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# Stepping Numbers

Given two integers ‘n’ and ‘m’, find all the stepping numbers in range [n, m]. A number is called stepping number if all adjacent digits have an absolute difference of 1. 321 is a Stepping Number while 421 is not.

Examples :

```Input : n = 0, m = 21
Output : 0 1 2 3 4 5 6 7 8 9 10 12 21

Input : n = 10, m = 15
Output : 10, 12```

Method 1: Brute force Approach
In this method, a brute force approach is used to iterate through all the integers from n to m and check if it’s a stepping number.

## C++

 `// 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 [N,M]` `    ``// 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;` `}`

## Java

 `// 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` `n,``int` `m)` `    ``{` `        ``// Iterate through all the numbers from [N,M]` `        ``// 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 [n,m]` `        ``displaySteppingNumbers(n,m);` `    ``}` `}`

## Python3

 `# 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 [N,M]` `    ``# 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`

## C#

 `// 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 [N,M] 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 [n,m]` `        ``displaySteppingNumbers(n, m);` `    ``}` `}`   `// This code is contributed by nitin mittal.`

## Javascript

 ``

Output

`0 1 2 3 4 5 6 7 8 9 10 12 21 `

Method 2: Using BFS/DFS

The idea is to use a Breadth First Search/Depth First Search traversal.

How to build the graph?
Every node in the graph represents a stepping number; there will be a directed edge from a node U to V if V can be transformed from U. (U and V are Stepping Numbers) A Stepping Number V can be transformed from U in following manner.
lastDigit refers to the last digit of U (i.e. U % 10)
An adjacent number V can be:

• U*10 + lastDigit + 1 (Neighbor A)
• U*10 + lastDigit – 1 (Neighbor B)

By applying above operations a new digit is appended to U, it is either lastDigit-1 or lastDigit+1, so that the new number V formed from U is also a Stepping Number.
Therefore, every Node will have at most 2 neighboring Nodes.
Edge Cases: When the last digit of U is 0 or 9

• Case 1: lastDigit is 0 : In this case only digit ‘1’ can be appended.
• Case 2: lastDigit is 9 : In this case only digit ‘8’ can be appended.

What will be the source/starting Node?

• Every single digit number is considered as a stepping Number, so bfs traversal for every digit will give all the stepping numbers starting from that digit.
• Do a bfs/dfs traversal for all the numbers from [0,9].

Note: For node 0, no need to explore neighbors during BFS traversal since it will lead to 01, 012, 010 and these will be covered by the BFS traversal starting from node 1.
Example to find all the stepping numbers from 0 to 21

```-> 0 is a stepping Number and it is in the range
so display it.
-> 1 is a Stepping Number, find neighbors of 1 i.e.,
10 and 12 and push them into the queue

How to get 10 and 12?
Here U is 1 and last Digit is also 1
V = 10 + 0 = 10 ( Adding lastDigit - 1 )
V = 10 + 2 = 12 ( Adding lastDigit + 1 )

Then do the same for 10 and 12 this will result into
101, 123, 121 but these Numbers are out of range.
Now any number transformed from 10 and 12 will result
into a number greater than 21 so no need to explore
their neighbors.

-> 2 is a Stepping Number, find neighbors of 2 i.e.
21, 23.
-> 23 is out of range so it is not considered as a
Stepping Number (Or a neighbor of 2)

The other stepping numbers will be 3, 4, 5, 6, 7, 8, 9.```

BFS based Solution:

## C++

 `// 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 [n,m]` `    ``displaySteppingNumbers(n,m);`   `    ``return` `0;` `}`

## Java

 `// 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` `n,``int` `m,``int` `num)` `    ``{` `        ``// Queue will contain all the stepping Numbers` `        ``Queue q = ``new` `LinkedList ();`   `        ``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 [n,m] 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` `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 code` `    ``public` `static` `void` `main(String args[])` `    ``{` `        ``int` `n = ``0``, m = ``21``;`   `        ``// Display Stepping Numbers in` `        ``// the range [n,m]` `        ``displaySteppingNumbers(n,m);` `    ``}` `}`

## Python3

 `# 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 [n,m]` `displaySteppingNumbers(n, m)`   `# This code is contributed by divyeshrabadiya07.`

## C#

 `// 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 [n,m] 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` `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 code` `    ``static` `public` `void` `Main ()` `    ``{` `        ``int` `n = 0, m = 21;` ` `  `        ``// Display Stepping Numbers in` `        ``// the range [n,m]` `        ``displaySteppingNumbers(n,m);` `    ``}` `}`   `// This code is contributed by avanitrachhadiya2155`

## Javascript

 ``

Output

`0 1 10 12 2 21 3 4 5 6 7 8 9 `

DFS based Solution:

## C++

 `// 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 [n,m] 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 [n,m]` `    ``displaySteppingNumbers(n,m);` `    ``return` `0;` `}`

## Java

 `// 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` `n,``int` `m,``int` `stepNum)` `    ``{` `        ``// If Stepping Number is in the` `        ``// range [n,m] 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 [n,m]` `        ``displaySteppingNumbers(n,m);` `    ``}` `}`

## Python3

 `# 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 [n,m] 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 [n,m]` `displaySteppingNumbers(n, m)`   `# This code is contributed by divyesh072019.`

## C#

 `// 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 [n,m] 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 [n,m]` `    ``displaySteppingNumbers(n,m);` `  ``}` `}`   `// This code is contributed by rag2127.`

## Javascript

 ``

Output

`0 1 10 12 2 21 3 4 5 6 7 8 9 `

Time Complexity:O(N log N)

Space Complexity:O(N),here N is the number of stepping numbers within the range.

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