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 3Explanation:
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> |
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)
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; } |
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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