Java Program for Range Queries for Frequencies of array elements
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.Â
Examples:Â
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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.Â
Below is the code of Naive approach:-Â
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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:Â
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)Â
Auxiliary space: O(1)
Please refer complete article on Range Queries for Frequencies of array elements for more details!
