# Array range queries for elements with frequency same as value

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.

## 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; ` `} ` |

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## 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

**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 method is O(Q * 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.

`/* 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; ` `} ` |

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**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.

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