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 subarrayint query(int L, int R){ return prefix[R] - prefix[L - 1];}// Driver codeint 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 twoimport 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 subarraystatic int query(int L, int R){ if (L == 0) return prefix[R]; return prefix[R] - prefix[L - 1];}// Driver codepublic 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 twoMAX = 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 subarraydef query(L, R): return prefix[R] - prefix[L - 1]# Driver codeif __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 twousing 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 subarraystatic int query(int L, int R){ if (L == 0) return prefix[R]; return prefix[R] - prefix[L - 1];}// Driver codepublic 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 twolet 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 subarrayfunction 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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