C++ Program for Range Queries for Frequencies of array elements

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:-

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; 
} 

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 the code of above approach

C++

// 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.
Please refer complete article on Range Queries for Frequencies of array elements for more details!

C++ Program For Adding Two Numbers Represented By Linked Lists- Set 2

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