Array range queries for elements with frequency same as value

Last Updated : 27 Apr, 2023

Given an array of N numbers, the task is to answer Q queries of the following type:

query(start, end) = Number of times a 
number x occurs exactly x times in a 
subarray from start to end

Examples:

Input : arr = {1, 2, 2, 3, 3, 3}
Query 1: start = 0, end = 1,
Query 2: start = 1, end = 1,
Query 3: start = 0, end = 2,
Query 4: start = 1, end = 3,
Query 5: start = 3, end = 5,
Query 6: start = 0, end = 5
Output : 1 0 2 1 1 3

Explanation:
In Query 1, Element 1 occurs once in subarray [1, 2];
In Query 2, No Element satisfies the required condition is subarray [2];
In Query 3, Element 1 occurs once and 2 occurs twice in subarray [1, 2, 2];
In Query 4, Element 2 occurs twice in subarray [2, 2, 3];
In Query 5, Element 3 occurs thrice in subarray [3, 3, 3];
In Query 6, Element 1 occurs once, 2 occurs twice and 3 occurs thrice in subarray [1, 2, 2, 3, 3, 3]

Method 1 (Brute Force): Calculate frequency of every element in the subarray under each query. If any number x has frequency x in the subarray covered under each query, we increment the counter.

Implementation:

C++

/* C++ Program to answer Q queries to 
   find number of times an element x 
   appears x times in a Query subarray */
#include <bits/stdc++.h>
using namespace std;
 
/* Returns the count of number x with
   frequency x in the subarray from 
   start to end */
int solveQuery(int start, int end, int arr[])
{
    // map for frequency of elements
    unordered_map<int, int> frequency;
 
    // store frequency of each element 
    // in arr[start; end]
    for (int i = start; i <= end; i++) 
        frequency[arr[i]]++;    
 
    // Count elements with same frequency
    // as value
    int count = 0;
    for (auto x : frequency) 
        if (x.first == x.second) 
            count++;    
    return count;
}
 
int main()
{
    int A[] = { 1, 2, 2, 3, 3, 3 };
    int n = sizeof(A) / sizeof(A[0]);
 
    // 2D array of queries with 2 columns
    int queries[][3] = { { 0, 1 },
                         { 1, 1 },
                         { 0, 2 },
                         { 1, 3 },
                         { 3, 5 },
                         { 0, 5 } };
 
    // calculating number of queries
    int q = sizeof(queries) / sizeof(queries[0]);
 
    for (int i = 0; i < q; i++) {
        int start = queries[i][0];
        int end = queries[i][1];
        cout << "Answer for Query " << (i + 1)
             << " = " << solveQuery(start,
             end, A) << endl;
    }
 
    return 0;
}

Java

/* Java Program to answer Q queries to 
find number of times an element x 
appears x times in a Query subarray */
import java.util.HashMap;
import java.util.Map;
 
class GFG 
{
 
 
/* Returns the count of number x with
frequency x in the subarray from 
start to end */
static int solveQuery(int start, int end, int arr[])
{
    // map for frequency of elements
    Map<Integer,Integer> mp = new HashMap<>();
 
    // store frequency of each element 
    // in arr[start; end]
    for (int i = start; i <= end; i++) 
        mp.put(arr[i],mp.get(arr[i]) == null?1:mp.get(arr[i])+1); 
 
    // Count elements with same frequency
    // as value
    int count = 0;
    for (Map.Entry<Integer,Integer> entry : mp.entrySet()) 
        if (entry.getKey() == entry.getValue()) 
            count++; 
    return count;
}
 
// Driver code
public static void main(String[] args) 
{
    int A[] = { 1, 2, 2, 3, 3, 3 };
    int n = A.length;
 
    // 2D array of queries with 2 columns
    int [][]queries = { { 0, 1 },
                        { 1, 1 },
                        { 0, 2 },
                        { 1, 3 },
                        { 3, 5 },
                        { 0, 5 } };
 
