Number of elements less than or equal to a number in a subarray : MO's Algorithm
Last Updated :
12 Jul, 2025
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
1
Explanation:
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 \sum_{i=0}^{X} freq[i]
To calculate the sum in step 7, we cannot do iteration because that would lead to O(N) time complexity per query so we will use sqrt decomposition technique to find the sum whose time 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;
}
import java.util.*;
public class GFG{
static int MAX = 100001;
static int SQRSIZE = 400;
// Variable to represent block size.
// This is made global so compare()
// of sort can use it.
static int query_blk_sz;
// Frequency array
// to keep count of elements
static int[] frequency = new int[MAX];
// Array which contains the frequency
// of a particular block
static int[] blocks = new int[SQRSIZE];
static int blk_sz;
// class to represent a
// query range
static class Query {
int L;
int R;
int x;
Query(int a,int b,int c)
{
L = a;
R = b;
x = c;
}
}
// Function used to get the block
// number of current a[i] i.e ind
static int getblocknumber(int ind) {
return ind / blk_sz;
}
// Function to get the answer
// of range [0, k] which uses the
// sqrt decomposition technique
static 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;
}
static 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]++;
}
static 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]--;
}
static void queryResults(int[] a, int n, Query[] q, int m) {
query_blk_sz = (int) Math.sqrt(m);
Arrays.sort(q, new Comparator<Query>() {
@Override
// 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.
public int 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;
}
});
// 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;
while (currR <= R) {
add(currR, a);
currR++;
}
while (currL > L) {
add(currL - 1, a);
currL--;
}
while (currR > R + 1) {
remove(currR - 1, a);
currR--;
}
while (currL < L) {
remove(currL, a);
currL++;
}
System.out.println("query[" + L + ", " + R + ", " + x + "] : " + getans(x));
}
}
public static void main(String[] args) {
int arr[] = { 2, 0, 3, 1, 4, 2, 5, 11 };
int n = arr.length;
blk_sz = (int)Math.sqrt(n);
Query q[] ={new Query(0,2,2),new Query(0,3,5),new Query(5,7,10)};
int m = q.length;
queryResults(arr,n,q,m);
}
}
// This code is contributed by anskalyan3
# Python equivalent
MAX = 100001
SQRSIZE = 400
# Variable to represent block size.
# This is made global so compare()
# of sort can use it.
query_blk_sz = 0
# Frequency array
# to keep count of elements
frequency = [0] * MAX
# Array which contains the frequency
# of a particular block
blocks = [0] * SQRSIZE
blk_sz = 0
# class to represent a
# query range
class Query:
def __init__(self, a, b, c):
self.L = a
self.R = b
self.x = c
# Function used to get the block
# number of current a[i] i.e ind
def getblocknumber(ind):
return ind // blk_sz
# Function to get the answer
# of range [0, k] which uses the
# sqrt decomposition technique
def getans(k):
ans = 0
left_blk, right_blk = 0, getblocknumber(k)
# If left block is equal to
# right block then we can traverse
# that block
if left_blk == right_blk:
for i in range(0, k+1):
ans += frequency[i]
else:
# Traversing first block in
# range
for i in range(0, (left_blk + 1) * blk_sz):
ans += frequency[i]
# Traversing completely overlapped
# blocks in range
for i in range(left_blk + 1, right_blk):
ans += blocks[i]
# Traversing last block in range
for i in range(right_blk * blk_sz, k+1):
ans += frequency[i]
return ans
def add(ind, a):
# Increment the frequency of a[ind]
# in the frequency array
frequency[a[ind]] += 1
# Get the block number of a[ind]
# to update the result in blocks
block_num = getblocknumber(a[ind])
blocks[block_num] += 1
def remove(ind, a):
# Decrement the frequency of
# a[ind] in the frequency array
frequency[a[ind]] -= 1
# Get the block number of a[ind]
# to update the result in blocks
block_num = getblocknumber(a[ind])
blocks[block_num] -= 1
def queryResults(a, n, q, m):
query_blk_sz = int(m ** 0.5)
q.sort(key = lambda x: (x.L // query_blk_sz, x.R))
# Initialize current L,
# current R and current result
currL = 0
currR = 0
for i in range(m):
# L and R values of the
# current range
L = q[i].L
R = q[i].R
x = q[i].x
while currR <= R:
add(currR, a)
currR += 1
while currL > L:
add(currL - 1, a)
currL -= 1
while currR > R + 1:
remove(currR - 1, a)
currR -= 1
while currL < L:
remove(currL, a)
currL += 1
print("query[{}, {}, {}] : {}".format(L, R, x, getans(x)))
def main():
global blk_sz
arr = [2, 0, 3, 1, 4, 2, 5, 11]
n = len(arr)
blk_sz = int(n ** 0.5)
q = [Query(0,2,2),Query(0,3,5),Query(5,7,10)]
m = len(q)
queryResults(arr,n,q,m)
if __name__ == "__main__":
main()
using System;
using System.Collections.Generic;
class GFG {
static int MAX = 100001;
static int SQRSIZE = 400;
// Variable to represent block size.
