// A C++ program to find all the Stepping Number in [n m] #include   using namespace std; // This function checks if an integer n is a Stepping Number bool isStepNum(int n) {  // Initialize prevDigit with -1  int prevDigit = -1;  // Iterate through all digits of n and compare difference  // between value of previous and current digits  while (n)  {  // Get Current digit  int curDigit = n % 10;  // Single digit is consider as a  // Stepping Number  if (prevDigit == -1)  prevDigit = curDigit;  else  {  // Check if absolute difference between  // prev digit and current digit is 1  if (abs(prevDigit - curDigit) != 1)  return false;  }  prevDigit = curDigit;  n /= 10;  }  return true; } // A brute force approach based function to find all // stepping numbers. void displaySteppingNumbers(int n int m) {  // Iterate through all the numbers from [NM]  // and check if it’s a stepping number.  for (int i=n; i<=m; i++)  if (isStepNum(i))  cout << i << ' '; } // Driver program to test above function int main() {  int n = 0 m = 21;  // Display Stepping Numbers in  // the range [n m]  displaySteppingNumbers(n m);  return 0; } 

// A Java program to find all the Stepping Number in [n m] class Main {  // This Method checks if an integer n  // is a Stepping Number  public static boolean isStepNum(int n)  {  // Initialize prevDigit with -1  int prevDigit = -1;  // Iterate through all digits of n and compare  // difference between value of previous and  // current digits  while (n > 0)  {  // Get Current digit  int curDigit = n % 10;  // Single digit is consider as a  // Stepping Number  if (prevDigit != -1)  {  // Check if absolute difference between  // prev digit and current digit is 1  if (Math.abs(curDigit-prevDigit) != 1)  return false;  }  n /= 10;  prevDigit = curDigit;  }  return true;  }  // A brute force approach based function to find all  // stepping numbers.  public static void displaySteppingNumbers(int nint m)  {  // Iterate through all the numbers from [NM]  // and check if it is a stepping number.  for (int i = n; i <= m; i++)  if (isStepNum(i))  System.out.print(i+ ' ');  }  // Driver code  public static void main(String args[])  {  int n = 0 m = 21;  // Display Stepping Numbers in the range [nm]  displaySteppingNumbers(nm);  } } 

# A Python3 program to find all the Stepping Number in [n m] # This function checks if an integer n is a Stepping Number def isStepNum(n): # Initialize prevDigit with -1 prevDigit = -1 # Iterate through all digits of n and compare difference # between value of previous and current digits while (n): # Get Current digit curDigit = n % 10 # Single digit is consider as a # Stepping Number if (prevDigit == -1): prevDigit = curDigit else: # Check if absolute difference between # prev digit and current digit is 1 if (abs(prevDigit - curDigit) != 1): return False prevDigit = curDigit n //= 10 return True # A brute force approach based function to find all # stepping numbers. def displaySteppingNumbers(n m): # Iterate through all the numbers from [NM] # and check if it’s a stepping number. for i in range(n m + 1): if (isStepNum(i)): print(i end = ' ') # Driver code if __name__ == '__main__': n m = 0 21 # Display Stepping Numbers in # the range [n m] displaySteppingNumbers(n m) # This code is contributed by mohit kumar 29 

// A C# program to find all  // the Stepping Number in [n m] using System; class GFG {  // This Method checks if an   // integer n is a Stepping Number  public static bool isStepNum(int n)  {  // Initialize prevDigit with -1  int prevDigit = -1;  // Iterate through all digits   // of n and compare difference   // between value of previous   // and current digits  while (n > 0)  {  // Get Current digit  int curDigit = n % 10;  // Single digit is considered   // as a Stepping Number  if (prevDigit != -1)  {  // Check if absolute difference   // between prev digit and current   // digit is 1  if (Math.Abs(curDigit -   prevDigit) != 1)  return false;  }  n /= 10;  prevDigit = curDigit;  }  return true;  }  // A brute force approach based   // function to find all stepping numbers.  public static void displaySteppingNumbers(int n   int m)  {  // Iterate through all the numbers   // from [NM] and check if it is   // a stepping number.  for (int i = n; i <= m; i++)  if (isStepNum(i))  Console.Write(i+ ' ');  }  // Driver code  public static void Main()  {  int n = 0 m = 21;  // Display Stepping Numbers   // in the range [nm]  displaySteppingNumbers(n m);  } } // This code is contributed by nitin mittal. 