    // calculating number of queries
    int q = queries.length;
 
    for (int i = 0; i < q; i++) 
    {
        int start = queries[i][0];
        int end = queries[i][1];
        System.out.println("Answer for Query " + (i + 1)
            + " = " + solveQuery(start,
            end, A));
    }
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python 3 Program to answer Q queries 
# to find number of times an element x 
# appears x times in a Query subarray
import math as mt
 
# Returns the count of number x with
# frequency x in the subarray from 
# start to end
def solveQuery(start, end, arr):
 
    # map for frequency of elements
    frequency = dict()
 
    # store frequency of each element 
    # in arr[start end]
    for i in range(start, end + 1):
 
 
        if arr[i] in frequency.keys():
            frequency[arr[i]] += 1
        else:
            frequency[arr[i]] = 1
                 
    # Count elements with same 
    # frequency as value
    count = 0
    for x in frequency: 
        if x == frequency[x]: 
            count += 1
    return count
 
# Driver code 
A = [1, 2, 2, 3, 3, 3 ]
n = len(A)
 
# 2D array of queries with 2 columns
queries = [[ 0, 1 ], [ 1, 1 ],
           [ 0, 2 ], [ 1, 3 ],
           [ 3, 5 ], [ 0, 5 ]]
 
# calculating number of queries
q = len(queries)
 
for i in range(q):
    start = queries[i][0]
    end = queries[i][1]
    print("Answer for Query ", (i + 1), 
          " = ", solveQuery(start,end, A))
           
# This code is contributed 
# by Mohit kumar 29

C#

// C# Program to answer Q queries to 
// find number of times an element x 
// appears x times in a Query subarray
using System;
using System.Collections.Generic;
 
class GFG
{
    /* Returns the count of number x with 
    frequency x in the subarray from 
    start to end */
    public static int solveQuery(int start, 
                                 int end, 
                                 int[] arr)
    {
 
        // map for frequency of elements 
        Dictionary<int, 
                   int> mp = new Dictionary<int, 
                                            int>();
 
        // store frequency of each element 
        // in arr[start; end] 
        for (int i = start; i <= end; i++)
        {
            if (mp.ContainsKey(arr[i]))
                mp[arr[i]]++;
            else
                mp.Add(arr[i], 1);
        }
 
        // Count elements with same frequency 
        // as value 
        int count = 0;
        foreach (KeyValuePair<int, 
                              int> entry in mp)
        {
            if (entry.Key == entry.Value)
                count++;
        }
        return count;
    }
 
    // Driver code 
    public static void Main(String[] args)
    {
        int[] A = { 1, 2, 2, 3, 3, 3 };
        int n = A.Length;
 
        // 2D array of queries with 2 columns 
        int[,] queries = {{ 0, 1 }, { 1, 1 },
                          { 0, 2 }, { 1, 3 },
                          { 3, 5 }, { 0, 5 }};
 
        // calculating number of queries 
        int q = queries.Length;
 
        for (int i = 0; i < q; i++)
        {
            int start = queries[i, 0];
            int end = queries[i, 1];
            Console.WriteLine("Answer for Query " + (i + 1) +
                              " = " + solveQuery(start, end, A));
        }
    }
}
 
// This code is contributed by
// sanjeev2552

Javascript

<script>
/* Javascript Program to answer Q queries to 
find number of times an element x 
appears x times in a Query subarray */
     
    /* Returns the count of number x with
frequency x in the subarray from 
start to end */
    function solveQuery(start,end,arr)
    {
        // map for frequency of elements
    let mp = new Map();
   
    // store frequency of each element 
    // in arr[start; end]
    for (let i = start; i <= end; i++) 
        mp.set(arr[i],mp.get(arr[i]) == null?1:mp.get(arr[i])+1); 
   
    // Count elements with same frequency
    // as value
    let count = 0;
    for (let [key, value] of mp.entries()) 
        if (key == value) 
            count++; 
    return count;
    }
     
