# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ implementation to find ` `// elements that are a power of two ` ` `  `#include ` `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)) ` `        ``prefix = 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); ` `    ``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)) ` `        ``prefix = 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 `

Output:

```2
1
```

Time Complexity: O(max(Q, N)) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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Improved By : offbeat, Code_Mech, chitranayal

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