Range Queries for Frequencies of array elements
Last Updated : 12 Sep, 2023
Given an array of n non-negative integers. The task is to find frequency of a particular element in the arbitrary range of array[]. The range is given as positions (not 0 based indexes) in array. There can be multiple queries of given type.
Examples:
Input : arr[] = {2, 8, 6, 9, 8, 6, 8, 2, 11}; left = 2, right = 8, element = 8 left = 2, right = 5, element = 6 Output : 3 1 The element 8 appears 3 times in arr[left-1..right-1] The element 6 appears 1 time in arr[left-1..right-1] Naive approach: is to traverse from left to right and update count variable whenever we find the element.
Below is the code of Naive approach:-
C++ // C++ program to find total count of an element // in a range #include<bits/stdc++.h> using namespace std; // Returns count of element in arr[left-1..right-1] int findFrequency(int arr[], int n, int left, int right, int element) { int count = 0; for (int i=left-1; i<=right; ++i) if (arr[i] == element) ++count; return count; } // Driver Code int main() { int arr[] = {2, 8, 6, 9, 8, 6, 8, 2, 11}; int n = sizeof(arr) / sizeof(arr[0]); // Print frequency of 2 from position 1 to 6 cout << "Frequency of 2 from 1 to 6 = " << findFrequency(arr, n, 1, 6, 2) << endl; // Print frequency of 8 from position 4 to 9 cout << "Frequency of 8 from 4 to 9 = " << findFrequency(arr, n, 4, 9, 8); return 0; } // JAVA Code to find total count of an element // in a range class GFG { // Returns count of element in arr[left-1..right-1] public static int findFrequency(int arr[], int n, int left, int right, int element) { int count = 0; for (int i = left - 1; i < right; ++i) if (arr[i] == element) ++count; return count; } /* Driver program to test above function */ public static void main(String[] args) { int arr[] = {2, 8, 6, 9, 8, 6, 8, 2, 11}; int n = arr.length; // Print frequency of 2 from position 1 to 6 System.out.println("Frequency of 2 from 1 to 6 = " + findFrequency(arr, n, 1, 6, 2)); // Print frequency of 8 from position 4 to 9 System.out.println("Frequency of 8 from 4 to 9 = " + findFrequency(arr, n, 4, 9, 8)); } } // This code is contributed by Arnav Kr. Mandal. # Python program to find total # count of an element in a range # Returns count of element # in arr[left-1..right-1] def findFrequency(arr, n, left, right, element): count = 0 for i in range(left - 1, right): if (arr[i] == element): count += 1 return count # Driver Code arr = [2, 8, 6, 9, 8, 6, 8, 2, 11] n = len(arr) # Print frequency of 2 from position 1 to 6 print("Frequency of 2 from 1 to 6 = ", findFrequency(arr, n, 1, 6, 2)) # Print frequency of 8 from position 4 to 9 print("Frequency of 8 from 4 to 9 = ", findFrequency(arr, n, 4, 9, 8)) # This code is contributed by Anant Agarwal. // C# Code to find total count // of an element in a range using System; class GFG { // Returns count of element // in arr[left-1..right-1] public static int findFrequency(int []arr, int n, int left, int right, int element) { int count = 0; for (int i = left - 1; i < right; ++i) if (arr[i] == element) ++count; return count; } // Driver Code public static void Main() { int []arr = {2, 8, 6, 9, 8, 6, 8, 2, 11}; int n = arr.Length; // Print frequency of 2 // from position 1 to 6 Console.WriteLine("Frequency of 2 from 1 to 6 = " + findFrequency(arr, n, 1, 6, 2)); // Print frequency of 8 // from position 4 to 9 Console.Write("Frequency of 8 from 4 to 9 = " + findFrequency(arr, n, 4, 9, 8)); } } // This code is contributed by Nitin Mittal. <?php // PHP program to find total count of // an element in a range // Returns count of element in // arr[left-1..right-1] function findFrequency(&$arr, $n, $left, $right, $element) { $count = 0; for ($i = $left - 1; $i <= $right; ++$i) if ($arr[$i] == $element) ++$count; return $count; } // Driver Code $arr = array(2, 8, 6, 9, 8, 6, 8, 2, 11); $n = sizeof($arr); // Print frequency of 2 from position 1 to 6 echo "Frequency of 2 from 1 to 6 = ". findFrequency($arr, $n, 1, 6, 2) ."\n"; // Print frequency of 8 from position 4 to 9 echo "Frequency of 8 from 4 to 9 = ". findFrequency($arr, $n, 4, 9, 8); // This code is contributed by ita_c ?> <script> // Javascript Code to find total count of an element // in a range // Returns count of element in arr[left-1..right-1] function findFrequency(arr,n,left,right,element) { let count = 0; for (let i = left - 1; i < right; ++i) if (arr[i] == element) ++count; return count; } /* Driver program to test above function */ let arr=[2, 8, 6, 9, 8, 6, 8, 2, 11]; let n = arr.length; // Print frequency of 2 from position 1 to 6 document.write("Frequency of 2 from 1 to 6 = " + findFrequency(arr, n, 1, 6, 2)+"<br>"); // Print frequency of 8 from position 4 to 9 document.write("Frequency of 8 from 4 to 9 = " + findFrequency(arr, n, 4, 9, 8)); // This code is contributed by rag2127 </script> OutputFrequency of 2 from 1 to 6 = 1 Frequency of 8 from 4 to 9 = 2
Time complexity of this approach is O(right - left + 1) or O(n)
Auxiliary space: O(1)
An Efficient approach is to use hashing. In C++, we can use unordered_map
- At first, we will store the position in map[] of every distinct element as a vector like that
int arr[] = {2, 8, 6, 9, 8, 6, 8, 2, 11}; map[2] = {1, 8} map[8] = {2, 5, 7} map[6] = {3, 6} ans so on...- As we can see that elements in map[] are already in sorted order (Because we inserted elements from left to right), the answer boils down to find the total count in that hash map[] using binary search like method.