    // Driver code
    let A=[1, 2, 2, 3, 3, 3];
    let n = A.length;
    // 2D array of queries with 2 columns
    let queries = [[ 0, 1 ], [ 1, 1 ],
           [ 0, 2 ], [ 1, 3 ],
           [ 3, 5 ], [ 0, 5 ]];
     
    // calculating number of queries
    let q = queries.length;
    for (let i = 0; i < q; i++) 
    {
        let start = queries[i][0];
        let end = queries[i][1];
        document.write("Answer for Query " + (i + 1)
            + " = " + solveQuery(start,
            end, A)+"<br>");
    }
 
// This code is contributed by unknown2108
</script>

Output

Answer for Query 1 = 1
Answer for Query 2 = 0
Answer for Query 3 = 2
Answer for Query 4 = 1
Answer for Query 5 = 1
Answer for Query 6 = 3

Time Complexity: O(Q * N)

Auxiliary Space: O(N)

Method 2 (Efficient):

We can solve this problem using the MO’s Algorithm.

We assign starting index, ending index and query number to each query, Each query takes the following form-

Starting Index(L): Starting Index of the subarray covered under the query;
Ending Index(R) : Ending Index of the subarray covered under the query;
Query Number(Index) : Since queries are sorted, this tells us original position of the query so that we answer the queries in the original order

Firstly, we divide the queries into blocks and sort the queries using a custom comparator.

Now we process the queries offline where we keep two pointers i.e. MO_RIGHT and MO_LEFT with each incoming query, we move these pointers forward and backward and insert and delete elements according to the starting and ending indices of the current query.
Let the current running answer be current_ans.

Whenever we insert an element we increment the frequency of the included element, if this frequency is equal to the element we just included, we increment the current_ans.If the frequency of this element becomes (current element + 1) this means that earlier this element was counted in the current_ans when it was equal to its frequency, thus we need to decrement current_ans in this case.

Whenever we delete/remove an element we decrement the frequency of the excluded element, if this frequency is equal to the element we just excluded, we increment the current_ans.If the frequency of this element becomes (current element – 1) this means that earlier this element was counted in the current_ans when it was equal to its frequency, thus we need to decrement current_ans in this case.

Implementation:

C++

/* C++ Program to answer Q queries to
   find number of times an element x 
   appears x times in a Query subarray */
#include <bits/stdc++.h>
using namespace std;
 
// Variable to represent block size. 
// This is made global so compare() 
// of sort can use it.
int block;
 
// Structure to represent a query range
struct Query {
    int L, R, index;
};
 
/* Function used to sort all queries 
   so that all queries of same block
   are arranged together and within 
   a block, queries are sorted in 
   increasing order of R values. */
bool compare(Query x, Query y)
{
    // Different blocks, sort by block.
    if (x.L / block != y.L / block)
        return x.L / block < y.L / block;
 
    // Same block, sort by R value
    return x.R < y.R;
}
 
/* Inserts element (x) into current range
   and updates current answer */
void add(int x, int& currentAns, 
         unordered_map<int, int>& freq)
{
 
    // increment frequency of this element
    freq[x]++;
 
    // if this element was previously 
    // contributing to the currentAns,
    // decrement currentAns
    if (freq[x] == (x + 1))
        currentAns--;
 
    // if this element has frequency 
    // equal to its value, increment
    // currentAns
    else if (freq[x] == x)
        currentAns++;
}
 
/* Removes element (x) from current 
   range btw L and R and updates 
   current Answer */
void remove(int x, int& currentAns, 
            unordered_map<int, int>& freq)
{
 
    // decrement frequency of this element
    freq[x]--;
 
    // if this element has frequency equal 
    // to its value, increment currentAns
    if (freq[x] == x)
        currentAns++;
 
    // if this element was previously 
    // contributing to the currentAns 
    // decrement currentAns
    else if (freq[x] == (x - 1)) 
        currentAns--;
}
 
