XOR Queries of a Subarray

class Solution {
public:
    vector<int> xorQueries(vector<int>& data, vector<vector<int>>& questions) {
        vector<int> output;
    
        for (int i = 1; i < data.size(); ++i) {
            data[i] ^= data[i - 1];
        }
    
        for (const auto& question : questions) {
            if (question[0] > 0) {
                output.push_back(data[question[0] - 1] ^ data[question[1]]);
            } else {
                output.push_back(data[question[1]]);
            }
        }
    
        return output;
    }
};

This C++ solution addresses the problem of executing XOR queries on a subarray. The function xorQueries uses a prefix XOR array to efficiently calculate the result for each query. Follow these steps to understand the approach implemented in the code:

1. Initialize a vector output to store the results of queries.
2. Convert the input data vector into a prefix XOR array. This transformation helps in retrieving the XOR of any subarray in constant time. For each index from 1 to the size of data, update data[i] with the XOR of data[i] and data[i-1].
3. Loop through each query in questions. Determine the result based on the range given:
   • If the start of the query range question[0] is greater than 0, calculate the XOR from data[question[0] - 1] to data[question[1]] using the prefix XOR property and push the result to output.
   • If question[0] is 0, directly push data[question[1]] to output as it represents the XOR from the beginning of the array up to question[1].
4. Return the vector output which contains the result for each query.

The use of a prefix XOR array enhances the efficiency of the solution, enabling each query result to be computed in constant time after the initial preprocessing step.

class Solution {
    
    public int[] xorQueries(int[] arr, int[][] queries) {
        List<Integer> resList = new ArrayList<>();
    
        // Creating prefix XOR representation for arr
        for (int i = 1; i < arr.length; i++) {
            arr[i] ^= arr[i - 1];
        }
    
        // Evaluate queries using the prefix XOR
        for (int[] query : queries) {
            if (query[0] > 0) {
                resList.add(arr[query[0] - 1] ^ arr[query[1]]);
            } else {
                resList.add(arr[query[1]]);
            }
        }
    
        return resList.stream().mapToInt(i -> i).toArray();
    }
}

The proposed Java solution revolves around solving the "XOR Queries of a Subarray" problem by utilizing the prefix XOR array concept. Here's a concise explanation of the approach adopted in the code:

1. Convert the input array into a prefix XOR array. This involves iterating through the array such that each element at index i becomes the XOR of all elements from the beginning of the array up to i. This is performed using an in-place modification for efficiency.

2. Process each query based on the prefix XOR array prepared in the previous step. Each query specifies a subarray using a start and end index. If the start index is more than 0, find the XOR for the subarray by XORing elements at positions defined by the query. If the start index is 0, directly take the value from the array since the XOR from start up to any index i in a prefix XOR array is directly stored at arr[i].

3. Convert the resultant list from query evaluations into an array of integers to match the expected output format.

This method is efficient, leveraging the properties of XOR and the cumulation nature of the prefix array to answer each query in constant time after the initial pre-processing step, resulting in a time complexity of O(N + Q) where N is the number of elements in the array, and Q is the number of queries.

class Solution:
    def computeXorQueries(self, data, query_list):
        query_results = []
    
        # Convert data list into a prefix XOR accumulation
        for idx in range(1, len(data)):
            data[idx] ^= data[idx - 1]
    
        # Handle each query with the accumulated XOR values
        for start, end in query_list:
            if start > 0:
                query_results.append(data[start - 1] ^ data[end])
            else:
                query_results.append(data[end])
    
        return query_results

The Python solution for XOR Queries of a Subarray involves leveraging the properties of XOR and prefix accumulation to efficiently answer multiple queries regarding the XOR of subarray elements. Follow the detailed steps included in the implementation:

1. Initiate an empty list query_results to store the results of each query.

2. Create a prefix XOR accumulation for the given list data. This step is crucial as it enables each element at index i in data to store the XOR of all elements from the start up to i. Modify the data list in-place starting from index 1 through the length of data. For each index, update the current element to be XORed with the previous element.

3. Process each query in query_list. Each query is defined by a pair of indices (start, end). For each query:

   • If start is greater than 0, compute the XOR from index start to end by XORing the elements at data[end] and data[start - 1] (because of the properties of prefix accumulation). Append the result to query_results.
   • If start equals 0, append the value of data[end] directly to query_results as this covers the XOR from the beginning up to the end.

4. Return the query_results as the output of the function.

This approach ensures that each query is handled in constant time, O(1), after the initial setup of the prefix XOR list, making the entire solution efficient, especially for scenarios with multiple queries.