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: 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
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 |
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 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
4
Time Complexity: O(Q*?N)
Space Complexity: O(N), since N extra space has been taken.