/* Utility Function to answer all queries
   and build the ans array in the original 
   order of queries */
void queryResultsUtil(int a[], Query q[], 
                        int ans[], int m)
{
 
    // map to store freq of each element
    unordered_map<int, int> freq;
 
    // Initialize current L, current R
    // and current sum
    int currL = 0, currR = 0;
    int currentAns = 0;
 
    // Traverse through all queries
    for (int i = 0; i < m; i++) {
        // L and R values of current range
        int L = q[i].L, R = q[i].R; 
        int index = q[i].index;
 
        // Remove extra elements of previous
        // range. For example if previous 
        // range is [0, 3] and current range 
        // is [2, 5], then a[0] and a[1] are 
        // removed
        while (currL < L) {
            remove(a[currL], currentAns, freq);
            currL++;
        }
 
        // Add Elements of current Range
        while (currL > L) {
            currL--;
            add(a[currL], currentAns, freq);
        }
        while (currR <= R) {
            add(a[currR], currentAns, freq);
            currR++;
        }
 
        // Remove elements of previous range.  For example
        // when previous range is [0, 10] and current range
        // is [3, 8], then a[9] and a[10] are Removed
        while (currR > R + 1) {
            currR--;
            remove(a[currR], currentAns, freq);
        }
 
        // Store current ans as the Query ans for
        // Query number index
        ans[index] = currentAns;
    }
}
 
/* Wrapper for queryResultsUtil() and outputs the
   ans array constructed by answering all queries */
void queryResults(int a[], int n, Query q[], int m)
{
    // Find block size
    block = (int)sqrt(n);
 
    // Sort all queries so that queries of same blocks
    // are arranged together.
    sort(q, q + m, compare);
 
    int* ans = new int[m];
    queryResultsUtil(a, q, ans, m);
 
    for (int i = 0; i < m; i++) {
        cout << "Answer for Query " << (i + 1)
             << " = " << ans[i] << endl;
    }
}
 
// Driver program
int main()
{
    int A[] = { 1, 2, 2, 3, 3, 3 };
 
    int n = sizeof(A) / sizeof(A[0]);
 
    // 2D array of queries with 2 columns
    Query queries[] = { { 0, 1, 0 },
                        { 1, 1, 1 },
                        { 0, 2, 2 },
                        { 1, 3, 3 },
                        { 3, 5, 4 },
                        { 0, 5, 5 } };
 
    // calculating number of queries
    int q = sizeof(queries) / sizeof(queries[0]);
 
    // Print result for each Query
    queryResults(A, n, queries, q);
 
    return 0;
}

Java

/* Java Program to answer Q queries to
   find number of times an element x 
   appears x times in a Query subarray */
 
import java.util.*;
 
public class GFG {
 
 
    // Variable to represent block size. 
    // This is made global so compare() 
    // of sort can use it.
    static int block;
 
    // Structure to represent a query range
    static class Query {
        int L, R, index;
 
        Query(int L, int R, int index) {
            this.L = L;
            this.R = R;
            this.index = index;
        }
    }
     
    /* Function used to sort all queries 
    so that all queries of 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) {
        // Different blocks, sort by block.
        if (x.L / block != y.L / block) {
            return x.L / block < y.L / block;
        }
        // Same block, sort by R value
        return x.R < y.R;
    }
     
    /* Inserts element (x) into current range
       and updates current answer */
    static void add(int x, int[] currentAns, HashMap<Integer, Integer> freq) {
         
         // increment frequency of this element
        freq.put(x, freq.getOrDefault(x, 0) + 1);
         
         // if this element was previously 
        // contributing to the currentAns,
        // decrement currentAns
        if (freq.get(x) == (x + 1)) {
            currentAns[0]--;
             
         
        // if this element has frequency 
        // equal to its value, increment
        // currentAns
        } else if (freq.get(x) == x) {
            currentAns[0]++;
        }
    }
     
    /* Removes element (x) from current 
    range btw L and R and updates 
    current Answer */
 
    static void remove(int x, int[] currentAns, HashMap<Integer, Integer> freq) {
         
        // decrement frequency of this element
        freq.put(x, freq.get(x) - 1);
         
