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

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. 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++

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// 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;
}

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Java

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// 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

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C#

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// 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

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Output:

2
1

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

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Improved By : offbeat, Code_Mech