Count of elements which are power of 2 in a given range subarray for Q queries

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Given an array arr[] consisting of N positive numbers and Q queries of the form [L, R], the task is to find the number of elements which are a power of two in a subarray [L, R] for each query.

Examples:

Input: arr[] = { 3, 8, 5, 2, 5, 10 }, Q = {{0, 4}, {3, 5}}
Output:
2
1
Explanation:
For Query 1, the subarray [3, 8, 5, 2, 5] has 2 elements which are a power of two, 8 and 2.
For Query 2, the subarray {2, 5, 10} has 1 element which are a power of two, 2.
Input: arr[] = { 1, 2, 3, 4, 5, 6 }, Q = {{0, 4}, {1, 5}}
Output:
3
2

Naive Approach: To solve the problem mentioned above the naive approach is that for all the Q queries, we can iterate through each L and R in the array and find the number of elements which are a power of two in a subarray [L, R].
Time Complexity: O(N * Q)

Efficient Approach:
To optimize the above method the idea here is to use a prefix sum array.

Initially, the prefix sum array contains 0 for all indices.
Iterate through the given array and set the prefix array for this index to 1 if the current array element is a power of two else leave it 0.
Now, obtain the prefix sum by adding the previous index prefix array value to compute the current index’s prefix sum. the prefix[i] will store the number of elements which are a power of two from 1 to i.
Once we have prefix array, We just need to return prefix[r] – prefix[l-1] for each query.

Below is the implementation of the above approach,

C++

// C++ implementation to find
// elements that are a power of two
 
#include <bits/stdc++.h>
using namespace std;
const int MAX = 10000;
 
// prefix[i] is going to store the
// number of elements which are a
// power of two till i (including i).
int prefix[MAX + 1];
 
bool isPowerOfTwo(int x)
{
    if (x && (!(x & (x - 1))))
        return true;
    return false;
}
 
// Function to find the maximum range
// whose sum is divisible by M.
void computePrefix(int n, int a[])
{
 
    // Calculate the prefix sum
    if (isPowerOfTwo(a[0]))
        prefix[0] = 1;
    for (int i = 1; i < n; i++) {
        prefix[i] = prefix[i - 1];
 
        if (isPowerOfTwo(a[i]))
            prefix[i]++;
    }
}
 
// Function to return the number of elements
// which are a power of two in a subarray
int query(int L, int R)
{
    return prefix[R] - prefix[L - 1];
}
 
// Driver code
int main()
{
    int A[] = { 3, 8, 5, 2, 5, 10 };
    int N = sizeof(A) / sizeof(A[0]);
    int Q = 2;
 
    computePrefix(N, A);
    cout << query(0, 4) << "\n";
    cout << query(3, 5) << "\n";
 
    return 0;
}

Java

// Java implementation to find
// elements that are a power of two
import java.util.*;
class GFG{
     
static final int MAX = 10000;
 
// prefix[i] is going to store the
// number of elements which are a
// power of two till i (including i).
static int[] prefix = new int[MAX + 1];
 
static boolean isPowerOfTwo(int x)
{
    if (x != 0 && ((x & (x - 1)) == 0))
        return true;
    return false;
}
 
// Function to find the maximum range
// whose sum is divisible by M.
static void computePrefix(int n, int a[])
{
 
    // Calculate the prefix sum
    if (isPowerOfTwo(a[0]))
        prefix[0] = 1;
    for (int i = 1; i < n; i++) 
    {
        prefix[i] = prefix[i - 1];
 
        if (isPowerOfTwo(a[i]))
            prefix[i]++;
    }
}
 
// Function to return the number of elements
// which are a power of two in a subarray
static int query(int L, int R)
{
    if (L == 0)
        return prefix[R];
 
    return prefix[R] - prefix[L - 1];
}
 
// Driver code
public static void main(String[] args)
{
    int A[] = { 3, 8, 5, 2, 5, 10 };
    int N = A.length;
    int Q = 2;
 
    computePrefix(N, A);
    System.out.println(query(0, 4));
    System.out.println(query(3, 5));
}
}
 
