## Friday, September 9, 2016

### Stepping Numbers - GeeksforGeeks

Stepping Numbers - GeeksforGeeks
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.

X. BFS
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.
`    ``// 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<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 [n,m] then display`
`            ``if` `(stepNum <= m && stepNum >= n)`
`            ``{`
`                ``System.out.print(stepNum + ``" "``);`
`            ``}`
`            ``// If Stepping Number is 0 or greater`
`            ``// then m ,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);`
`    ``}`

X. DFS
`    ``// 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);`
`    ``}`

X.
`    ``// This Method checks if an integer n`
`    ``// is a Stepping Number`
`    ``public` `static` `boolean` `isStepNum(``int` `n)`
`    ``{`
`        ``// Initalize 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+ ``" "``);`
`    ``}`