Given an array **arr[]** of **N** elements and two integers **A** to **B**, the task is to answer Q queries each having two integers **L **and **R.** For each query, find the number of elements in the subarray **arr[L…R]** which lies within the range **A** to **B** (inclusive).

**Examples:**

Input:arr[] = {7, 3, 9, 13, 5, 4}, A = 4, B = 7

query = {1, 5}

Output:2

Explanation:

Only 5 and 4 lies within 4 to 7

in the subarray {3, 9, 13, 5, 4}

Therefore, the count of such elements is 2.

Input:arr[] = {0, 1, 2, 3, 4, 5, 6, 7}, A = 1, B = 5

query = {3, 5}

Output:3

Explanation:

All the elements 3, 4 and 5 lies within

the range 1 to 5 in the subarray {3, 4, 5}.

Therefore, the count of such elements is 3.

**Prerequisites:** MO’s algorithm, SQRT Decomposition

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

- Group the queries into mutiple chunks where each chunk contains the values of starting range in (0 to √N – 1), (√N to 2x√N – 1) and so on. Sort the queries within a chunk in incresing order of
**R**. - Process all queries one by one in a way that every query uses result computed in the previous query.
- Maintain the frequency array that will count the frequency of arr[i] as they appear in the range [L, R].

For example:arr[] = [3, 4, 6, 2, 7, 1], L = 0, R = 4 and A = 1, B = 6

Initially frequency array is initialized to 0 i.e freq[]=[0….0]

Step 1:Add arr[0] and increment its frequency as freq[arr[0]]++ i.e freq[3]++

and freq[]=[0, 0, 0, 1, 0, 0, 0, 0]

Step 2:Add arr[1] and increment freq[arr[1]]++ i.e freq[4]++

and freq[]=[0, 0, 0, 1, 1, 0, 0, 0]

Step 3:Add arr[2] and increment freq[arr[2]]++ i.e freq[6]++

and freq[]=[0, 0, 0, 1, 1, 0, 1, 0]

Step 4:Add arr[3] and increment freq[arr[3]]++ i.e freq[2]++

and freq[]=[0, 0, 1, 1, 1, 0, 1, 0]

Step 5:Add arr[4] and increment freq[arr[4]]++ i.e freq[7]++

and freq[]=[0, 0, 1, 1, 1, 0, 1, 1]

Step 6:Now we need to find the numbers of elements between A and B.

Step 7:The answer is equal toTo calculate the sum in

step 7,we cannot do iteration because that would lead to O(N) time complexity per query so we will usesqrt decomposition techniqueto find the sum whosetime complexity is O(√N) per query.

Below is the implementation of the above approach:

## C++

`// C++ implementation to find the ` `// values in the range A to B ` `// in a subarray of L to R ` ` ` `#include <bits/stdc++.h> ` ` ` `using` `namespace` `std; ` ` ` `#define MAX 100001 ` `#define SQRSIZE 400 ` ` ` `// Variable to represent block size. ` `// This is made global so compare() ` `// of sort can use it. ` `int` `query_blk_sz; ` ` ` `// Structure to represent a ` `// query range ` `struct` `Query { ` ` ` `int` `L; ` ` ` `int` `R; ` `}; ` ` ` `// Frequency array ` `// to keep count of elements ` `int` `frequency[MAX]; ` ` ` `// Array which contains the frequency ` `// of a particular block ` `int` `blocks[SQRSIZE]; ` ` ` `// Block size ` `int` `blk_sz; ` ` ` `// 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) ` `{ ` ` ` `if` `(x.L / query_blk_sz != y.L / query_blk_sz) ` ` ` `return` `x.L / query_blk_sz < y.L / query_blk_sz; ` ` ` ` ` `return` `x.R < y.R; ` `} ` ` ` `// Function used to get the block ` `// number of current a[i] i.e ind ` `int` `getblocknumber(` `int` `ind) ` `{ ` ` ` `return` `(ind) / blk_sz; ` `} ` ` ` `// Function to get the answer ` `// of range [0, k] which uses the ` `// sqrt decompostion technique ` `int` `getans(` `int` `A, ` `int` `B) ` `{ ` ` ` `int` `ans = 0; ` ` ` `int` `left_blk, right_blk; ` ` ` `left_blk = getblocknumber(A); ` ` ` `right_blk = getblocknumber(B); ` ` ` ` ` `// If left block is equal to ` ` ` `// rigth block then we can traverse ` ` ` `// that block ` ` ` `if` `(left_blk == right_blk) { ` ` ` `for` `(` `int` `i = A; i <= B; i++) ` ` ` `ans += frequency[i]; ` ` ` `} ` ` ` `else` `{ ` ` ` `// Traversing first block in ` ` ` `// range ` ` ` `for` `(` `int` `i = A; i < (left_blk + 1) * blk_sz; i++) ` ` ` `ans += frequency[i]; ` ` ` ` ` `// Traversing completely overlapped ` ` ` `// blocks in range ` ` ` `for` `(` `int` `i = left_blk + 1; ` ` ` `i < right_blk; i++) ` ` ` `ans += blocks[i]; ` ` ` ` ` `// Traversing last block in range ` ` ` `for` `(` `int` `i = right_blk * blk_sz; ` ` ` `i <= B; i++) ` ` ` `ans += frequency[i]; ` ` ` `} ` ` ` `return` `ans; ` `} ` ` ` `void` `add(` `int` `ind, ` `int` `a[]) ` `{ ` ` ` `// Increment the frequency of a[ind] ` ` ` `// in the frequency array ` ` ` `frequency[a[ind]]++; ` ` ` ` ` `// Get the block number of a[ind] ` ` ` `// to update the result in blocks ` ` ` `int` `block_num = getblocknumber(a[ind]); ` ` ` ` ` `blocks[block_num]++; ` `} ` `void` `remove` `(` `int` `ind, ` `int` `a[]) ` `{ ` ` ` `// Decrement the frequency of ` ` ` `// a[ind] in the frequency array ` ` ` `frequency[a[ind]]--; ` ` ` ` ` `// Get the block number of a[ind] ` ` ` `// to update the result in blocks ` ` ` `int` `block_num = getblocknumber(a[ind]); ` ` ` ` ` `blocks[block_num]--; ` `} ` `void` `queryResults(` `int` `a[], ` `int` `n, ` ` ` `Query q[], ` `int` `m, ` `int` `A, ` `int` `B) ` `{ ` ` ` ` ` `// Initialize the block size ` ` ` `// for queries ` ` ` `query_blk_sz = ` `sqrt` `(m); ` ` ` ` ` `// 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 the ` ` ` `// 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++; ` ` ` `} ` ` ` `printf` `(` `"%d\n"` `, getans(A, B)); ` ` ` `} ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` ` ` `int` `arr[] = { 2, 0, 3, 1, 4, 2, 5, 11 }; ` ` ` `int` `N = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` ` ` `int` `A = 1, B = 5; ` ` ` `blk_sz = ` `sqrt` `(N); ` ` ` `Query Q[] = { { 0, 2 }, { 0, 3 }, { 5, 7 } }; ` ` ` ` ` `int` `M = ` `sizeof` `(Q) / ` `sizeof` `(Q[0]); ` ` ` ` ` `// Answer the queries ` ` ` `queryResults(arr, N, Q, M, A, B); ` ` ` `return` `0; ` `} ` |

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**Output:**

2 3 2