Count of odd and even parity elements in subarray using MO’s algorithm

Given an array arr consisting of N elements and Q queries represented by L and R denoting a range, the task is to print the count of odd and even parity elements in the subarray [L, R].

Examples:

Input:
arr[]=[5, 2, 3, 1, 4, 8, 10]
Q=2
1 3
0 4
Output:
2 1
3 2

Explanation:
In query 1, odd parity elements in subarray [1:3] are 2 and 1 and even parity element is 3.
In query 2, odd parity elements in subarray [0:4] are 2, 1 and 4 and even parity elements are 5 and 3.

Input:
arr[] = { 13, 17, 12, 10, 18, 19, 15, 7, 9, 6 }
Q=3
1 5
0 7
2 9
Output:
1 4
3 5
2 6
Explanation:
In query 1, odd parity element in subarray [1:4] is 19 and even parity elements are 17,12,10 and 18.
In query 2, odd parity elements in subarray [0:7] are 13, 19 and 7 and even parity elements are 17,12,10,18 and 15.
In query 3, odd parity elements in subarray [2:6] are 19 and 7 and even parity elements are 12,10,18, 15, 9 and 6.

Approach:
The idea of MO’s algorithm is to pre-process all queries so that result of one query can be used in the next query.

1. Sort all queries in a way that queries with L values from 0 to √n – 1 are put together, followed by queries from √n to 2×√n – 1, and so on. All queries within a block are sorted in increasing order of R values.
2. Count the odd parity elements and then calculate the even parity elements as (R-L+1- odd parity elements)
3. Process all queries one by one and increase the count of odd parity elements and store the result in the structure.
• Let count_oddP store the count of odd parity elements in previous query.
• Remove extra elements of previous query and add new elements for the current query. For example, if previous query was [0, 8] and the current query is [3, 9], then remove the elements arr[0], arr[1] and arr[2] and add arr[9].
4. In order to display the results, sort the queries in the order they were provided.

Adding elements()

• If the current element has odd parity then increase the count of count_oddP.

Removing elements()

• If the current element has odd parity then decrease the count of count_oddP.

Below code is the implementation of the above approach:

C++
// C++ program to count odd and
// even parity elements in subarray
// using MO's algorithm

#include <bits/stdc++.h>
using namespace std;

#define MAX 100000

// Variable to represent block size.
// This is made global so compare()
// of sort can use it.
int block;

// Structure to represent a query range
struct Query {
    // Starting index
    int L;
    // Ending index
    int R;
    // Index of query
    int index;
    // Count of odd
    // parity elements
    int odd;
    // Count of even
    // parity elements
    int even;
};

// To store the count of
// odd parity elements
int count_oddP;

// Function used to sort all queries so that
// all queries of the same block are arranged
// together and within a block, queries are
// sorted in increasing order of R values.
bool compare(Query x, Query y)
{
    // Different blocks, sort by block.
    if (x.L / block != y.L / block)
        return x.L / block < y.L / block;

    // Same block, sort by R value
    return x.R < y.R;
}

// Function used to sort all queries in order of their
// index value so that results of queries can be printed
// in same order as of input
bool compare1(Query x, Query y)
{
    return x.index < y.index;
}

// Function to Add elements
// of current range
void add(int currL, int a[])
{
    // _builtin_parity(x)returns true(1)
    // if the number has odd parity else
    // it returns false(0) for even parity.
    if (__builtin_parity(a[currL]))
        count_oddP++;
}

// Function to remove elements
// of previous range
void remove(int currR, int a[])
{
    // _builtin_parity(x)returns true(1)
    // if the number has odd parity else
    // it returns false(0) for even parity.
    if (__builtin_parity(a[currR]))
        count_oddP--;
}

// Function to generate the result of queries
void queryResults(int a[], int n, Query q[],
                int m)
{

    // Initialize number of odd parity
    // elements to 0
    count_oddP = 0;

    // Find block size
    block = (int)sqrt(n);

    // Sort all queries so that queries of
    // same blocks are arranged together.
    sort(q, q + m, compare);

    // Initialize current L, current R and
    // current result
    int currL = 0, currR = 0;

    for (int i = 0; i < m; i++) {
        // L and R values of current range
        int L = q[i].L, R = q[i].R;

        // Add Elements of current range
        while (currR <= R) {
            add(currR, a);
            currR++;
        }
        while (currL > L) {
            add(currL - 1, a);
            currL--;
        }

        // Remove element of previous range
        while (currR > R + 1)

        {
            remove(currR - 1, a);
            currR--;
        }
        while (currL < L) {
            remove(currL, a);
            currL++;
        }

        q[i].odd = count_oddP;
        q[i].even = R - L + 1 - count_oddP;
    }
}
// Function to display the results of
// queries in their initial order
void printResults(Query q[], int m)
{
    sort(q, q + m, compare1);
    for (int i = 0; i < m; i++) {
        cout << q[i].odd << " "
               << q[i].even << endl;
    }
}

// Driver Code
int main()
{

    int arr[] = { 5, 2, 3, 1, 4, 8, 10, 12 };
    int n = sizeof(arr) / sizeof(arr[0]);

    Query q[] = { { 1, 3, 0, 0, 0 },
                  { 0, 4, 1, 0, 0 },
                  { 4, 7, 2, 0, 0 } };

    int m = sizeof(q) / sizeof(q[0]);

    queryResults(arr, n, q, m);

    printResults(q, m);

    return 0;
}
Output:

2 1
3 2
2 2

Time Complexity: O(Q × √n)