Range Queries for Frequencies of array elements

brahmajit-mohapatra-f8fe5582 · 4 June 2021 06:44

Hello Everyone,

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++ 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++ 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