Recursive program to print all numbers less than N which consist of digits 1 or 3 only Last Updated : 11 Jul, 2025 Comments Improve Suggest changes Like Article Like Report Given an integer N, the task is to print all the numbers ? N which have their digits as only 1 or 3.Examples: Input: N = 10 Output: 3 1 Input: N = 20 Output: 13 11 3 1 Approach: First, check if the number is greater than 0. If yes then proceed further, else program is terminated.Check for the presence of digits 1 or 3 at each place of the number.If we find 1 or 3 at every place of the number then print the number. Now, check for the next number by using a recursive call for a number one less than the current number. Below is the implementation of the above approach: C++ // C++ implementation of the approach #include <iostream> using namespace std; // Recursive function to print the desired numbers void printNumbers(int N) { // Bool variable to track whether each digit of // the number fulfills the given condition bool flag = 1; // Creating a copy of the number int x = N; // Checking if the number has a positive value if (N > 0) { // Loop to iterate through digits // of the number until every digit // fulfills the given condition while (x > 0 && flag == 1) { // Get last digit int digit = x % 10; // Updating value of flag to be 0 if // the digit is neither 1 nor 3 if (digit != 1 && digit != 3) flag = 0; // Eliminate last digit x = x / 10; } // If N consists of digits 1 or 3 only if (flag == 1) cout << N << " "; // Recursive call for the next number printNumbers(N - 1); } } // Driver code int main() { int N = 20; printNumbers(N); return 0; } Java // Java implementation of the above approach class GFG { // Recursive function to print the desired numbers static void printNumbers(int N) { // flag variable to track whether each digit of // the number fulfills the given condition int flag = 1; // Creating a copy of the number int x = N; // Checking if the number has a positive value if (N > 0) { // Loop to iterate through digits // of the number until every digit // fulfills the given condition while (x > 0 && flag == 1) { // Get last digit int digit = x % 10; // Updating value of flag to be 0 if // the digit is neither 1 nor 3 if (digit != 1 && digit != 3) { flag = 0; } // Eliminate last digit x = x / 10; } // If N consists of digits 1 or 3 only if (flag == 1) { System.out.print(N + " "); } // Recursive call for the next number printNumbers(N - 1); } } // Driver code public static void main(String[] args) { int N = 20; printNumbers(N); } } // This code is contributed by PrinciRaj1992 Python3 # Python 3 implementation of the approach # Recursive function to print the # desired numbers def printNumbers(N): # Bool variable to track whether each digit # of the number fulfills the given condition flag = 1 # Creating a copy of the number x = N # Checking if the number has a # positive value if (N > 0): # Loop to iterate through digits # of the number until every digit # fulfills the given condition while (x > 0 and flag == 1): # Get last digit digit = x % 10 # Updating value of flag to be 0 if # the digit is neither 1 nor 3 if (digit != 1 and digit != 3): flag = 0 # Eliminate last digit x = x // 10 # If N consists of digits 1 or 3 only if (flag == 1): print(N, end = " ") # Recursive call for the next number printNumbers(N - 1) # Driver code if __name__ == '__main__': N = 20 printNumbers(N) # This code is contributed by # Surendra_Gangwar C# // C# implementation of the above approach using System; class GFG { // Recursive function to print the desired numbers static void printNumbers(int N) { // flag variable to track whether each digit of // the number fulfills the given condition int flag = 1; // Creating a copy of the number int x = N; // Checking if the number has a positive value if (N > 0) { // Loop to iterate through digits // of the number until every digit // fulfills the given condition while (x > 0 && flag == 1) { // Get last digit int digit = x % 10; // Updating value of flag to be 0 if // the digit is neither 1 nor 3 if (digit != 1 && digit != 3) flag = 0; // Eliminate last digit x = x / 10; } // If N consists of digits 1 or 3 only if (flag == 1) Console.Write(N + " "); // Recursive call for the next number printNumbers(N - 1); } } // Driver code public static void Main() { int N = 20; printNumbers(N); } } // This code is contributed by Ryuga PHP <?php // PHP implementation of the above approach // Recursive function to print the // desired numbers function printNumbers($N) { // Bool variable to track whether each // digit of the number fulfills the // given condition $flag = 1; // Creating a copy of the number $x = $N; // Checking if the number has a // positive value if ($N > 0) { // Loop to iterate through digits // of the number until every digit // fulfills the given condition while ((int)$x > 0 && $flag == 1) { // Get last digit $digit = $x % 10; // Updating value of flag to be 0 // if the digit is neither 1 nor 3 if ($digit != 1 && $digit != 3) $flag = 0; // Eliminate last digit $x = $x / 10; } // If N consists of digits 1 or 3 only if ($flag == 1) { echo $N ; echo " "; } // Recursive call for the next number printNumbers($N - 1); } } // Driver code $N = 20; printNumbers($N); // This code is contributed // by Arnab Kundu ?> JavaScript <script> // JavaScript implementation of the above approach // Recursive function to print the desired numbers function printNumbers(N) { // flag variable to track whether each digit of // the number fulfills the given condition let flag = 1; // Creating a copy of the number let x = N; // Checking if the number has a positive value if (N > 0) { // Loop to iterate through digits // of the number until every digit // fulfills the given condition while (x > 0 && flag == 1) { // Get last digit let digit = x % 10; // Updating value of flag to be 0 if // the digit is neither 1 nor 3 if (digit != 1 && digit != 3) flag = 0; // Eliminate last digit x = parseInt(x / 10, 10); } // If N consists of digits 1 or 3 only if (flag == 1) document.write(N + " "); // Recursive call for the next number printNumbers(N - 1); } } let N = 20; printNumbers(N); </script> Output: 13 11 3 1 Time complexity: O(N*logN)Auxiliary space: O(N) for recursive stack space. Note that the idea of this post to explain a recursive solution there exist a better approach to solve this problem. We can use queue to solve this efficiently. Please refer Count of Binary Digit numbers smaller than N for details of efficient approach. Comment More infoAdvertise with us Next Article Analysis of Algorithms L lakshaygupta2807 Follow Improve Article Tags : Technical Scripter Recursion C++ Programs DSA Practice Tags : Recursion Similar Reads Basics & PrerequisitesLogic Building ProblemsLogic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. It’s the heart of coding, enabling programmers to think, reason, and arrive at smart solutions just like we do.Here are some tips for improving your programming logic: Understand the pro 2 min read Analysis of AlgorithmsAnalysis of Algorithms is a fundamental aspect of computer science that involves evaluating performance of algorithms and programs. 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