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# 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!

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