<script>  // A Javascript program to find all the Stepping Number in [n m]    // This function checks if an integer n is a Stepping Number  function isStepNum(n)  {  // Initialize prevDigit with -1  let prevDigit = -1;  // Iterate through all digits of n and compare difference  // between value of previous and current digits  while (n > 0)  {  // Get Current digit  let curDigit = n % 10;  // Single digit is consider as a  // Stepping Number  if (prevDigit == -1)  prevDigit = curDigit;  else  {  // Check if absolute difference between  // prev digit and current digit is 1  if (Math.abs(prevDigit - curDigit) != 1)  return false;  }  prevDigit = curDigit;  n = parseInt(n / 10 10);  }  return true;  }  // A brute force approach based function to find all  // stepping numbers.  function displaySteppingNumbers(n m)  {  // Iterate through all the numbers from [NM]  // and check if it’s a stepping number.  for (let i = n; i <= m; i++)  if (isStepNum(i))  document.write(i + ' ');  }  let n = 0 m = 21;    // Display Stepping Numbers in  // the range [n m]  displaySteppingNumbers(n m);    // This code is contributed by mukesh07. </script> 

// A C++ program to find all the Stepping Number from N=n // to m using BFS Approach #include   using namespace std; // Prints all stepping numbers reachable from num // and in range [n m] void bfs(int n int m int num) {  // Queue will contain all the stepping Numbers  queue<int> q;  q.push(num);  while (!q.empty())  {  // Get the front element and pop from the queue  int stepNum = q.front();  q.pop();  // If the Stepping Number is in the range  // [n m] then display  if (stepNum <= m && stepNum >= n)  cout << stepNum << ' ';  // If Stepping Number is 0 or greater than m  // no need to explore the neighbors  if (num == 0 || stepNum > m)  continue;  // Get the last digit of the currently visited  // Stepping Number  int lastDigit = stepNum % 10;  // There can be 2 cases either digit to be  // appended is lastDigit + 1 or lastDigit - 1  int stepNumA = stepNum * 10 + (lastDigit- 1);  int stepNumB = stepNum * 10 + (lastDigit + 1);  // If lastDigit is 0 then only possible digit  // after 0 can be 1 for a Stepping Number  if (lastDigit == 0)  q.push(stepNumB);  //If lastDigit is 9 then only possible  //digit after 9 can be 8 for a Stepping  //Number  else if (lastDigit == 9)  q.push(stepNumA);  else  {  q.push(stepNumA);  q.push(stepNumB);  }  } } // Prints all stepping numbers in range [n m] // using BFS. void displaySteppingNumbers(int n int m) {  // For every single digit Number 'i'  // find all the Stepping Numbers  // starting with i  for (int i = 0 ; i <= 9 ; i++)  bfs(n m i); } //Driver program to test above function int main() {  int n = 0 m = 21;  // Display Stepping Numbers in the  // range [nm]  displaySteppingNumbers(nm);  return 0; } 

// A Java program to find all the Stepping Number in // range [n m] import java.util.*; class Main {  // Prints all stepping numbers reachable from num  // and in range [n m]  public static void bfs(int nint mint num)  {  // Queue will contain all the stepping Numbers  Queue<Integer> q = new LinkedList<Integer> ();  q.add(num);  while (!q.isEmpty())  {  // Get the front element and pop from  // the queue  int stepNum = q.poll();  // If the Stepping Number is in  // the range [nm] then display  if (stepNum <= m && stepNum >= n)  {  System.out.print(stepNum + ' ');  }  // If Stepping Number is 0 or greater  // then m no need to explore the neighbors  if (stepNum == 0 || stepNum > m)  continue;  // Get the last digit of the currently  // visited Stepping Number  int lastDigit = stepNum % 10;  // There can be 2 cases either digit  // to be appended is lastDigit + 1 or  // lastDigit - 1  int stepNumA = stepNum * 10 + (lastDigit- 1);  int stepNumB = stepNum * 10 + (lastDigit + 1);  // If lastDigit is 0 then only possible  // digit after 0 can be 1 for a Stepping  // Number  if (lastDigit == 0)  q.add(stepNumB);  // If lastDigit is 9 then only possible  // digit after 9 can be 8 for a Stepping  // Number  else if (lastDigit == 9)  q.add(stepNumA);  else  {  q.add(stepNumA);  q.add(stepNumB);  }  }  }  // Prints all stepping numbers in range [n m]  // using BFS.  public static void displaySteppingNumbers(int nint m)  {  // For every single digit Number 'i'  // find all the Stepping Numbers  // starting with i  for (int i = 0 ; i <= 9 ; i++)  bfs(n m i);  }  //Driver code  public static void main(String args[])  {  int n = 0 m = 21;  // Display Stepping Numbers in  // the range [nm]  displaySteppingNumbers(nm);  } } 

