# 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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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, arr and arr and add arr.
4. In order to display the results, sort the queries in the order they were provided.

• 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 ` `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); ` ` `  `    ``Query q[] = { { 1, 3, 0, 0, 0 },  ` `                  ``{ 0, 4, 1, 0, 0 },  ` `                  ``{ 4, 7, 2, 0, 0 } }; ` ` `  `    ``int` `m = ``sizeof``(q) / ``sizeof``(q); ` ` `  `    ``queryResults(arr, n, q, m); ` ` `  `    ``printResults(q, m); ` `     `  `    ``return` `0; ` `} `

Output:

```2 1
3 2
2 2
```

Time Complexity: O(Q × √n) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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