Leetcode 1310: XOR Queries of a Subarray

Input: The input consists of an array of positive integers and an array of queries, where each query contains two integers representing a range in the array.

Example: Input: arr = [2, 5, 7, 10], queries = [[0,2],[1,3],[2,3],[0,1]]

Constraints:

• 1 <= arr.length, queries.length <= 3 * 10^4

• 1 <= arr[i] <= 10^9

• queries[i].length == 2

• 0 <= lefti <= righti < arr.length

Output: The output is an array of integers where each element is the XOR result of the corresponding query.

Example: Output: [4, 5, 13, 7]

Constraints:

• Each element of the output corresponds to the XOR result for the respective query.

Goal: Calculate the XOR of elements between indices `lefti` and `righti` for each query efficiently.

Steps:

• Use prefix XOR array to precompute XOR values for the array.

• For each query, calculate the XOR from the prefix array using the formula: `prefix[righti + 1] ^ prefix[lefti]`.

Goal: The solution should handle up to 30,000 queries and array elements efficiently.

Steps:

• Ensure that the algorithm runs in optimal time for the input size constraints.

Assumptions:

• The array and queries are well-formed and follow the given constraints.

• Input: Input: arr = [2, 5, 7, 10], queries = [[0,2],[1,3],[2,3],[0,1]]

• Explanation: The XOR values for queries are computed by applying the XOR operation on the array elements between indices `lefti` and `righti`.

Approach: We can solve the problem efficiently by using a prefix XOR array, which allows us to compute the XOR for any range in constant time.

Observations:

• The XOR of a range of elements can be derived from a prefix XOR array.

• Precompute the XOR for all elements up to index `i` in a prefix array. Then, for each query, the result can be derived in constant time.

Steps:

• Initialize a prefix XOR array where `prefix[i]` stores the XOR of elements from index 0 to `i-1`.

• For each query, use the formula `prefix[righti + 1] ^ prefix[lefti]` to compute the XOR for the range.

Empty Inputs:

• If the array is empty, there are no queries to process.

Large Inputs:

• For large arrays and queries, ensure that the solution handles up to 30,000 elements efficiently.

Special Values:

• If the elements in the array are all the same, the XOR result for any range may be predictable.

Constraints:

• The solution must handle arrays with up to 30,000 elements and queries efficiently.

vector<int> xorQueries(vector<int>& arr, vector<vector<int>>& q) {
    int n = arr.size();
    vector<int> xr(n +1,0);
    for(int i = 1; i < n +1; i++)
        xr[i] = xr[i - 1] ^ arr[i - 1];
    vector<int> res;
    for(int i = 0; i < q.size(); i++) {
        auto &v = q[i];
        res.push_back(xr[v[1]+1] ^ xr[v[0]]);
     }
    return res;
}

1 : Function Declaration

vector<int> xorQueries(vector<int>& arr, vector<vector<int>>& q) {

The function 'xorQueries' is defined with two parameters: a vector 'arr' which contains the array elements and a 2D vector 'q' which holds the queries. Each query specifies a range for which the XOR is computed.

2 : Variable Initialization

    int n = arr.size();

The variable 'n' is initialized to the size of the input array 'arr'. This value is used to manage array bounds during the processing of queries.

3 : Vector Initialization

    vector<int> xr(n +1,0);

A vector 'xr' is initialized with size 'n+1' to store the cumulative XOR values. The extra space is to handle the prefix sum logic efficiently.

4 : Loop Control

    for(int i = 1; i < n +1; i++)

This for loop starts from index 1 and iterates through the array to compute the cumulative XOR for each element.

5 : XOR Operation

        xr[i] = xr[i - 1] ^ arr[i - 1];

In each iteration, the cumulative XOR at index 'i' is calculated by XORing the previous cumulative value with the current element of the array. This builds a prefix XOR array.

6 : Vector Initialization

    vector<int> res;

An empty vector 'res' is initialized to store the results of each query.

7 : Loop Control

    for(int i = 0; i < q.size(); i++) {

This loop iterates through each query in the 2D vector 'q'. Each query specifies a range of indices for which the XOR operation needs to be performed.

8 : Query Processing

        auto &v = q[i];

The current query is extracted by referencing the query vector 'v', which contains two indices representing the range for XOR computation.

9 : XOR Operation

        res.push_back(xr[v[1]+1] ^ xr[v[0]]);

The XOR for the range specified by the query is calculated using the prefix XOR array 'xr'. The result is pushed to the 'res' vector.

10 : Return Statement

    return res;

The function returns the 'res' vector, which contains the results of all the XOR queries.

Best Case: O(n) - The best case occurs when all queries are processed after building the prefix XOR array.

Average Case: O(n + m) - Where n is the length of the array and m is the number of queries.

Worst Case: O(n + m) - The worst case involves processing all queries with the prefix XOR array.

Description: The time complexity is O(n + m), where n is the length of the array and m is the number of queries.

Best Case: O(n) - The best case still requires space for the prefix XOR array.

Worst Case: O(n) - The space complexity is O(n) due to the storage of the prefix XOR array.

Description: The space complexity is O(n), where n is the length of the array.

Leetcode 1310: XOR Queries of a Subarray

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