# A Python3 program to find all the Stepping Number from N=n # to m using BFS Approach # Prints all stepping numbers reachable from num # and in range [n m] def bfs(n m num) : # Queue will contain all the stepping Numbers q = [] q.append(num) while len(q) > 0 : # Get the front element and pop from the queue stepNum = q[0] q.pop(0); # If the Stepping Number is in the range # [n m] then display if (stepNum <= m and stepNum >= n) : print(stepNum end = ' ') # If Stepping Number is 0 or greater than m # no need to explore the neighbors if (num == 0 or stepNum > m) : continue # Get the last digit of the currently visited # Stepping Number lastDigit = stepNum % 10 # There can be 2 cases either digit to be # appended is lastDigit + 1 or lastDigit - 1 stepNumA = stepNum * 10 + (lastDigit- 1) stepNumB = stepNum * 10 + (lastDigit + 1) # If lastDigit is 0 then only possible digit # after 0 can be 1 for a Stepping Number if (lastDigit == 0) : q.append(stepNumB) #If lastDigit is 9 then only possible #digit after 9 can be 8 for a Stepping #Number elif (lastDigit == 9) : q.append(stepNumA) else : q.append(stepNumA) q.append(stepNumB) # Prints all stepping numbers in range [n m] # using BFS. def displaySteppingNumbers(n m) : # For every single digit Number 'i' # find all the Stepping Numbers # starting with i for i in range(10) : bfs(n m i) # Driver code n m = 0 21 # Display Stepping Numbers in the # range [nm] displaySteppingNumbers(n m) # This code is contributed by divyeshrabadiya07. 

// A C# program to find all the Stepping Number in // range [n m] using System; using System.Collections.Generic; public class GFG {    // Prints all stepping numbers reachable from num  // and in range [n m]  static void bfs(int n int m int num)  {    // Queue will contain all the stepping Numbers  Queue<int> q = new Queue<int>();  q.Enqueue(num);  while(q.Count != 0)  {    // Get the front element and pop from  // the queue  int stepNum = q.Dequeue();    // If the Stepping Number is in  // the range [nm] then display  if (stepNum <= m && stepNum >= n)  {  Console.Write(stepNum + ' ');  }    // If Stepping Number is 0 or greater  // then m no need to explore the neighbors  if (stepNum == 0 || stepNum > m)  continue;    // Get the last digit of the currently  // visited Stepping Number  int lastDigit = stepNum % 10;    // There can be 2 cases either digit  // to be appended is lastDigit + 1 or  // lastDigit - 1  int stepNumA = stepNum * 10 + (lastDigit- 1);  int stepNumB = stepNum * 10 + (lastDigit + 1);    // If lastDigit is 0 then only possible  // digit after 0 can be 1 for a Stepping  // Number  if (lastDigit == 0)  q.Enqueue(stepNumB);    // If lastDigit is 9 then only possible  // digit after 9 can be 8 for a Stepping  // Number  else if (lastDigit == 9)  q.Enqueue(stepNumA);  else  {  q.Enqueue(stepNumA);  q.Enqueue(stepNumB);  }  }  }    // Prints all stepping numbers in range [n m]  // using BFS.  static void displaySteppingNumbers(int nint m)  {  // For every single digit Number 'i'  // find all the Stepping Numbers  // starting with i  for (int i = 0 ; i <= 9 ; i++)  bfs(n m i);  }    // Driver code  static public void Main ()  {  int n = 0 m = 21;    // Display Stepping Numbers in  // the range [nm]  displaySteppingNumbers(nm);  } } // This code is contributed by avanitrachhadiya2155 

