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:-
Java
// 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. |
Output:
Frequency 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)
Please refer complete article on Range Queries for Frequencies of array elements for more details!
