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+ " ");
    }
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