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Javascript Program for Range Queries for Frequencies of array elements

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  • Last Updated : 06 Jan, 2022

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:- 
 

Javascript




<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>

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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