Range Queries to count elements lying in a given Range : MO’s Algorithm

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 multiple 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 increasing 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 to 
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.

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 decomposition 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
    // right 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;
}

Java

// Java implementation to find the
// values in the range A to B
// in a subarray of L to R
import java.util.*;
import java.lang.Math;

public class GFG {

  public static int MAX=100001;
  public static int SQRSIZE=400;

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

  // Frequency array
  // to keep count of elements
  public static int[] frequency = new int[MAX];

  // Array which contains the frequency
  // of a particular block
  public static int[] blocks = new int[SQRSIZE];

  // Block size
  public static 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.

  static Comparator<int[]> arrayComparator = new Comparator<int[]>() {
    @Override
    public int compare(int[] x, int[] y) {
      if (x[0] / query_blk_sz != y[0] / query_blk_sz)
        return Integer.compare(x[0] / query_blk_sz, y[0] / query_blk_sz);

      return Integer.compare(x[1],y[1]);
    }
  };

  // Function used to get the block
  // number of current a[i] i.e ind
  public static int getblocknumber(int ind)
  {
    return (ind) / blk_sz;
  }

  // Function to get the answer
  // of range [0, k] which uses the
  // sqrt decomposition technique
  public static 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
    // right 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;
  }

  public 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]++;
  }
  public 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]--;
  }
  public static void queryResults(int a[], int n,
                                  int[][] q, int m, int A, int B)
  {

    // Initialize the block size
    // for queries
    query_blk_sz = (int)Math.sqrt(m);

    // Sort all queries so that queries
    // of same blocks are arranged
    // together.
    Arrays.sort(q,arrayComparator);

    // 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][0], R = q[i][1];

      // 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++;
      }
      System.out.println(getans(A, B));
    }
  }

  // Driver code
  public static void main(String[] args) {
    int arr[] = { 2, 0, 3, 1, 4, 2, 5, 11 };
    int N = arr.length;

    int A = 1, B = 5;
    blk_sz = (int) Math.sqrt(N);
    int[][] Q = { { 0, 2 }, { 0, 3 }, { 5, 7 } };

    int M = Q.length;

    // Answer the queries
    queryResults(arr, N, Q, M, A, B);
  }
}

// This code is contributed
// by Shubham Singh

Python3

# Python implementation to find the
# values in the range A to B
# in a subarray of L to R
import math
from functools import cmp_to_key
MAX = 100001
SQRSIZE = 400

# Variable to represent block size.
# This is made global so compare()
# of sort can use it.
query_blk_sz = 1

# Frequency array
# to keep count of elements
frequency = [0] * MAX

# Array which contains the frequency
# of a particular block
blocks = [0]*SQRSIZE

# Block size
blk_sz = 0

# 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.


def compare(x, y):
    if ((x[0] // query_blk_sz) != (y[0] // query_blk_sz)):
        return x[0] // query_blk_sz < y[0] // query_blk_sz

    return x[1] < y[1]


# 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(A, B):
    ans = 0

    left_blk = getblocknumber(A)
    right_blk = getblocknumber(B)

    # If left block is equal to
    # right block then we can traverse
    # that block
    if (left_blk == right_blk):
        for i in range(A, B+1):
            ans += frequency[i]

    else:
        # Traversing first block in
        # range
        i = A
        while(i < (left_blk+1)*blk_sz):
            ans += frequency[i]
            i += 1

        # Traversing completely overlapped
        # blocks in range
        i = (int)(left_blk + 1)
        while(i < right_blk):
            ans += blocks[i]
            i += 1

        # Traversing last block in range
        i = (int)(right_blk * blk_sz)
        while(i <= B):
            ans += frequency[i]
            i += 1

    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[(int)(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[(int)(block_num)] -= 1


def queryResults(a, n, q, m, A, B):

    # Initialize the block size
    # for queries
    query_blk_sz = (int)(math.sqrt(m))

    # Sort all queries so that queries
    # of same blocks are arranged
    # together.
    q.sort(key=cmp_to_key(compare))

    # Initialize current L,
    # current R and current result
    currL = 0
    currR = 0

    for i in range(0, m):

        # L and R values of the
        # current range

        L = q[i][0]
        R = q[i][1]

        # Add Elements of current
        # range
        while (currR <= R):
            add(currR, a)
            currR += 1

        while (currL > L):
            add(currL - 1, a)
            currL -= 1

        # Remove element of previous
        # range
        while (currR > R + 1):
            remove(currR - 1, a)
            currR -= 1

        while (currL < L):
            remove(currL, a)
            currL += 1

        print(getans(A, B))


# Driver code
arr = [2, 0, 3, 1, 4, 2, 5, 11]
N = len(arr)

A = 1
B = 5
blk_sz = (int)(math.sqrt(N))

Q = [[0, 2], [0, 3], [5, 7]]

M = len(Q)

# Answer the queries
queryResults(arr, N, Q, M, A, B)

# This code is contributed by rj113to.

Output:

2
3
2