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

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 Codeint 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  
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 .  
 
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 codeint 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!
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