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++

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/* 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.

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/* 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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Improved By : mohit kumar 29