// This is made global so compare()
// of sort can use it.
static int query_blk_sz;
// Frequency array
// to keep count of elements
static int[] frequency = new int[MAX];
// Array which contains the frequency
// of a particular block
static int[] blocks = new int[SQRSIZE];
static int blk_sz;
// class to represent a
// query range
public class Query {
public int L;
public int R;
public int x;
public Query(int a, int b, int c)
{
L = a;
R = b;
x = c;
}
}
// Function used to get the block
// number of current a[i] i.e ind
static int getblocknumber(int ind)
{
return ind / blk_sz;
}
// Function to get the answer
// of range [0, k] which uses the
// sqrt decomposition technique
static 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;
}
static 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]++;
}
static 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]--;
}
static void queryResults(int[] a, int n, Query[] q,
int m)
{
query_blk_sz = (int)Math.Sqrt(m);
Array.Sort(q, new Comparison<Query>((x, 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;
}));
// 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;
while (currR <= R) {
add(currR, a);
currR++;
}
while (currL > L) {
add(currL - 1, a);
currL--;
}
while (currR > R + 1) {
remove(currR - 1, a);
currR--;
}
while (currL < L) {
remove(currL, a);
currL++;
}
Console.WriteLine("query[" + L + ", " + R + ", "
+ x + "] : " + getans(x));
}
}
// Driver code
static void Main(string[] args)
{
int[] arr = { 2, 0, 3, 1, 4, 2, 5, 11 };
int n = arr.Length;
blk_sz = (int)Math.Sqrt(n);
Query[] q
= { new Query(0, 2, 2), new Query(0, 3, 5),
new Query(5, 7, 10) };
int m = q.Length;
// Function call
queryResults(arr, n, q, m);
}
}
// This code is contributed by phasing17
const MAX = 100001;
const SQRSIZE = 400;
let query_blk_sz = 0;
let frequency = new Array(MAX).fill(0);
let blocks = new Array(SQRSIZE).fill(0);
let blk_sz = 0;
class Query {
constructor(a, b, c) {
this.L = a;
this.R = b;
this.x = c;
}
}
function getblocknumber(ind) {
return Math.floor(ind / blk_sz);
}
function getans(k) {
let ans = 0;
let left_blk = 0;
let right_blk = getblocknumber(k);
if (left_blk == right_blk) {
for (let i = 0; i <= k; i++) {
ans += frequency[i];
}
} else {
for (let i = 0; i <= (left_blk + 1) * blk_sz - 1; i++) {
ans += frequency[i];
}
for (let i = left_blk + 1; i <= right_blk - 1; i++) {
ans += blocks[i];
}
for (let i = right_blk * blk_sz; i <= k; i++) {
ans += frequency[i];
}
}
return ans;
}
function add(ind, a) {
frequency[a[ind]]++;
let block_num = getblocknumber(a[ind]);
blocks[block_num]++;
}
function remove(ind, a) {
frequency[a[ind]]--;
let block_num = getblocknumber(a[ind]);
blocks[block_num]--;
}
function queryResults(a, n, q, m) {
query_blk_sz = Math.floor(Math.sqrt(m));
q.sort((a, b) => {
return a.L / query_blk_sz - b.L / query_blk_sz || a.R - b.R;
});
let currL = 0;
let currR = 0;
for (let i = 0; i < m; i++) {
let L = q[i].L;
let R = q[i].R;
let x = q[i].x;
while (currR <= R) {
add(currR, a);
currR++;
}
while (currL > L) {
add(currL - 1, a);
currL--;
}
while (currR > R + 1) {
remove(currR - 1, a);
currR--;
}
while (currL < L) {
remove(currL, a);
currL++;
}
console.log(`query[${L}, ${R}, ${x}] : ${getans(x)}`);
}
}
function main() {
let arr = [2, 0, 3, 1, 4, 2, 5, 11];
let n = arr.length;
blk_sz = Math.floor(Math.sqrt(n));
let q = [new Query(0, 2, 2), new Query(0, 3, 5), new Query(5, 7, 10)];
let m = q.length;
queryResults(arr, n, q, m);
}
main();
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).
Space Complexity: O(N + sqrt(N) + M), where N is the size of the input array, M is the number of queries, and sqrt(N) is the size of the blocks used in the sqrt decomposition technique.
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