algoadvance

Given an integer n and an integer start.

Define an array nums where nums[i] = start + 2*i (0-indexed) and n == nums.length.

Your task is to return the bitwise XOR of all elements of nums.

Example:

• Input: n = 5, start = 0
• Output: 8
• Explanation:
  • nums = [0, 2, 4, 6, 8]
  • Result is 0 ^ 2 ^ 4 ^ 6 ^ 8 which is 8.

Clarifying Questions

1. Constraints:
   • 1 <= n <= 1000
   • 0 <= start <= 1000
2. Output:
   • Single integer which is the bitwise XOR of the array elements.
3. Special Cases:
   • n = 1, here the result will be just start.
   • Consistency of the start, i.e., always non-negative integers for simplification purposes.

Strategy

1. Initialize Result:
   • Initialize a variable to store the XOR result, say result = 0.
2. Iterate Through the Array:
   • Loop through the range [0, n-1].
   • Calculate nums[i] = start + 2*i.
   • Perform the XOR operation on the variable result with nums[i].
3. Return Result:
   • After the loop, the result variable will hold the desired XOR value.

Code Implementation

def xorOperation(n: int, start: int) -> int:
    result = 0
    for i in range(n):
        result ^= start + 2 * i
    return result

# Example usage
print(xorOperation(5, 0))  # Output: 8
print(xorOperation(4, 3))  # Output: 8
print(xorOperation(1, 7))  # Output: 7
print(xorOperation(10, 5)) # Output: 2

Time Complexity

The time complexity of this algorithm is O(n) because we are iterating through the array only once, performing constant-time operations for each element.

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