Find length of the longest consecutive path from a given starting character

Given a matrix of characters. Find length of the longest path from a given character, such that all characters in the path are consecutive to each other, i.e., every character in path is next to previous in alphabetical order. It is allowed to move in all 8 directions from a cell.

example:

for mat[][]={{a,c,d},
                     {h,b,e},
                     {i,g,f}} 
and input character 'e' o/p will be 5

At first this problem is looking easy (Indeed this is) by simply finding the next character via doing a 8 way recursion, this is doable this way but we can do it more efficiently by using dynamic programing, there can be some subproblem while we are searching for a path so we will store that paths in our dp matrix and will use it in between.

simple solution in java 

class SearchPath

{

    public static final int R=3;

    public static final int C=3;

    // matrix for tracking purpose 

    public static boolean [][] visited;

    

    //input matrix

    public static char [][] mat ={ {'a','c','d'},{ 'h','b','a'},{ 'i','g','f'}};

    

    // just storing all 8 position for iterative recurs 

    public static int [] x = {1,0,1,1,-1,-1,0,-1};

    public static int [] y = {0,1,1,-1,1,-1,-1,0};

    

    //dp storage

    public static int [][]dp;

    

    public static int max(int a, int b)

    {

        return(a>b?a:b);

    } 

    //will check if and i,j pair in matrix valid or not

    public static boolean isValid(int i, int j)

    {

        if(i<0 || j<0 || i>=R || j >=C)

            return false;

        else

            return true;

    }

    

    public static int getpath_util(char m, int r, int c)

    {

        if(!isValid(r,c) || visited[r][c] ||  m+1 != mat[r][c])

            return 0;

        if(dp[r][c]!= -1)

        {

            return dp[r][c];

        }

        

        // make this visited so in subsequent call it will not be taken into account

        

        visited[r][c] = true;

        int ans=0;

        for(int k=0; k<8; k++)

        {

            ans= max(ans, 1+getpath_util(mat[r][c], x[k]+r, y[k]+c));

        }  

        

        // unset this

        visited [r][c] = false;

        

        return dp[r][c] = ans;

        

    }

    public static void getpath(char m)

    {

        int ans=0;

        for(int i=0; i<R; i++)

        {

            for(int k=0; k<C;k++)

            {

                if(mat[i][k] == m)

                {

                    for(int j=0; j<8; j++)

                    {   

                        ans = max(ans, 1+getpath_util(m, i+x[j], k+y[j]));

                    }       

                }

            }

        }   

        System.out.println(ans);

    }

    public static void main(String[]args)

    { 

        visited  = new boolean[R][C];

        dp = new int [R][C];

        for(int i=0; i<R;i++)

            for(int k=0; k<C; k++)

                dp[i][k]=-1;

        for(int i=0; i<R;i++)

            for(int k=0; k<C; k++)

                visited[i][k]=false;

        getpath('a');

        getpath('e');

        getpath('b');

        getpath('f'); 

    }

}

Source:- gfg

Merge two sorted linked list (Recursive Approach)

Merge is one of those nice recursive problems where the recursive solution code is much cleaner than the iterative code

Node *Merge(Node * head1, Node *head2)
{
        // Base conditions

       if(! head1 || !head2)
       {
              if(!head1)
                   return head2;
              else
                   return head1;
     
        }
        else
        {

                 if(Node1.data <= Node2.data)
                 {

                        Node1.next = Merge(Node1.next, Node2)
                        return Node1;

                 }
                 else
                 {
       
                       Node2.next= Merge(Node1, Node2.next);
                       return Node2;
                 }
        }

}

[Uber][LeetCode] Trapping Rain Water

Problem: Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it can trap after raining.

Example:

Input: height = [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6
Explanation: The above elevation map (black section) is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped.

Input: height = [4,2,0,3,2,5]
Output: 9

Approach: Basically if we see the problem closely, we can easily figure out that the trapped water amount at any bar 'i' is as follows:

water(i) = Min (max height among all the bar in left of i, max height among all the bars in right of i ) - height(i)

We just need the sum of all of these amount to get the answer. We can achieve it using brute force approach but let's try to do better:

we can pre calculate the max height from left for every i and store in an array say leftMax and we can do the same from the right and store it in the array say rightMax. Now the calculation is something like this:

• FOR i = 1 TO n
• water += ( Min(leftMax[i], rightMax[i]) - height[i])

This approach will solve the problem in linear time but if you see it is taking extra space. Let's try to reduce it. The intuition is if till leftMax is greater than rightMax, we just calculate according to rightMax only right? (see above, we take the min only). Same thing we can say about when rightMax is greater than leftMax.

Now we can use two pointer approach one from left (index = 0) and one from right(index = n - 1) and can follow the following steps:

• WHILE left < right
• IF height[left] < height[right]
• IF height[left] > leftMax //. can't trap water 
• leftMax = height[left]
• ELSE
• water += leftMax - height[left]
• ++left;
• ELSE
• // same for right
• --right

Implementation in C#:

Approach 1:

    public int Trap(int[] height) 

    {

        int length = height?.Length ?? 0;

        if (length <= 1)

        {

            return 0;

        }

        int[] leftMax = new int[length];

        leftMax[0] = height[0];

        for (int i = 1; i < length; ++i)

        {

            leftMax[i] = Math.Max(leftMax[i - 1], height[i]);

        }

        int[] rightMax = new int[length];

        rightMax[length - 1] = height[length - 1];

        for (int i = length - 2; i >=0; --i)

        {

            rightMax[i] = Math.Max(rightMax[i + 1], height[i]);

        }

        int sum = 0;

        for (int i = 1; i < length - 1; ++i)

        {

            sum += (Math.Min(leftMax[i], rightMax[i]) - height[i]);

        }

        return sum;

    }

Approach 2:

    public int Trap(int[] height)

    {

        int length = height?.Length ?? 0;

        if (length <= 2)

        {

            return 0;

        }

        int water = 0;

        int leftMax = 0, rightMax = 0;

        int left = 0, right = length - 1;

        while (left < right)

        {

            if (height[left] < height[right])

            {

                if (height[left] <= leftMax)

                {

                    water += leftMax - height[left];

                }

                else

                {

                    leftMax = height[left];

                }

                ++left;

            }

            else

            {

                if (height[right] <= rightMax)

                {

                    water += rightMax - height[right];

                }

                else

                {

                    rightMax = height[right];

                }

                --right;

            }

        }

        return water;

    }

Complexity: O(n)