Print numbers in the range 1 to n having bits in alternate pattern Last Updated : 15 Jun, 2022 Comments Improve Suggest changes Like Article Like Report Given a positive integer n. The problem is to print the numbers in the range 1 to n having bits in alternate pattern. Here alternate pattern means that the set and unset bits in the number occur in alternate order. For example- 5 has an alternate pattern i.e. 101.Examples: Input : n = 10 Output : 1 2 5 10 Input : n = 50 Output : 1 2 5 10 21 42 Recommended PracticeNumbers having alternate bits in range 1 to nTry It! Method 1 (Naive Approach): Generate all the numbers in the range 1 to n and for each generated number check whether it has bits in alternate pattern. Time Complexity is of O(n).Method 2 (Efficient Approach): Algorithm: printNumHavingAltBitPatrn(n) Initialize curr_num = 1 print curr_num while (1) curr_num <<= 1 if n < curr_num then break print curr_num curr_num = ((curr_num) << 1) ^ 1 if n < curr_num then break print curr_num CPP // C++ implementation to print numbers in the range // 1 to n having bits in alternate pattern #include <bits/stdc++.h> using namespace std; // function to print numbers in the range 1 to n // having bits in alternate pattern void printNumHavingAltBitPatrn(int n) { // first number having bits in alternate pattern int curr_num = 1; // display cout << curr_num << " "; // loop until n < curr_num while (1) { // generate next number having alternate // bit pattern curr_num <<= 1; // if true then break if (n < curr_num) break; // display cout << curr_num << " "; // generate next number having alternate // bit pattern curr_num = ((curr_num) << 1) ^ 1; // if true then break if (n < curr_num) break; // display cout << curr_num << " "; } } // Driver program to test above int main() { int n = 50; printNumHavingAltBitPatrn(n); return 0; } Java // Java implementation to print numbers in the range // 1 to n having bits in alternate pattern import java.io.*; import java.util.*; class GFG { public static void printNumHavingAltBitPatrn(int n) { // first number having bits in alternate pattern int curr_num = 1, i = 1; // display System.out.print(curr_num + " "); // loop until n < curr_num while (i!=0) { i++; // generate next number having alternate // bit pattern curr_num <<= 1; // if true then break if (n < curr_num) break; // display System.out.print(curr_num + " "); // generate next number having alternate // bit pattern curr_num = ((curr_num) << 1) ^ 1; // if true then break if (n < curr_num) break; // display System.out.print(curr_num + " "); } } public static void main (String[] args) { int n = 50; printNumHavingAltBitPatrn(n); } } // Code Contributed by Mohit Gupta_OMG <(0_o)> Python3 # Python3 program for count total # zero in product of array # function to print numbers in the range # 1 to nhaving bits in alternate pattern def printNumHavingAltBitPatrn(n): # first number having bits in # alternate pattern curr_num = 1 # display print (curr_num) # loop until n < curr_num while (1) : # generate next number having # alternate bit pattern curr_num = curr_num << 1; # if true then break if (n < curr_num): break; # display print( curr_num ) # generate next number having # alternate bit pattern curr_num = ((curr_num) << 1) ^ 1; # if true then break if (n < curr_num): break # display print( curr_num ) # Driven code n = 50 printNumHavingAltBitPatrn(n) # This code is contributed by "rishabh_jain". C# // C# implementation to print numbers in the range // 1 to n having bits in alternate pattern using System; class GFG { // function to print numbers in the range 1 to n // having bits in alternate pattern public static void printNumHavingAltBitPatrn(int n) { // first number having bits in alternate pattern int curr_num = 1, i = 1; // display Console.Write(curr_num + " "); // loop until n < curr_num while (i!=0) { // generate next number having alternate // bit pattern curr_num <<= 1; // if true then break if (n < curr_num) break; // display Console.Write(curr_num + " "); // generate next number having alternate // bit pattern curr_num = ((curr_num) << 1) ^ 1; // if true then break if (n < curr_num) break; // display Console.Write(curr_num + " "); } } // Driver code public static void Main () { int n = 50; printNumHavingAltBitPatrn(n); } } // This code is contributed by Sam007. PHP <?php // php implementation to print // numbers in the range // 1 to n having bits in // alternate pattern // function to print numbers // in the range 1 to n // having bits in alternate // pattern function printNumHavingAltBitPatrn($n) { // first number having bits // in alternate pattern $curr_num = 1; // display echo $curr_num." "; // loop until n < curr_num while (1) { // generate next number // having alternate // bit pattern $curr_num <<= 1; // if true then break if ($n < $curr_num) break; // display echo $curr_num." "; // generate next number // having alternate // bit pattern $curr_num = (($curr_num) << 1) ^ 1; // if true then break if ($n < $curr_num) break; // display echo $curr_num." "; } } // Driver code $n = 50; printNumHavingAltBitPatrn($n); // This code is contributed by mits ?> JavaScript <script> // Javascript implementation to print numbers in the range // 1 to n having bits in alternate pattern // function to print numbers in the range 1 to n // having bits in alternate pattern function printNumHavingAltBitPatrn(n) { // first number having bits in alternate pattern var curr_num = 1; // display document.write(curr_num + " "); // loop until n < curr_num while (true) { // generate next number having alternate // bit pattern curr_num <<= 1; // if true then break if (n < curr_num) break; // display document.write(curr_num + " "); // generate next number having alternate // bit pattern curr_num = ((curr_num) << 1) ^ 1; // if true then break if (n < curr_num) break; // display document.write(curr_num + " "); } } // Driver program to test above var n = 50; printNumHavingAltBitPatrn(n); </script> Output: 1 2 5 10 21 42 Time Complexity: O(log n)Space Complexity: O(1) Comment More infoAdvertise with us A ayushjauhari14 Follow Improve Article Tags : Bit Magic DSA Practice Tags : Bit Magic Similar Reads Basics & PrerequisitesLogic Building ProblemsLogic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. 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