<script> // A Javascript program to find all // the Stepping Number in // range [n m]    // Prints all stepping numbers   // reachable from num  // and in range [n m]  function bfs(nmnum)  {  // Queue will contain all the   // stepping Numbers  let q = [];    q.push(num);    while (q.length!=0)  {  // Get the front element and pop from  // the queue  let stepNum = q.shift();    // If the Stepping Number is in  // the range [nm] then display  if (stepNum <= m && stepNum >= n)  {  document.write(stepNum + ' ');  }    // If Stepping Number is 0 or greater  // then m no need to explore the neighbors  if (stepNum == 0 || stepNum > m)  continue;    // Get the last digit of the currently  // visited Stepping Number  let lastDigit = stepNum % 10;    // There can be 2 cases either digit  // to be appended is lastDigit + 1 or  // lastDigit - 1  let stepNumA = stepNum * 10 + (lastDigit- 1);  let stepNumB = stepNum * 10 + (lastDigit + 1);    // If lastDigit is 0 then only possible  // digit after 0 can be 1 for a Stepping  // Number  if (lastDigit == 0)  q.push(stepNumB);    // If lastDigit is 9 then only possible  // digit after 9 can be 8 for a Stepping  // Number  else if (lastDigit == 9)  q.push(stepNumA);    else  {  q.push(stepNumA);  q.push(stepNumB);  }  }  }    // Prints all stepping numbers in range [n m]  // using BFS.  function displaySteppingNumbers(nm)  {  // For every single digit Number 'i'  // find all the Stepping Numbers  // starting with i  for (let i = 0 ; i <= 9 ; i++)  bfs(n m i);  }      // Driver code  let n = 0 m = 21;    // Display Stepping Numbers in  // the range [nm]  displaySteppingNumbers(nm);      // This code is contributed by unknown2108   </script> 

// A C++ program to find all the Stepping Numbers // in range [n m] using DFS Approach #include   using namespace std; // Prints all stepping numbers reachable from num // and in range [n m] void dfs(int n int m int stepNum) {  // If Stepping Number is in the  // range [nm] then display  if (stepNum <= m && stepNum >= n)  cout << stepNum << ' ';  // If Stepping Number is 0 or greater  // than m then return  if (stepNum == 0 || stepNum > m)  return ;  // Get the last digit of the currently  // visited Stepping Number  int lastDigit = stepNum % 10;  // There can be 2 cases either digit  // to be appended is lastDigit + 1 or  // lastDigit - 1  int stepNumA = stepNum*10 + (lastDigit-1);  int stepNumB = stepNum*10 + (lastDigit+1);  // If lastDigit is 0 then only possible  // digit after 0 can be 1 for a Stepping  // Number  if (lastDigit == 0)  dfs(n m stepNumB);  // If lastDigit is 9 then only possible  // digit after 9 can be 8 for a Stepping  // Number  else if(lastDigit == 9)  dfs(n m stepNumA);  else  {  dfs(n m stepNumA);  dfs(n m stepNumB);  } } // Method displays all the stepping // numbers in range [n m] void displaySteppingNumbers(int n int m) {  // For every single digit Number 'i'  // find all the Stepping Numbers  // starting with i  for (int i = 0 ; i <= 9 ; i++)  dfs(n m i); } //Driver program to test above function int main() {  int n = 0 m = 21;  // Display Stepping Numbers in  // the range [nm]  displaySteppingNumbers(nm);  return 0; } 

// A Java program to find all the Stepping Numbers // in range [n m] using DFS Approach import java.util.*; class Main {  // Method display's all the stepping numbers  // in range [n m]  public static void dfs(int nint mint stepNum)  {  // If Stepping Number is in the  // range [nm] then display  if (stepNum <= m && stepNum >= n)  System.out.print(stepNum + ' ');  // If Stepping Number is 0 or greater  // than m then return  if (stepNum == 0 || stepNum > m)  return ;  // Get the last digit of the currently  // visited Stepping Number  int lastDigit = stepNum % 10;  // There can be 2 cases either digit  // to be appended is lastDigit + 1 or  // lastDigit - 1  int stepNumA = stepNum*10 + (lastDigit-1);  int stepNumB = stepNum*10 + (lastDigit+1);  // If lastDigit is 0 then only possible  // digit after 0 can be 1 for a Stepping  // Number  if (lastDigit == 0)  dfs(n m stepNumB);  // If lastDigit is 9 then only possible  // digit after 9 can be 8 for a Stepping  // Number  else if(lastDigit == 9)  dfs(n m stepNumA);  else  {  dfs(n m stepNumA);  dfs(n m stepNumB);  }  }  // Prints all stepping numbers in range [n m]  // using DFS.  public static void displaySteppingNumbers(int n int m)  {  // For every single digit Number 'i'  // find all the Stepping Numbers  // starting with i  for (int i = 0 ; i <= 9 ; i++)  dfs(n m i);  }  // Driver code  public static void main(String args[])  {  int n = 0 m = 21;  // Display Stepping Numbers in  // the range [nm]  displaySteppingNumbers(nm);  } } 