            // if this element has frequency equal 
               // to its value, increment currentAns
        if (freq.get(x) == x) {
            currentAns[0]++;
             
            // if this element was previously 
            // contributing to the currentAns 
            // decrement currentAns
        } else if (freq.get(x) == (x - 1)) {
            currentAns[0]--;
        }
    }
     
    /* Utility Function to answer all queries
    and build the ans array in the original 
    order of queries */
    static void queryResultsUtil(int[] a, Query[] q, int[] ans, int m) {
         
         
        // map to store freq of each element
        HashMap<Integer, Integer> freq = new HashMap<>();
             
        // Initialize current L, current R
        // and current sum
        int currL = 0, currR = 0;
        int[] currentAns = {0};
 
        // Traverse through all queries
        for (int i = 0; i < m; i++) {
 
            int L = q[i].L, R = q[i].R;
            int index = q[i].index;
             
            // Remove extra elements of previous
            // range. For example if previous 
            // range is [0, 3] and current range 
            // is [2, 5], then a[0] and a[1] are 
            // removed
            while (currL < L) {
                remove(a[currL], currentAns, freq);
                currL++;
            }
             
            // Add Elements of current Range
            while (currL > L) {
                currL--;
                add(a[currL], currentAns, freq);
            }
 
            while (currR <= R) {
                add(a[currR], currentAns, freq);
                currR++;
            }
             
            // Remove elements of previous range.  For example
            // when previous range is [0, 10] and current range
            // is [3, 8], then a[9] and a[10] are Removed
            while (currR > R + 1) {
                currR--;
                remove(a[currR], currentAns, freq);
            }
             
            // Store current ans as the Query ans for
            // Query number index
            ans[index] = currentAns[0];
        }
    }
     
     
    /* Wrapper for queryResultsUtil() and outputs the
   ans array constructed by answering all queries */
 
    static void queryResults(int[] a, int n, Query[] q, int m) {
        // Find block size
        block = (int) Math.sqrt(n);
 
        Arrays.sort(q, new Comparator<Query>() {
    @Override
    public int compare(Query x, Query y) {
        // Different blocks, sort by block.
        if (x.L / block != y.L / block)
            return Integer.compare(x.L / block, y.L / block);
 
        // Same block, sort by R value
        return Integer.compare(x.R, y.R);
    }
});
 
        int[] ans = new int[m];
        queryResultsUtil(a, q, ans, m);
 
        for (int i = 0; i < m; i++) {
            System.out.println("Answer for Query " + (i + 1) + " = " + ans[i]);
        }
    }
 
    public static void main(String[] args) {
        int[] A = {1, 2, 2, 3, 3, 3};
        int n = A.length;
 
        // 2D array of queries with 2 columns
        Query[] queries = {
            new Query(0, 1, 0),
            new Query(1, 1, 1),
            new Query(0, 2, 2),
            new Query(1, 3, 3),
            new Query(3, 5, 4),
            new Query(0, 5, 5)
        };
 
        // calculating number of queries
        int q = queries.length;
 
        // Print result for each Query
        queryResults(A, n, queries, q);
    }
}
 
//this code is contributed by bhardwajji

Python3

# /* python Program to answer Q queries to
#    find number of times an element x 
#    appears x times in a Query subarray */
from functools import cmp_to_key
import math
 
# Variable to represent block size. 
# This is made global so compare() 
# of sort can use it.
block = 0
 
# Structure to represent a query range
class Query:
 
    def __init__(self, L, R, index):
        self.L = L
        self.R = R
        self.index = index
     
 
 
# /* Function used to sort all queries 
# so that all queries of same block
# are arranged together and within 
# a block, queries are sorted in 
# increasing order of R values. */
def Compare(x, y):
        
    # Different blocks, sort by block.
    if (block != 0 and x.L / block != y.L / block):
        return x.L / block - y.L / block
 
