1310. XOR Queries of a Subarray

Given the array arr of positive integers and the array queries where queries[i] = [Li, Ri], for each query i compute the XOR of elements from Li to Ri (that is, arr[Li] xor arr[Li+1] xor ... xor arr[Ri] ). Return an array containing the result for the given queries.

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 <= 3 * 10^4
• 1 <= arr[i] <= 10^9
• 1 <= queries.length <= 3 * 10^4
• queries[i].length == 2
• 0 <= queries[i][0] <= queries[i][1] < arr.length

This question is very important for placement point of view.

STEPS :

Example : A = [1,3,4,8] and queries = [[0,1],[1,2],[0,3],[3,3]]
STEP 1
inplace XOR values of array A
A[0] = 1 since starting from index 1 so A[1] remains same
A[1] = 1^3
A[2] = 1^3^4
A[3] = 1^3^4^8

STEP 2
Iterate through queries and compare

Case 1 : If start value is zero then just return inplace value of A[end]
In given example : [0,1] and [0,3]

value for [0,1] is A[1] i.e 1^3

and for[0,3] is [A[3] i.e 1^3^4^8

Case 2 : If start >= 1 then just return inplace values of A[start-1]^A[end]
In given example : [1,2] and [3,3]

so value for [1,2] is A[0]^A[2] i.e 1^(1^3^4) which equals 3^4

value for [3,3] is A[2]^A[3] i.e (1^3^4)^(1^3^4^8) which equals 8

hence all values are : [1 ^ 3, 3 ^ 4, 1^ 3^ 4^ 8, 8]
which equals

v = [2, 7, 14, 8]

T = O(n) & S = O(1)

vector<int> ans; // create an answer vector to store output
        
        // STEP 1
        for(int i = 1; i < A.size(); i++) {  // in-place in array A
            A[i] = A[i]^A[i-1];   // store in-place XOR till current index 
        }
        // STEP 2
        for(int i = 0; i < queries.size(); i++) {
            int start = queries[i][0]; // first element of i th queries
            int end = queries[i][1];   // second element of i th queries
            
            if(start == 0) {   // if first element of i th queries is zero then just push in-place stored value at index end in array A
                ans.push_back(A[end]);
            }
            else {
                ans.push_back(A[start-1]^A[end]);  // if first element is not zero then push inplace value XOR first-1 th inplace value
            }
        }
        return ans;