# Number of elements less than or equal to a number in a subarray : MO’s Algorithm

Given an array **arr** of size **N** and **Q **queries of the form L, R and X, the task is to print the number of elements less than or equal to X in the subarray represented by L to R.

**Prerequisites:** MO’s Algorithm, Sqrt Decomposition

**Examples:**

Input:arr[] = {2, 3, 4, 5} Q = {{0, 3, 5}, {0, 2, 2}}Output:4 1Explanation:Number of elements less than or equal to 5 in arr[0..3] is 4 (all elements) Number of elements less than or equal to 2 in arr[0..2] is 1 (only 2)

**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.

Let **arr[0…n-1]** be input array and **Q[0..m-1]** be an array of queries.

- Sort all queries in a way that queries with L values from
**0 to √n – 1**are put together, then all queries from**√n to 2×√n – 1**, and so on. All queries within a block are sorted in increasing order of R values. - Process all queries one by one in a way that every query uses result computed in the previous query.
- We will 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 X=5

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 less than or equal to X(here X=5).Step 7– The answer is equal to

To calculate thesum 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.

## C++

`// C++ program to answer queries to ` `// count number of elements smaller ` `// than or equal to x.` `#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;` ` ` `int` `x;` `};` `// 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 decomposition technique` `int` `getans(` `int` `k)` `{` ` ` `int` `ans = 0;` ` ` `int` `left_blk, right_blk;` ` ` `left_blk = 0;` ` ` `right_blk = getblocknumber(k);` ` ` `// If left block is equal to` ` ` `// right block then we can traverse` ` ` `// that block` ` ` `if` `(left_blk == right_blk) {` ` ` `for` `(` `int` `i = 0; i <= k; i++)` ` ` `ans += frequency[i];` ` ` `}` ` ` `else` `{` ` ` `// Traversing first block in ` ` ` `// range` ` ` `for` `(` `int` `i = 0; 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 <= k; 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)` `{` ` ` `// 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,` ` ` `x = q[i].x;` ` ` `// 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` `(` `"query[%d, %d, %d] : %d\n"` `, ` ` ` `L, R, x, getans(x));` ` ` `}` `}` `// Driver code` `int` `main()` `{` ` ` `int` `arr[] = { 2, 0, 3, 1, 4, 2, 5, 11 };` ` ` `int` `N = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `blk_sz = ` `sqrt` `(N);` ` ` `Query Q[] = { { 0, 2, 2 }, { 0, 3, 5 },` ` ` `{ 5, 7, 10 } };` ` ` `int` `M = ` `sizeof` `(Q) / ` `sizeof` `(Q[0]);` ` ` ` ` `// Answer the queries` ` ` `queryResults(arr, N, Q, M);` ` ` `return` `0;` `}` |

**Output:**

query[0, 2, 2] : 2 query[0, 3, 5] : 4 query[5, 7, 10] : 2

**Time Complexity:** **O(Q × √N)**.

It takes O(Q × √N) time for MO’s algorithm and O(Q × √N) time for sqrt decomposition technique to answer the sum of freq[0]+….freq[k], therefore total time complexity is O(Q × √N + Q × √N) which is O(Q × √N).