- In C++ we can use lower_bound which will returns an iterator pointing to the first element in the range [first, last] which has a value not less than 'left'. and upper_bound returns an iterator pointing to the first element in the range [first,last) which has a value greater than 'right'.
- After that we just need to subtract the upper_bound() and lower_bound() result to get the final answer. For example, suppose if we want to find the total count of 8 in the range from [1 to 6], then the map[8] of lower_bound() function will return the result 0 (pointing to 2) and upper_bound() will return 2 (pointing to 7), so we need to subtract the both the result like 2 - 0 = 2 .
Below is the code of above approach
C++ // C++ program to find total count of an element #include<bits/stdc++.h> using namespace std; unordered_map< int, vector<int> > store; // Returns frequency of element in arr[left-1..right-1] int findFrequency(int arr[], int n, int left, int right, int element) { // Find the position of first occurrence of element int a = lower_bound(store[element].begin(), store[element].end(), left) - store[element].begin(); // Find the position of last occurrence of element int b = upper_bound(store[element].begin(), store[element].end(), right) - store[element].begin(); return b-a; } // Driver code int main() { int arr[] = {2, 8, 6, 9, 8, 6, 8, 2, 11}; int n = sizeof(arr) / sizeof(arr[0]); // Storing the indexes of an element in the map for (int i=0; i<n; ++i) store[arr[i]].push_back(i+1); //starting index from 1 // Print frequency of 2 from position 1 to 6 cout << "Frequency of 2 from 1 to 6 = " << findFrequency(arr, n, 1, 6, 2) <<endl; // Print frequency of 8 from position 4 to 9 cout << "Frequency of 8 from 4 to 9 = " << findFrequency(arr, n, 4, 9, 8); return 0; } // Java program to find total count of an element import java.util.*; public class GFG { static HashMap<Integer, ArrayList<Integer> > store; static int lower_bound(ArrayList<Integer> a, int low, int high, int key) { if (low > high) { return low; } int mid = low + (high - low) / 2; if (key <= a.get(mid)) { return lower_bound(a, low, mid - 1, key); } return lower_bound(a, mid + 1, high, key); } static int upper_bound(ArrayList<Integer> a, int low, int high, int key) { if (low > high || low == a.size()) return low; int mid = low + (high - low) / 2; if (key >= a.get(mid)) { return upper_bound(a, mid + 1, high, key); } return upper_bound(a, low, mid - 1, key); } // Returns frequency of element in arr[left-1..right-1] static int findFrequency(int arr[], int n, int left, int right, int element) { // Find the position of first occurrence of element int a = lower_bound(store.get(element), 0, store.get(element).size(), left); // Find the position of last occurrence of element int b = upper_bound(store.get(element), 0, store.get(element).size(), right); return b - a; } // Driver code public static void main(String[] args) { int arr[] = { 2, 8, 6, 9, 8, 6, 8, 2, 11 }; int n = arr.length; // Storing the indexes of an element in the map store = new HashMap<>(); for (int i = 0; i < n; ++i) { if (!store.containsKey(arr[i])) store.put(arr[i], new ArrayList<>()); store.get(arr[i]).add( i + 1); // starting index from 1 } // Print frequency of 2 from position 1 to 6 System.out.println( "Frequency of 2 from 1 to 6 = " + findFrequency(arr, n, 1, 6, 2)); // Print frequency of 8 from position 4 to 9 System.out.println( "Frequency of 8 from 4 to 9 = " + findFrequency(arr, n, 4, 9, 8)); } } // This code is contributed by Karandeep1234 # Python3 program to find total count of an element from collections import defaultdict as dict from bisect import bisect_left as lower_bound from bisect import bisect_right as upper_bound store = dict(list) # Returns frequency of element # in arr[left-1..right-1] def findFrequency(arr, n, left, right, element): # Find the position of # first occurrence of element a = lower_bound(store[element], left) # Find the position of # last occurrence of element b = upper_bound(store[element], right) return b - a # Driver code arr = [2, 8, 6, 9, 8, 6, 8, 2, 11] n = len(arr) # Storing the indexes of # an element in the map for i in range(n): store[arr[i]].append(i + 1) # Print frequency of 2 from position 1 to 6 print("Frequency of 2 from 1 to 6 = ", findFrequency(arr, n, 1, 6, 2)) # Print frequency of 8 from position 4 to 9 print("Frequency of 8 from 4 to 9 = ", findFrequency(arr, n, 4, 9, 8)) # This code is contributed by Mohit Kumar // C# program to find total count of an element using System; using System.Collections; using System.Collections.Generic; public class GFG { static Dictionary<int, List<int> > store; static int lower_bound(List<int> a, int low, int high, int key) { if (low > high) { return low; } int mid = low + (high - low) / 2; if (key <= a[mid]) { return lower_bound(a, low, mid - 1, key); } return lower_bound(a, mid + 1, high, key); } static int upper_bound(List<int> a, int low, int high, int key) { if (low > high || low == a.Count) return low; int mid = low + (high - low) / 2; if (key >= a[mid]) { return upper_bound(a, mid + 1, high, key); } return upper_bound(a, low, mid - 1, key); } // Returns frequency of element in arr[left-1..right-1] static int findFrequency(int[] arr, int n, int left, int right, int element) { // Find the position of first occurrence of element int a = lower_bound(store[element], 0, store[element].Count, left); // Find the position of last occurrence of element int b = upper_bound(store[element], 0, store[element].Count, right); return b - a; } // Driver code public static void Main(string[] args) { int[] arr = { 2, 8, 6, 9, 8, 6, 8, 2, 11 }; int n = arr.Length; // Storing the indexes of an element in the map store = new Dictionary<int, List<int> >(); for (int i = 0; i < n; ++i) { if (!store.ContainsKey(arr[i])) store.Add(arr[i], new List<int>()); store[arr[i]].Add(i + 1); // starting index from 1 } // Print frequency of 2 from position 1 to 6 Console.WriteLine("Frequency of 2 from 1 to 6 = " + findFrequency(arr, n, 1, 6, 2)); // Print frequency of 8 from position 4 to 9 Console.WriteLine("Frequency of 8 from 4 to 9 = " + findFrequency(arr, n, 4, 9, 8)); } } // This code is contributed by Karandeep1234 var store = null; function lower_bound(a, low, high, key) { if (low > high) { return low; } var mid = low + parseInt((high - low) / 2); if (key <= a[mid]) { return lower_bound(a, low, mid - 1, key); } return lower_bound(a, mid + 1, high, key); } function upper_bound(a, low, high, key) { if (low > high || low == a.length) { return low; } var mid = low + parseInt((high - low) / 2); if (key >= a[mid]) { return upper_bound(a, mid + 1, high, key); } return upper_bound(a, low, mid - 1, key); } // Returns frequency of element in arr[left-1..right-1] function findFrequency(arr, n, left, right, element) { // Find the position of first occurrence of element var a = lower_bound(store.get(element), 0, store.get(element).length, left); // Find the position of last occurrence of element var b = upper_bound(store.get(element), 0, store.get(element).length, right); return b - a; } // Driver code var arr = [2, 8, 6, 9, 8, 6, 8, 2, 11]; var n = arr.length; // Storing the indexes of an element in the map store = new Map(); var i=0; for (i; i < n; ++i) { if (!store.has(arr[i])) { store.set(arr[i],new Array()); } (store.get(arr[i]).push(i + 1) > 0); } // Print frequency of 2 from position 1 to 6 console.log("Frequency of 2 from 1 to 6 = " + findFrequency(arr, n, 1, 6, 2)); // Print frequency of 8 from position 4 to 9 console.log("Frequency of 8 from 4 to 9 = " + findFrequency(arr, n, 4, 9, 8)); // This code is contributed by sourabhdalal0001. OutputFrequency of 2 from 1 to 6 = 1 Frequency of 8 from 4 to 9 = 2
This approach will be beneficial if we have a large number of queries of an arbitrary range asking the total frequency of particular element.
Time complexity: O(log N) for single query.
Auxiliary Space: O(N)
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