// This code is contributed by offbeat

Python3

# Python3 implementation to find
# elements that are a power of two
MAX = 10000
 
# prefix[i] is going to store the
# number of elements which are a
# power of two till i (including i).
prefix = [0] * (MAX + 1)
 
def isPowerOfTwo(x):
     
    if (x and (not (x & (x - 1)))):
        return True
    return False
 
# Function to find the maximum range
# whose sum is divisible by M.
def computePrefix(n, a):
 
    # Calculate the prefix sum
    if (isPowerOfTwo(a[0])):
        prefix[0] = 1
         
    for i in range(1, n):
        prefix[i] = prefix[i - 1]
 
        if (isPowerOfTwo(a[i])):
            prefix[i] += 1
 
# Function to return the number of elements
# which are a power of two in a subarray
def query(L, R):
     
    return prefix[R] - prefix[L - 1]
 
# Driver code
if __name__ == "__main__":
     
    A = [ 3, 8, 5, 2, 5, 10 ]
    N = len(A)
    Q = 2
 
    computePrefix(N, A)
    print(query(0, 4))
    print(query(3, 5))
 
# This code is contributed by chitranayal

C#

// C# implementation to find
// elements that are a power of two
using System;
class GFG{
     
static int MAX = 10000;
 
// prefix[i] is going to store the
// number of elements which are a
// power of two till i (including i).
static int[] prefix = new int[MAX + 1];
 
static bool isPowerOfTwo(int x)
{
    if (x != 0 && ((x & (x - 1)) == 0))
        return true;
    return false;
}
 
// Function to find the maximum range
// whose sum is divisible by M.
static void computePrefix(int n, int []a)
{
 
    // Calculate the prefix sum
    if (isPowerOfTwo(a[0]))
        prefix[0] = 1;
    for (int i = 1; i < n; i++) 
    {
        prefix[i] = prefix[i - 1];
 
        if (isPowerOfTwo(a[i]))
            prefix[i]++;
    }
}
 
// Function to return the number of elements
// which are a power of two in a subarray
static int query(int L, int R)
{
    if (L == 0)
        return prefix[R];
 
    return prefix[R] - prefix[L - 1];
}
 
// Driver code
public static void Main()
{
    int []A = { 3, 8, 5, 2, 5, 10 };
    int N = A.Length;
 
    computePrefix(N, A);
    Console.WriteLine(query(0, 4));
    Console.WriteLine(query(3, 5));
}
}
 
// This code is contributed by Code_Mech

Javascript

<script>
 
// Javascript implementation to find
// elements that are a power of two
 
let MAX = 10000;
   
// prefix[i] is going to store the
// number of elements which are a
// power of two till i (including i).
let prefix = Array.from({length: MAX + 1}, (_, i) => 0); 
   
function isPowerOfTwo(x)
{
    if (x != 0 && ((x & (x - 1)) == 0))
        return true;
    return false;
}
   
// Function to find the maximum range
// whose sum is divisible by M.
function computePrefix(n, a)
{
   
    // Calculate the prefix sum
    if (isPowerOfTwo(a[0]))
        prefix[0] = 1;
    for (let i = 1; i < n; i++) 
    {
        prefix[i] = prefix[i - 1];
   
        if (isPowerOfTwo(a[i]))
            prefix[i]++;
    }
}
   
// Function to return the number of elements
// which are a power of two in a subarray
function query(L, R)
{
    if (L == 0)
        return prefix[R];
   
    return prefix[R] - prefix[L - 1];
}
 
// Driver Code
     
    let A = [ 3, 8, 5, 2, 5, 10 ];
    let N = A.length;
   
    computePrefix(N, A);
    document.write(query(0, 4) + "<br/>");
    document.write(query(3, 5));
     
</script>

Output:

2
1

Time Complexity: O(max(Q, N))

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