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 C++ code of Naive approach:-

// 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; }

**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<>

- 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...

- 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.
- 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'.
- 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.

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