XOR of numbers that appeared even number of times in given Range

Given an array of numbers of size N and Q queries. Each query or a range can be represented by L (LeftIndex) and R(RightIndex). Find the XOR-sum of the numbers that appeared even number of times in the given range.

Prerequisite : Queries for number of distinct numbers in given range. | Segment Tree for range query

Examples : 

Input : arr[] = { 1, 2, 1, 3, 3, 2, 3 }
        Q = 5
        L = 3,  R = 6
        L = 3,  R = 4
        L = 0,  R = 2
        L = 0,  R = 6
        L = 0,  R = 4
Output : 0
         3
         1
         3
         2

Explanation of above example: 

In Query 1, there are no numbers which appeared even number of times. 
Hence the XOR-sum is 0. 
In Query 2, {3} appeared even number of times. XOR-sum is 3. 
In Query 3, {1} appeared even number of times. XOR-sum is 1. 
In Query 4, {1, 2} appeared even number of times. XOR-sum is 1 xor 2 = 3. 
In Query 5, {1, 3} appeared even number of times. XOR-sum is 1 xor 3 = 2.

Segment Trees or Binary Indexed Trees can be used to solve this problem efficiently.

Approach : 

Firstly, it is easy to note that the answer for the query is the XOR-sum of all elements in the query range xor-ed with XOR-sum of distinct elements in the query range (since taking XOR of an element with itself results into a null value). Find the XOR-sum of all numbers in query range using prefix XOR-sums. 

To find the XOR-sum of distinct elements in range : Number of distinct elements in a subarray of given range. 
Now, returning back to our main problem, just change the assignment BIT[i] = 1 to BIT[i] = arr_i and count the XOR-sum instead of sum. 

Below is the implementation using Binary Indexed Trees in CPP

CPP

// CPP Program to Find the XOR-sum
// of elements that appeared even
// number of times within a range
#include <bits/stdc++.h>
using namespace std;

/* structure to store queries
   L --> Left Bound of Query
   R --> Right Bound of Query
   idx --> Query Number */
struct que {
    int L, R, idx;
};

// cmp function to sort queries
// according to R
bool cmp(que a, que b)
{
    if (a.R != b.R)
        return a.R < b.R;
    else
        return a.L < b.L;
}

/*  N  --> Number of elements present in
    input array. BIT[0..N] --> Array that
    represents Binary Indexed Tree*/

// Returns XOR-sum of arr[0..index]. This
// function assumes that the array is
// preprocessed and partial sums of array
// elements are stored in BIT[].
int getSum(int BIT[], int index)
{
    // Initialize result
    int xorSum = 0;

    // index in BITree[] is 1 more than
    // the index in arr[]
    index = index + 1;

    // Traverse ancestors of BIT[index]
    while (index > 0)
    {
        // Take XOR of current element
        // of BIT to xorSum
        xorSum ^= BIT[index];

        // Move index to parent node
        // in getSum View
        index -= index & (-index);
    }
    return xorSum;
}

// Updates a node in Binary Index Tree
// (BIT) at given index in BIT.  The
// given value 'val' is xored to BIT[i]
// and all of its ancestors in tree.
void updateBIT(int BIT[], int N,
               int index, int val)
{
    // index in BITree[] is 1 more than
    // the index in arr[]
    index = index + 1;

    // Traverse all ancestors and
    // take xor with 'val'
    while (index <= N)
    {
        // Take xor with 'val' to
        // current node of BIT
        BIT[index] ^= val;

        // Update index to that of
        // parent in update View
        index += index & (-index);
    }
}

// Constructs and returns a Binary Indexed
// Tree for given array of size N.
int* constructBITree(int arr[], int N)
{
    // Create and initialize BITree[] as 0
    int* BIT = new int[N + 1];

    for (int i = 1; i <= N; i++)
        BIT[i] = 0;

    return BIT;
}

// Function to answer the Queries
void answeringQueries(int arr[], int N,
        que queries[], int Q, int BIT[])
{
    // Creating an array to calculate
    // prefix XOR sums
    int* prefixXOR = new int[N + 1];

    // map for coordinate compression
    // as numbers can be very large but we
    // have limited space
    map<int, int> mp;

    for (int i = 0; i < N; i++) {

        // If A[i] has not appeared yet
        if (!mp[arr[i]])
            mp[arr[i]] = i;

        // calculate prefixXOR sums
        if (i == 0)
            prefixXOR[i] = arr[i];
        else
            prefixXOR[i] =
                prefixXOR[i - 1] ^ arr[i];
    }

    // Creating an array to store the
    // last occurrence of arr[i]
    int lastOcc[1000001];
    memset(lastOcc, -1, sizeof(lastOcc));

    // sort the queries according to comparator
    sort(queries, queries + Q, cmp);

    // answer for each query
    int res[Q];

    // Query Counter
    int j = 0;

    for (int i = 0; i < Q; i++)
    {
        while (j <= queries[i].R)
        {
            // If last visit is not -1 update
            // arr[j] to set null by taking
            // xor with itself at the idx
            // equal lastOcc[mp[arr[j]]]
            if (lastOcc[mp[arr[j]]] != -1)
                updateBIT(BIT, N,
                      lastOcc[mp[arr[j]]], arr[j]);

            // Setting lastOcc[mp[arr[j]]] as j and
            // updating the BIT array accordingly
            updateBIT(BIT, N, j, arr[j]);
            lastOcc[mp[arr[j]]] = j;
            j++;
        }

        // get the XOR-sum of all elements within
        // range using precomputed prefix XORsums
        int allXOR = prefixXOR[queries[i].R] ^
                     prefixXOR[queries[i].L - 1];

        // get the XOR-sum of distinct elements
        // within range using BIT query function
        int distinctXOR = getSum(BIT, queries[i].R) ^
                          getSum(BIT, queries[i].L - 1);

        // store the final answer at the numbered query
        res[queries[i].idx] = allXOR ^ distinctXOR;
    }

    // Output the result
    for (int i = 0; i < Q; i++)
        cout << res[i] << endl;
}

// Driver program to test above functions
int main()
{
    int arr[] = { 1, 2, 1, 3, 3, 2, 3 };
    int N = sizeof(arr) / sizeof(arr[0]);

    int* BIT = constructBITree(arr, N);

    // structure of array for queries
    que queries[5];

    // Initializing values (L, R, idx) to queries
    queries[0].L = 3;
    queries[0].R = 6, queries[0].idx = 0;
    queries[1].L = 3;
    queries[1].R = 4, queries[1].idx = 1;
    queries[2].L = 0;
    queries[2].R = 2, queries[2].idx = 2;
    queries[3].L = 0;
    queries[3].R = 6, queries[3].idx = 3;
    queries[4].L = 0;
    queries[4].R = 4, queries[4].idx = 4;

    int Q = sizeof(queries) / sizeof(queries[0]);

    // answer Queries
    answeringQueries(arr, N, queries, Q, BIT);

    return 0;
}

Output:

0
3
1
3
2

Complexity Analysis:

• Time Complexity: O(Q * Log(N)), where N is the size of array, Q is the total number of queries.
• Space complexity: O(N) where N is size of array.