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

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:
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:
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 MOâ€™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:

1. 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.
2. Process all queries one by one in a way that every query uses result computed in the previous query.
3. 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[]=[0â€¦.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 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 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 rangestruct Query {    int L;    int R;}; // Frequency array// to keep count of elementsint frequency[MAX]; // Array which contains the frequency// of a particular blockint blocks[SQRSIZE]; // Block sizeint 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 indint getblocknumber(int ind){    return (ind) / blk_sz;} // Function to get the answer// of range [0, k] which uses the// sqrt decomposition techniqueint 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 codeint 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 Rimport math MAX = 100001SQRSIZE = 400# Structure to represent a query rangeclass Query:    def __init__(self, l, r):        self.L = l        self.R = r  # Frequency array# to keep count of elementsfrequency = [0] * MAXblocks = [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 inddef getblocknumber(ind):    return ind // blk_sz # Function to get the answer# of range [0, k] which uses the# sqrt decomposition techniquedef 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 codeif __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

## C#

 using System; class GFG{    static readonly int MAX = 100001;    static readonly 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    class Query    {        public int L;        public 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 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    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.        Array.Sort(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++;            }            Console.WriteLine(getans(A, B));        }    }     // Driver code    public static void Main()    {        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);    }}

## Javascript

 // Declare constantsconst MAX = 100001;const SQRSIZE = 400; // Declare empty arrayslet frequency = new Array(MAX).fill(0);let blocks = new Array(SQRSIZE).fill(0); // Define Query classclass Query {  constructor(l, r) {    this.L = l;    this.R = r;  }} // Function to compare two Query objectsfunction 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 indexfunction getblocknumber(ind) {  return Math.floor(ind / blk_sz);} // Function to get the answerfunction 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 indexfunction add(ind, a) {  frequency[a[ind]]++;  let block_num = getblocknumber(a[ind]);  blocks[block_num]++;} // Function to remove the frequency of a given indexfunction remove(ind, a) {  frequency[a[ind]]--;  let block_num = getblocknumber(a[ind]);  blocks[block_num]--;} // Function to get the query resultsfunction 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 querieslet 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 resultsqueryResults(arr, N, Q, M, A, B);

Output
4

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

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