterns.

Solution Approach

The implementation of the solution uses a simple iterative approach, leveraging the properties of the XOR operation. The reference solution provided in Python demonstrates the approach clearly.

The primary algorithmic pattern used here is iteration with a single loop. We exploit two key insights: first, we know exactly how many times to iterate since we are given n, the size of the array nums. Second, the XOR operation allows us to combine elements one by one without needing to store the entire array in memory.

Here is a step-by-step breakdown of the code implementation:

  1. Initialize ans to 0. This serves as an accumulation variable to store the cumulative XOR result.
  2. Iterate over a range from 0 to n - 1. The range function generates a sequence of numbers, allowing us to calculate each array element without an actual array.
    • Within the loop, update ans by XORing its current value with start + 2 * i, which represents the current element of the array. In code: ans ^= start + 2 * i.
  3. After completing the loop, all elements have been XORed into ans.
  4. Finally, return ans, which now contains the cumulative XOR of all elements as per the array nums[i] = start + 2 * i.

The reason no additional data structures are used here is due to the nature of the XOR operation, which is both associative and commutative. As a result, we don't need to remember past values once they've been incorporated into ans.

The code uses bit manipulation (^ operator in Python) to perform the XOR operation. Bit manipulation is a powerful technique in programming that allows for efficient computation, often with lower memory usage and faster execution, especially suitable for such bitwise arithmetic tasks.

The simplicity of the approach lies in the fact that the problem requires no more than applying the given formula iteratively and combining the results using the XOR operation. No complex data structures or algorithms are needed. The final result is directly returned after the loop’s completion.

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

What is the worst case running time for finding an element in a binary tree (not necessarily binary search tree) of size n?

Example Walkthrough

Let's illustrate the solution approach using a small example. Consider n = 4 and start = 3. Based on the problem description, we need to create an array nums such that:

  • nums[0] = start + 2 * 0
  • nums[1] = start + 2 * 1
  • nums[2] = start + 2 * 2
  • nums[3] = start + 2 * 3

Calculating the above values based on the given formula:

  • nums[0] = 3 + 2 * 0 = 3
  • nums[1] = 3 + 2 * 1 = 5
  • nums[2] = 3 + 2 * 2 = 7
  • nums[3] = 3 + 2 * 3 = 9

Now, we apply the bitwise XOR operation to all elements in this array. The goal is to perform the XOR operation in a cumulative manner.

  • Start with an initial value of ans = 0.
  • Perform ans ^= nums[0], hence ans = 0 ^ 3 = 3 (since XOR with zero gives the number itself).
  • Update ans with ans ^= nums[1], so ans = 3 ^ 5.
    • The binary representation of 3 is 011.
    • The binary representation of 5 is 101.
    • Performing XOR on these yields 110, which is binary for 6. Now, ans = 6.
  • Update ans again with ans ^= nums[2], so ans = 6 ^ 7.
    • The binary representation of 6 is 110.
    • The binary representation of 7 is 111.
    • Performing XOR on these yields 001, which is binary for 1. Now, ans = 1.
  • Finally, update ans a last time with ans ^= nums[3], so ans = 1 ^ 9.
    • The binary representation of 1 is 001.
    • The binary representation of 9 is 1001.
    • Performing XOR on these (after aligning the bits) yields 1000, which is binary for 8. Now, ans = 8.

The XOR of the entire array nums is 8. Therefore, given n = 4 and start = 3, the function would return 8. This example demonstrates how the iterative approach can be used to calculate the cumulative XOR without explicitly creating the array nums.

Solution Implementation

1class Solution:
2    def xor_operation(self, n: int, start: int) -> int:
3        # Initialize the answer to zero
4        answer = 0
6        # Loop from 0 to n-1
7        for i in range(n):
8            # Perform XOR operation between the current answer and the new element in the series
9            answer ^= start + 2 * i
11        # Return the final answer after performing all XOR operations
12        return answer
14# The Solution class contains a method xor_operation which takes two arguments:
15# n - the number of elements in the array
16# start - the first element in the array where the array is defined by the formula start + 2 * i for each element i in the range [0, n).
17# The method applies the XOR operation cumulatively over the entire array to find the single resultant value and returns it.
1// Class Solution contains a method to perform XOR operations on a sequence of numbers.
2class Solution {
4    // The xorOperation method takes the number of elements 'n' and the 'start' value of the series as parameters.
5    public int xorOperation(int n, int start) {
6        // Initialize 'result' that will hold the cumulative XOR value.
7        int result = 0;
9        // Loop over the series from 0 to 'n-1'.
10        for (int i = 0; i < n; ++i) {
11            // Perform XOR of 'result' with each element in the sequence.
12            // Element value is calculated as 'start + 2*i'.
13            result ^= start + 2 * i;
14        }
16        // Return the final XOR result after processing all elements.
17        return result;
18    }
1class Solution {
3    // Function to compute the XOR of all numbers in the generated array.
4    int xorOperation(int n, int start) {
5        int xor_result = 0; // Initialize result variable to store the final XOR
7        // Loop through the sequence to compute the XOR
8        for (int i = 0; i < n; ++i) {
10            // Each element in the sequence is the start value plus twice the index
11            // XOR it with the current result
12            xor_result ^= start + 2 * i;
13        }
15        // Return the final XOR result of the sequence
16        return xor_result;
17    }
1// Variable to hold the XOR result
2let xorResult: number = 0;
4// Function to compute the XOR of all numbers in the generated array.
5function xorOperation(n: number, start: number): number {
6    // Initialize the XOR result to zero for each call of the function
7    xorResult = 0;
9    // Loop through the numbers from 0 to n-1 to compute the XOR of the generated elements
10    for (let i = 0; i < n; i++) {
11        // The current element in the sequence is calculated by adding start with twice the index
12        // XOR the current element with the accumulated result stored in xorResult
13        xorResult ^= start + 2 * i;
14    }
16    // Return the XOR result of all the elements in the array
17    return xorResult;

Time and Space Complexity

  • The time complexity of the code is O(n), where n is the number of elements we are performing the XOR operation on. This is because there is a single loop in the function xorOperation that iterates n times, and the XOR operation itself is a constant-time operation.

  • The space complexity of the code is O(1), implying that the space required for computation does not depend on the input size n. The function maintains a single integer ans that is used to accumulate the result, hence no additional space proportional to n is utilized.

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

Fast Track Your Learning with Our Quick Skills Quiz:

Which data structure is used in a depth first search?

Recommended Readings

Got a question? Ask the Monster Assistant anything you don't understand.

Still not clear? Ask in the Forum,  Discord or Submit the part you don't understand to our editors.