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:-
PHP
<?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) ." "; // 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 ?> |
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!
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