# A Python3 program to find all the Stepping Numbers # in range [n m] using DFS Approach # Prints all stepping numbers reachable from num # and in range [n m] def dfs(n m stepNum) : # If Stepping Number is in the # range [nm] then display if (stepNum <= m and stepNum >= n) : print(stepNum end = ' ') # If Stepping Number is 0 or greater # than m then return if (stepNum == 0 or stepNum > m) : return # Get the last digit of the currently # visited Stepping Number lastDigit = stepNum % 10 # There can be 2 cases either digit # to be appended is lastDigit + 1 or # lastDigit - 1 stepNumA = stepNum * 10 + (lastDigit - 1) stepNumB = stepNum * 10 + (lastDigit + 1) # If lastDigit is 0 then only possible # digit after 0 can be 1 for a Stepping # Number if (lastDigit == 0) : dfs(n m stepNumB) # If lastDigit is 9 then only possible # digit after 9 can be 8 for a Stepping # Number elif(lastDigit == 9) : dfs(n m stepNumA) else : dfs(n m stepNumA) dfs(n m stepNumB) # Method displays all the stepping # numbers in range [n m] def displaySteppingNumbers(n m) : # For every single digit Number 'i' # find all the Stepping Numbers # starting with i for i in range(10) : dfs(n m i) n m = 0 21 # Display Stepping Numbers in # the range [nm] displaySteppingNumbers(n m) # This code is contributed by divyesh072019. 

// A C# program to find all the Stepping Numbers // in range [n m] using DFS Approach using System; public class GFG {  // Method display's all the stepping numbers  // in range [n m]  static void dfs(int n int m int stepNum)  {  // If Stepping Number is in the  // range [nm] then display  if (stepNum <= m && stepNum >= n)  Console.Write(stepNum + ' ');  // If Stepping Number is 0 or greater  // than m then return  if (stepNum == 0 || stepNum > m)  return ;  // Get the last digit of the currently  // visited Stepping Number  int lastDigit = stepNum % 10;  // There can be 2 cases either digit  // to be appended is lastDigit + 1 or  // lastDigit - 1  int stepNumA = stepNum*10 + (lastDigit - 1);  int stepNumB = stepNum*10 + (lastDigit + 1);  // If lastDigit is 0 then only possible  // digit after 0 can be 1 for a Stepping  // Number  if (lastDigit == 0)  dfs(n m stepNumB);  // If lastDigit is 9 then only possible  // digit after 9 can be 8 for a Stepping  // Number  else if(lastDigit == 9)  dfs(n m stepNumA);  else  {  dfs(n m stepNumA);  dfs(n m stepNumB);  }  }  // Prints all stepping numbers in range [n m]  // using DFS.  public static void displaySteppingNumbers(int n int m)  {  // For every single digit Number 'i'  // find all the Stepping Numbers  // starting with i  for (int i = 0 ; i <= 9 ; i++)  dfs(n m i);  }  // Driver code  static public void Main ()  {  int n = 0 m = 21;  // Display Stepping Numbers in  // the range [nm]  displaySteppingNumbers(nm);  } } // This code is contributed by rag2127. 

<script> // A Javascript program to find all the Stepping Numbers // in range [n m] using DFS Approach // Method display's all the stepping numbers  // in range [n m] function dfs(n m stepNum) {  // If Stepping Number is in the  // range [nm] then display  if (stepNum <= m && stepNum >= n)  document.write(stepNum + ' ');    // If Stepping Number is 0 or greater  // than m then return  if (stepNum == 0 || stepNum > m)  return ;    // Get the last digit of the currently  // visited Stepping Number  let lastDigit = stepNum % 10;    // There can be 2 cases either digit  // to be appended is lastDigit + 1 or  // lastDigit - 1  let stepNumA = stepNum*10 + (lastDigit-1);  let stepNumB = stepNum*10 + (lastDigit+1);    // If lastDigit is 0 then only possible  // digit after 0 can be 1 for a Stepping  // Number  if (lastDigit == 0)  dfs(n m stepNumB);    // If lastDigit is 9 then only possible  // digit after 9 can be 8 for a Stepping  // Number  else if(lastDigit == 9)  dfs(n m stepNumA);  else  {  dfs(n m stepNumA);  dfs(n m stepNumB);  } } // Prints all stepping numbers in range [n m]  // using DFS. function displaySteppingNumbers(n m) {  // For every single digit Number 'i'  // find all the Stepping Numbers  // starting with i  for (let i = 0 ; i <= 9 ; i++)  dfs(n m i); } // Driver code let n = 0 m = 21;   // Display Stepping Numbers in // the range [nm] displaySteppingNumbers(nm); // This code is contributed by ab2127 </script>