    # Same block, sort by R value
    return x.R- y.R
 
# /* Inserts element (x) into current range
#    and updates current answer */
def add(x, currentAns, freq):
 
     # increment frequency of this element
    if(x in freq):
        freq[x] = freq[x] + 1
    else:
        freq[x] =  1
 
    #  if this element was previously 
    # contributing to the currentAns,
    # decrement currentAns
    if (freq[x] == (x + 1)):
        currentAns[0] = currentAns[0] - 1
 
 
     # if this element has frequency 
     # equal to its value, increment
     # currentAns
    elif(freq[x] == x):
        currentAns[0] = currentAns[0] + 1
         
 
# /* Removes element (x) from current 
# range btw L and R and updates 
# current Answer */
 
def remove(x, currentAns, freq):
 
    # decrement frequency of this element
    freq[x] = freq[x] - 1;
 
    # if this element has frequency equal 
    # to its value, increment currentAns
    if (freq[x] == x):
        currentAns[0] = currentAns[0] + 1
 
        #  if this element was previously 
        # contributing to the currentAns 
        #  decrement currentAns
    elif(freq[x] == (x-1)):
        currentAns[0] = currentAns[0] - 1
 
# /* Utility Function to answer all queries
# and build the ans array in the original 
# order of queries */
def queryResultsUtil(a, q, ans, m):
 
 
    # map to store freq of each element
    freq = {}
 
    # Initialize current L, current R
    # and current sum
    currL = 0
    currR = 0
    currentAns = [0]
 
    # Traverse through all queries
    for i in range(m):
 
        L = q[i].L
        R = q[i].R
        index = q[i].index
 
        # Remove extra elements of previous
        # range. For example if previous 
        # range is [0, 3] and current range 
        # is [2, 5], then a[0] and a[1] are 
        # removed
        while (currL < L):
            remove(a[currL], currentAns, freq)
            currL = currL + 1
 
        # Add Elements of current Range
        while (currL > L):
            currL = currL - 1
            add(a[currL], currentAns, freq)
 
        while (currR <= R):
            add(a[currR], currentAns, freq)
            currR = currR + 1
 
        # Remove elements of previous range.  For example
        # when previous range is [0, 10] and current range
        # is [3, 8], then a[9] and a[10] are Removed
        while (currR > R + 1):
            currR = currR - 1
            remove(a[currR], currentAns, freq)
 
        # Store current ans as the Query ans for
        # Query number index
        ans[index] = currentAns[0]
         
 
 
# /* Wrapper for queryResultsUtil() and outputs the
# ans array constructed by answering all queries */
 
def queryResults(a, n, q, m):
    # Find block size
    block = math.floor(math.sqrt(n))
 
    
    sorted(q, key=cmp_to_key(Compare))
 
    ans = [0]*m
    queryResultsUtil(a, q, ans, m)
 
    for i in range(m):
        print("Answer for Query " ,(i + 1) ," = " ,ans[i])
 
 
A = [1, 2, 2, 3, 3, 3]
n = len(A)
 
# 2D array of queries with 2 columns
queries = [
    Query(0, 1, 0),
    Query(1, 1, 1),
    Query(0, 2, 2),
    Query(1, 3, 3),
    Query(3, 5, 4),
    Query(0, 5, 5)
]
 
# calculating number of queries
q = len(queries)
 
# Print result for each Query
queryResults(A, n, queries, q)
 
# this code is contributed by Arushi Jindal. 

Javascript

/* Javascript Program to answer Q queries to
   find number of times an element x 
   appears x times in a Query subarray */
 
 
 
// Variable to represent block size. 
// This is made global so compare() 
// of sort can use it.
let block = 0;
 
// Structure to represent a query range
class Query {
 
    constructor(L, R, index) {
        this.L = L;
        this.R = R;
        this.index = index;
    }
}
 
/* Function used to sort all queries 
so that all queries of same block
are arranged together and within 
a block, queries are sorted in 
increasing order of R values. */
function Compare(x, y){
        
