Introduction

In this blog post, we will explore how to solve the LeetCode problem "1310" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an array arr of positive integers. You are also given the array queries where queries[i] = [lefti, righti]. For each query i compute the XOR of elements from lefti to righti (that is, arr[lefti] XOR arr[lefti + 1] XOR ... XOR arr[righti] ). Return an array answer where answer[i] is the answer to the ith query. Example 1: Input: arr = [1,3,4,8], queries = [[0,1],[1,2],[0,3],[3,3]] Output: [2,7,14,8] Explanation: The binary representation of the elements in the array are: 1 = 0001 3 = 0011 4 = 0100 8 = 1000 The XOR values for queries are: [0,1] = 1 xor 3 = 2 [1,2] = 3 xor 4 = 7 [0,3] = 1 xor 3 xor 4 xor 8 = 14 [3,3] = 8 Example 2: Input: arr = [4,8,2,10], queries = [[2,3],[1,3],[0,0],[0,3]] Output: [8,0,4,4] Constraints: 1 <= arr.length, queries.length <= 3 * 104 1 <= arr[i] <= 109 queries[i].length == 2 0 <= lefti <= righti < arr.length

Explanation

Here's a breakdown of the approach and the Python code:

• Prefix XOR Array: Create a prefix XOR array where each element prefix_xor[i] stores the XOR of all elements from arr[0] to arr[i].
• Query Calculation: For each query [left, right], calculate the XOR sum from left to right using the prefix XOR array. If left > 0, the XOR sum is prefix_xor[right] ^ prefix_xor[left-1]. If left == 0, the XOR sum is simply prefix_xor[right].

• Result: Store the results of each query in an answer array.

• Runtime Complexity: O(n + m), where n is the length of arr and m is the number of queries.

• Storage Complexity: O(n)

Code

    def xor_queries(arr, queries):
    """
    Calculates the XOR of elements within given ranges in an array.

    Args:
        arr (list of int): An array of positive integers.
        queries (list of list of int): A list of queries, where each query is a list [left, right].

    Returns:
        list of int: An array containing the XOR results for each query.
    """

    prefix_xor = [0] * len(arr)
    prefix_xor[0] = arr[0]

    for i in range(1, len(arr)):
        prefix_xor[i] = prefix_xor[i - 1] ^ arr[i]

    answer = []
    for left, right in queries:
        if left == 0:
            answer.append(prefix_xor[right])
        else:
            answer.append(prefix_xor[right] ^ prefix_xor[left - 1])

    return answer