Queries for elements having values within the range A to B using MO’s Algorithm

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Prerequisites: MO’s algorithm, SQRT Decomposition
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[] = {3, 4, 6, 2, 7, 1}, A = 1, B = 6, query = {0, 4}
Output: 4
Explanation:
All 3, 4, 6, 2 lies within 1 to 6 in the subarray {3, 4, 6, 2}
Therefore, the count of such elements is 4.

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.

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 $\sum_{i=A}^B freq[i]$

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 square root 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[] = { 3, 4, 6, 2, 7, 1 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    int A = 1, B = 6;
    blk_sz = sqrt(N);
    Query Q[] = { { 0, 4 } };
 
    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.Arrays;
 
public class GFG {
 
    static final int MAX = 100001;
    static final int SQRSIZE = 400;
 
    // Variable to represent block size.
    // This is made global so compare()
    // of sort can use it.
    static int query_blk_sz;
 
    // Structure to represent a query range
    static class Query {
        int L;
        int R;
 
        public Query(int l, int r)
        {
            L = l;
            R = r;
        }
    }
 
    // 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];
 
    // Block size
    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 boolean 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
    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 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;
    }
 
    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, 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.parallelSort(q, (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;
 
            // 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[] = { 3, 4, 6, 2, 7, 1 };
        int N = arr.length;
 
        int A = 1, B = 6;
        blk_sz = (int)Math.sqrt(N);
        Query Q[] = { new Query(0, 4) };
 
        int M = Q.length;
 
        // Answer the queries
        queryResults(arr, N, Q, M, A, B);
    }
}
 
// This code is contributed by Lovely Jain

Python3

# Python implementation to find the
# values in the range A to B
# in a subarray of L to R
import math
 
MAX = 100001
SQRSIZE = 400
# Structure to represent a query range
class Query:
    def __init__(self, l, r):
        self.L = l
        self.R = r
 
 
# Frequency array
# to keep count of elements
frequency = [0] * MAX
blocks = [0] * SQRSIZE
 
 
 
# 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.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
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:
        for i in range(A, (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, B+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, A, B):
    # Initialize the block size
    # for queries
    global blk_sz, query_blk_sz
    query_blk_sz = int(math.sqrt(m))
    # Sort all queries so that queries
    # of same blocks are arranged
    # together
    q.sort(key=lambda x: (x.L // query_blk_sz, x.R))
    # Initialize current L,
    # current R and current result
    currL, currR = 0, 0
    for i in range(m):
        # L and R values of the
        # current range
        L, R = q[i].L, q[i].R
        # 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
if __name__ == '__main__':
    arr = [3, 4, 6, 2, 7, 1]
    N = len(arr)
    A, B = 1, 6
    blk_sz = int(math.sqrt(N))
    Q = [Query(0, 4)]
    M = len(Q)
    queryResults(arr, N, Q, M, A, B)
     
     
# This code is contributed by Shivhack999 

Javascript

// Declare constants
const MAX = 100001;
const SQRSIZE = 400;
 
// Declare empty arrays
let frequency = new Array(MAX).fill(0);
let blocks = new Array(SQRSIZE).fill(0);
 
// Define Query class
class Query {
  constructor(l, r) {
    this.L = l;
    this.R = r;
  }
}
 
// Function to compare two Query objects
function compare(x, y) {
  // Compare the left values of the queries
  if (Math.floor(x.L / query_blk_sz) !== Math.floor(y.L / query_blk_sz)) {
    return Math.floor(x.L / query_blk_sz) < Math.floor(y.L / query_blk_sz);
  }
  // Compare the right values of the queries
  return x.R < y.R;
}
 
// Function to get the block number of a given index
function getblocknumber(ind) {
  return Math.floor(ind / blk_sz);
}
 
// Function to get the answer
function getans(A, B) {
  let ans = 0;
  let left_blk = getblocknumber(A);
  let right_blk = getblocknumber(B);
 
  // If the left and right blocks are the same, 
  // add the frequencies between A and B
  if (left_blk === right_blk) {
    for (let i = A; i <= B; i++) {
      ans += frequency[i];
    }
  // If the left and right blocks are different, 
  // add the frequencies between A and the end of the left block, 
  // add all the frequencies of blocks between the left and right block, 
  // and add the frequencies between the start of the right block and B
  } else {
    for (let i = A; i <= (left_blk + 1) * blk_sz - 1; i++) {
      ans += frequency[i];
    }
    for (let i = left_blk + 1; i < right_blk; i++) {
      ans += blocks[i];
    }
    for (let i = right_blk * blk_sz; i <= B; i++) {
      ans += frequency[i];
    }
  }
  return ans;
}
 
// Function to add the frequency of a given index
function add(ind, a) {
  frequency[a[ind]]++;
  let block_num = getblocknumber(a[ind]);
  blocks[block_num]++;
}
 
// Function to remove the frequency of a given index
function remove(ind, a) {
  frequency[a[ind]]--;
  let block_num = getblocknumber(a[ind]);
  blocks[block_num]--;
}
 
// Function to get the query results
function queryResults(a, n, q, m, A, B) {
  // Define block sizes
  let blk_sz, query_blk_sz;
  query_blk_sz = Math.floor(Math.sqrt(m));
  // Sort the queries
  q.sort(compare);
  // Initialize left and right indices
  let currL = 0,
    currR = 0;
  // Iterate through queries
  for (let i = 0; i < m; i++) {
    // Get the left and right indices of the query
    let L = q[i].L,
      R = q[i].R;
    // Add the frequency of all indices between the current 
    // right index and the right index of the query
    while (currR <= R) {
      add(currR, a);
      currR++;
    }
    // Add the frequency of all indices between the current left index 
    // and the left index of the query
    while (currL > L) {
      add(currL - 1, a);
      currL--;
    }
    // Remove the frequency of all indices between the current 
    // right index and the right index of the query
    while (currR > R + 1) {
      remove(currR - 1, a);
      currR--;
    }
    // Remove the frequency of all indices between the current left 
    // index and the left index of the query
    while (currL < L) {
      remove(currL, a);
      currL++;
    }
    // Log the answer
    console.log(getans(A, B));
  }
}
 
// Declare an array, the left and right indices, and an array of queries
let arr = [3, 4, 6, 2, 7, 1];
let N = arr.length;
let A = 1,
  B = 6;
let blk_sz = Math.floor(Math.sqrt(N));
let Q = [new Query(0, 4)];
let M = Q.length;
 
// Log the query results
queryResults(arr, N, Q, M, A, B);

Output:

Time Complexity: O(Q*?N)
Space Complexity: O(N), since N extra space has been taken.

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