    // Different blocks, sort by block.
    if (x.L / block != y.L / block)
        return x.L / block - y.L / block;
 
    // Same block, sort by R value
    return x.R- y.R;
 
}
 
/* Inserts element (x) into current range
   and updates current answer */
function add(x, currentAns, freq) {
 
     // increment frequency of this element
    if(x in freq){
        freq[x] = freq[x] + 1;
    }
    else{
        freq[x] =  1;
    }
 
     // if this element was previously 
    // contributing to the currentAns,
    // decrement currentAns
    if (freq[x] == (x + 1)) {
        currentAns[0]--;
 
 
    // if this element has frequency 
    // equal to its value, increment
    // currentAns
    } else if (freq[x] == x) {
        currentAns[0]++;
    }
}
 
/* Removes element (x) from current 
range btw L and R and updates 
current Answer */
 
function remove(x, currentAns, freq) {
 
    // decrement frequency of this element
    freq[x] = freq[x] - 1;
 
        // if this element has frequency equal 
           // to its value, increment currentAns
    if (freq[x] == x) {
        currentAns[0]++;
 
        // if this element was previously 
        // contributing to the currentAns 
        // decrement currentAns
    } else if (freq[x] == (x - 1)) {
        currentAns[0]--;
    }
}
 
/* Utility Function to answer all queries
and build the ans array in the original 
order of queries */
function queryResultsUtil(a, q, ans, m) {
 
 
    // map to store freq of each element
    let freq = {}
 
    // Initialize current L, current R
    // and current sum
    let currL = 0, currR = 0;
    let currentAns = [0];
 
    // Traverse through all queries
    for (let i = 0; i < m; i++) {
 
        let L = q[i].L, R = q[i].R;
        let index = q[i].index;
 
        // Remove extra elements of previous
        // range. For example if previous 
        // range is [0, 3] and current range 
        // is [2, 5], then a[0] and a[1] are 
        // removed
        while (currL < L) {
            remove(a[currL], currentAns, freq);
            currL++;
        }
 
        // Add Elements of current Range
        while (currL > L) {
            currL--;
            add(a[currL], currentAns, freq);
        }
 
        while (currR <= R) {
            add(a[currR], currentAns, freq);
            currR++;
        }
 
        // Remove elements of previous range.  For example
        // when previous range is [0, 10] and current range
        // is [3, 8], then a[9] and a[10] are Removed
        while (currR > R + 1) {
            currR--;
            remove(a[currR], currentAns, freq);
        }
 
        // Store current ans as the Query ans for
        // Query number index
        ans[index] = currentAns[0];
    }
}
 
 
/* Wrapper for queryResultsUtil() and outputs the
ans array constructed by answering all queries */
 
function queryResults(a, n, q, m) {
    // Find block size
    block = Math.floor(Math.sqrt(n));
 
    q.sort(Compare);
 
    let ans = new Array(m);
    queryResultsUtil(a, q, ans, m);
 
    for (let i = 0; i < m; i++) {
        console.log("Answer for Query " + (i + 1) + " = " + ans[i]);
    }
}
 
 
let A = [1, 2, 2, 3, 3, 3];
let n = A.length;
 
// 2D array of queries with 2 columns
let queries = [
    new Query(0, 1, 0),
    new Query(1, 1, 1),
    new Query(0, 2, 2),
    new Query(1, 3, 3),
    new Query(3, 5, 4),
    new Query(0, 5, 5)
];
 
// calculating number of queries
let q = queries.length;
 
// Print result for each Query
queryResults(A, n, queries, q);
 
//this code is contributed by Arushi Jindal. 

