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.

[sourcecode language=”CPP”]
/* 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;
}
[/sourcecode]

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.

[sourcecode language=”CPP”]
/* 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;
}
[/sourcecode]

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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