Given a number n, print first n positive integers with exactly two set bits in their binary representation.

Examples:

Input: n = 3 Output: 3 5 6 The first 3 numbers with two set bits are 3 (0011), 5 (0101) and 6 (0110) Input: n = 5 Output: 3 5 6 9 10 12

A **Simple Solution** is to consider all positive integers one by one starting from 1. For every number, check if it has exactly two sets bits. If a number has exactly two set bits, print it and increment count of such numbers.

An **Efficient Solution** is to directly generate such numbers. If we clearly observe the numbers, we can rewrite them as given below pow(2,1)+pow(2,0), pow(2,2)+pow(2,0), pow(2,2)+pow(2,1), pow(2,3)+pow(2,0), pow(2,3)+pow(2,1), pow(2,3)+pow(2,2), ………

All numbers can be generated in increasing order according to higher of two set bits. The idea is to fix higher of two bits one by one. For current higher set bit, consider all lower bits and print the formed numbers.

## C++

// C++ program to print first n numbers // with exactly two set bits #include <iostream> using namespace std; // Prints first n numbers with two set bits void printTwoSetBitNums(int n) { // Initialize higher of two sets bits int x = 1; // Keep reducing n for every number // with two set bits. while (n > 0) { // Consider all lower set bits for // current higher set bit int y = 0; while (y < x) { // Print current number cout << (1 << x) + (1 << y) << " "; // If we have found n numbers n--; if (n == 0) return; // Consider next lower bit for current // higher bit. y++; } // Increment higher set bit x++; } } // Driver code int main() { printTwoSetBitNums(4); return 0; }

## Java

// Java program to print first n numbers // with exactly two set bits import java.io.*; class GFG { // Function to print first n numbers with two set bits static void printTwoSetBitNums(int n) { // Initialize higher of two sets bits int x = 1; // Keep reducing n for every number // with two set bits while (n > 0) { // Consider all lower set bits for // current higher set bit int y = 0; while (y < x) { // Print current number System.out.print(((1 << x) + (1 << y)) +" "); // If we have found n numbers n--; if (n == 0) return; // Consider next lower bit for current // higher bit. y++; } // Increment higher set bit x++; } } // Driver program public static void main (String[] args) { int n = 4; printTwoSetBitNums(n); } } // This code is contributed by Pramod Kumar

Output:

3 5 6 9

Time Complexity : O(n)

This article is contributed by **Bharath Reddy Appareddy**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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