C#

/* C# Program to answer Q queries to
   find number of times an element x 
   appears x times in a Query subarray */
 
using System;
using System.Collections.Generic;
 
public class GFG {
// Variable to represent block size. 
// This is made global so compare() 
// of sort can use it.
    static int block;
 
 
/* Inserts element (x) into current range
   and updates current answer *//* Removes element (x) from current 
   range btw L and R and updates 
   current Answer */
    static void add(int x, int[] currentAns,
                    Dictionary<int, int> freq)
    {
        freq.TryGetValue(x, out int value);
        freq[x] = value + 1;
        if (freq[x] == x + 1) {
            currentAns[0]--;
        }
        else if (freq[x] == x) {
            currentAns[0]++;
        }
    }
/* Removes element (x) from current 
   range btw L and R and updates 
   current Answer */
    static void remove(int x, int[] currentAns,
                       Dictionary<int, int> freq)
    {
        freq[x]--;
        if (freq[x] == x) {
            currentAns[0]++;
        }
        else if (freq[x] == x - 1) {
            currentAns[0]--;
        }
    }
// Structure to represent a query range
    class Query {
        public int L;
        public int R;
        public int index;
 
        public Query(int L, int R, int index)
        {
            this.L = L;
            this.R = R;
            this.index = index;
        }
    }
    /* Utility Function to answer all queries
   and build the ans array in the original 
   order of queries */
static void queryResultsUtil(int[] a, Query[] q,
                                 int[] ans, int m)
    {
         // map to store freq of each element
        var freq = new Dictionary<int, int>();
        // Initialize current L, current R
    // and current sum
        int currL = 0, currR = 0;
        int[] currentAns = { 0 };
 // Traverse through all queries
        for (int i = 0; i < m; i++) {
            int L = q[i].L, R = q[i].R;
            int index = q[i].index;
 // Remove extra elements of previous
        // range. For example if previous 
        // range is [0, 3] and current range 
        // is [2, 5], then a[0] and a[1] are 
        // removed
            while (currL < L) {
                remove(a[currL], currentAns, freq);
                currL++;
            }
     // Add Elements of current Range
            while (currL > L) {
                currL--;
                add(a[currL], currentAns, freq);
            }
 
            while (currR <= R) {
                add(a[currR], currentAns, freq);
                currR++;
            }
    // Remove elements of previous range.  For example
        // when previous range is [0, 10] and current range
        // is [3, 8], then a[9] and a[10] are Removed
            while (currR > R + 1) {
                currR--;
                remove(a[currR], currentAns, freq);
            }
 
            ans[index] = currentAns[0];
        }
    }
     
    /* Wrapper for queryResultsUtil() and outputs the
   ans array constructed by answering all queries */
 
    static void queryResults(int[] a, int n, Query[] q,
                             int m)
    {
        block = (int)Math.Sqrt(n);
 
        Array.Sort(q, (x, y) => {
            /* Function used to sort all queries 
   so that all queries of same block
   are arranged together and within 
   a block, queries are sorted in 
   increasing order of R values. */
            if (x.L / block != y.L / block) {
                return x.L / block - y.L / block;
            }
            return x.R - y.R;
        });
 
        int[] ans = new int[m];
        queryResultsUtil(a, q, ans, m);
 
        for (int i = 0; i < m; i++) {
            Console.WriteLine("Answer for Query " + (i + 1)
                              + " = " + ans[i]);
        }
    }
 
     
    public static void Main(string[] args)
    {
        int[] A = { 1, 2, 2, 3, 3, 3 };
        int n = A.Length;
 
        Query[] q
            = { new Query(0, 1, 0), new Query(1, 1, 1),
                new Query(0, 2, 2), new Query(1, 3, 3),
                new Query(3, 5, 4), new Query(0, 5, 5) };
        int m = q.Length;
 
        queryResults(A, n, q, m);
    }
}

Output

Answer for Query 1 = 1
Answer for Query 2 = 0
Answer for Query 3 = 2
Answer for Query 4 = 1
Answer for Query 5 = 1
Answer for Query 6 = 3

Time Complexity of this approach using MO’s Algorithm is O(Q * sqrt(N) * logA) where logA is the complexity to insert an element A into the unordered_map for each query.

Auxiliary Space: O(N+Q)