Count of elements having odd number of divisors in index range [L, R] for Q queries

Given an array arr[] of N positive integers and the number of queries Q, each query contains two numbers L and R. The task is to count the number of elements in the array having odd number of divisors from index L to R.

Examples:

Input: arr[] = [2, 4, 5, 6, 9], Q = 3, Query[][] = { {0, 2}, {1, 3}, {1, 4} } 
Output: 1 1 2 
Explanation:
1st query: in 2 4 5 only 4 has an odd number of divisors. 
2nd query: in 4 5 6 only 4 has an odd number of divisors. 
3rd query: in 4 5 6 9 only 4, 9 has an odd number of divisors.

Input: arr[] = [1, 16, 5, 4, 9], Q = 2, Query[][] = { {1, 3}, {0, 2} } 
Output: 2 1 

Naive Approach: The naive approach is to iterate over the array from L to R for each query and count the element in the range [L, R] having odd numbers of divisors. If yes then count that element for that query.

Time Complexity: O(Q * N * sqrt(N)) 
Auxiliary Space: O(1)

Efficient Approach: We can observe that the number of divisors is odd only in case of perfect squares. Hence the best solution is to check if the given number is a perfect square or not in the range [L, R]. Below are the steps: 

1. Initialize the array dp[] of size N with value 0.
2. Traverse the given array arr[] and if any element in the it is a perfect square the update the value at that index in dp[] as 1.
3. To calculate the answer for each query efficiently we will precompute the answer.
4. We will do the prefix sum of the array dp[] and for each query in the range [L, R] the answer will be given by: 

OddDivisorCount(L, R) = DP[R] - DP[L-1]

Below is the implementation of the above approach: 

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function count the number of elements
// having odd number of divisors
void OddDivisorsCount(
    int n, int q, int a[],
    vector<pair<int, int> > Query)
{
    // Initialise dp[] array
    int DP[n] = { 0 };

    // Precomputation
    for (int i = 0; i < n; i++) {
        int x = sqrt(a[i]);

        if (x * x == a[i])
            DP[i] = 1;
    }

    // Find the Prefix Sum
    for (int i = 1; i < n; i++) {
        DP[i] = DP[i - 1] + DP[i];
    }

    int l, r;

    // Iterate for each query
    for (int i = 0; i < q; i++) {
        l = Query[i].first;
        r = Query[i].second;

        // Find the answer for each query
        if (l == 0) {
            cout << DP[r] << endl;
        }
        else {
            cout << DP[r] - DP[l - 1]
                 << endl;
        }
    }
}

// Driver Code
int main()
{
    int N = 5;
    int Q = 3;

    // Given array arr[]
    int arr[] = { 2, 4, 5, 6, 9 };

    // Given Query
    vector<pair<int, int> > Query
        Query
        = { { 0, 2 }, { 1, 3 }, { 1, 4 } };

    // Function Call
    OddDivisorsCount(N, Q, arr, Query);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

static class pair
{
    int first, second;
    public pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}

// Function count the number of elements
// having odd number of divisors
static void OddDivisorsCount(int n, int q,
                             int a[],
                             pair []Query)
{

    // Initialise dp[] array
    int DP[] = new int[n];

    // Precomputation
    for(int i = 0; i < n; i++)
    {
       int x = (int)Math.sqrt(a[i]);

       if (x * x == a[i])
           DP[i] = 1;
    }

    // Find the Prefix Sum
    for(int i = 1; i < n; i++)
    {
       DP[i] = DP[i - 1] + DP[i];
    }

    int l, r;

    // Iterate for each query
    for(int i = 0; i < q; i++)
    {
       l = Query[i].first;
       r = Query[i].second;

       // Find the answer for each query
       if (l == 0)
       {
           System.out.print(DP[r] + "\n");
       }
       else
       {
           System.out.print(DP[r] -
                            DP[l - 1] + "\n");
       }
    }
}

// Driver Code
public static void main(String[] args)
{
    int N = 5;
    int Q = 3;

    // Given array arr[]
    int arr[] = { 2, 4, 5, 6, 9 };

    // Given Query
    pair []Query = { new pair(0, 2),
                     new pair(1, 3),
                     new pair(1, 4) };

    // Function Call
    OddDivisorsCount(N, Q, arr, Query);
}
}

// This code is contributed by amal kumar choubey

Python3

# Python3 program for the above approach
import math

# Function count the number of elements
# having odd number of divisors
def OddDivisorsCount(n, q, a, Query):

    # Initialise dp[] array
    DP = [0 for i in range(n)]

    # Precomputation
    for i in range(n):
        x = int(math.sqrt(a[i]));

        if (x * x == a[i]):
            DP[i] = 1;

    # Find the Prefix Sum
    for i in range(1, n):
        DP[i] = DP[i - 1] + DP[i];

    l = 0
    r = 0

    # Iterate for each query
    for i in range(q):
        l = Query[i][0];
        r = Query[i][1];

        # Find the answer for each query
        if (l == 0):
            print(DP[r])
        else:
            print(DP[r] - DP[l - 1])

# Driver code
if __name__=="__main__":

    N = 5;
    Q = 3;

    # Given array arr[]
    arr = [ 2, 4, 5, 6, 9 ]

    # Given Query
    Query = [ [ 0, 2 ],
              [ 1, 3 ],
              [ 1, 4 ] ]

    # Function call
    OddDivisorsCount(N, Q, arr, Query);

# This code is contributed by rutvik_56

C#

// C# program for the above approach
using System;

class GFG{
class pair
{
    public int first, second;
    public pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}

// Function count the number of elements
// having odd number of divisors
static void OddDivisorsCount(int n, int q,
                             int []a,
                             pair []Query)
{

    // Initialise []dp array
    int []DP = new int[n];

    // Precomputation
    for(int i = 0; i < n; i++)
    {
       int x = (int)Math.Sqrt(a[i]);
       if (x * x == a[i])
           DP[i] = 1;
    }

    // Find the Prefix Sum
    for(int i = 1; i < n; i++)
    {
       DP[i] = DP[i - 1] + DP[i];
    }

    int l, r;

    // Iterate for each query
    for(int i = 0; i < q; i++)
    {
       l = Query[i].first;
       r = Query[i].second;

       // Find the answer for each query
       if (l == 0)
       {
           Console.Write(DP[r] + "\n");
       }
       else
       {
           Console.Write(DP[r] -
                         DP[l - 1] + "\n");
       }
    }
}

// Driver Code
public static void Main(String[] args)
{
    int N = 5;
    int Q = 3;

    // Given array []arr
    int []arr = { 2, 4, 5, 6, 9 };

    // Given Query
    pair []Query = { new pair(0, 2),
                     new pair(1, 3),
                     new pair(1, 4) };

    // Function Call
    OddDivisorsCount(N, Q, arr, Query);
}
}

// This code is contributed by gauravrajput1

Output:

1
1
2

Time complexity:

• Precomputation: O(N) 
• For each query: O(1) 
  

Auxiliary Space: O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.