# 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. For example:-

```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:-

## C++

```// C++ program to find total count of an element
// in a range
#include<bits/stdc++.h>
using namespace std;

// Returns count of element in arr[left-1..right-1]
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 Code
int main()
{
int arr[] = {2, 8, 6, 9, 8, 6, 8, 2, 11};
int n = sizeof(arr) / sizeof(arr[0]);

// Print frequency of 2 from position 1 to 6
cout << "Frequency of 2 from 1 to 6 = "
<< findFrequency(arr, n, 1, 6, 2) << endl;

// Print frequency of 8 from position 4 to 9
cout << "Frequency of 8 from 4 to 9 = "
<< findFrequency(arr, n, 4, 9, 8);

return 0;
}
```

## 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)

An Efficient approach is to use hashing. In C++, we can use unordered_map<>

1. At first, we will store the position in map[] of every distinct element as a vector like that
```  int arr[] = {2, 8, 6, 9, 8, 6, 8, 2, 11};
map[2] = {1, 8}
map[8] = {2, 5, 7}
map[6] = {3, 6}
ans so on...```
2. As we can see that elements in map[] are already in sorted order (Because we inserted elements from left to right), the answer boils down to find the total count in that hash map[] using binary search like method.
3. In C++ we can use lower_bound which will returns an iterator pointing to the first element in the range [first, last] which has a value not less than 'left'. and upper_bound returns an iterator pointing to the first element in the range [first,last) which has a value greater than 'right'.
4. After that we just need to subtract the upper_bound() and lower_bound() result to get the final answer. For example, suppose if we want to find the total count of 8 in the range from [1 to 6], then the map[8] of lower_bound() function will return the result 0 (pointing to 2) and upper_bound() will return 2 (pointing to 7), so we need to subtract the both the result like 2 – 0 = 2 .

Below is C++ code of above approach

```// C++ program to find total count of an element
#include<bits/stdc++.h>
using namespace std;

unordered_map< int, vector<int> > store;

// Returns frequency of element in arr[left-1..right-1]
int findFrequency(int arr[], int n, int left,
int right, int element)
{
// Find the position of first occurrence of element
int a = lower_bound(store[element].begin(),
store[element].end(),
left)
- store[element].begin();

// Find the position of last occurrence of element
int b = upper_bound(store[element].begin(),
store[element].end(),
right)
- store[element].begin();

return b-a;
}

// Driver code
int main()
{
int arr[] = {2, 8, 6, 9, 8, 6, 8, 2, 11};
int n = sizeof(arr) / sizeof(arr[0]);

// Storing the indexes of an element in the map
for (int i=0; i<n; ++i)
store[arr[i]].push_back(i+1); //starting index from 1

// Print frequency of 2 from position 1 to 6
cout << "Frequency of 2 from 1 to 6 = "
<< findFrequency(arr, n, 1, 6, 2) <<endl;

// Print frequency of 8 from position 4 to 9
cout << "Frequency of 8 from 4 to 9 = "
<< findFrequency(arr, n, 4, 9, 8);

return 0;
}
```

Output:

```Frequency of 2 from 1 to 6 = 1
Frequency of 8 from 4 to 9 = 2
```

This approach will be beneficial if we have a large number of queries of an arbitrary range asking the total frequency of particular element.

Time complexity: O(log N) for single query.

This article is contributed by Shubham Bansal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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