GeeksForGeeks Print all combinations of balanced parentheses

Question reference

http://www.geeksforgeeks.org/print-all-combinations-of-balanced-parentheses/

Write a function to generate all possible n pairs of balanced parentheses. 

For example, if n=1
{}
for n=2
{}{}
{{}}

I understood implementation from below  references

http://www.geeksforgeeks.org/print-all-combinations-of-balanced-parentheses/

https://discuss.leetcode.com/topic/8724/easy-to-understand-java-backtracking-solution/2

Solution in Java

View point is that the no of "(" is equal to ")"

package generate_all_paranthesis;

import java.util.*;

public class Solution1 {

public ArrayList<String> generateParenthesis(int a) 

{

    ArrayList<String> res = new ArrayList<String>();

    if(a==0)

    {

    return res;

    }    

    backtracking(0,0,a,new StringBuilder(),res);

    return res;

}

public static void backtracking(int open,int close,int a,StringBuilder temp,ArrayList<String> res)

{

if(close==a)

{

res.add(temp.toString());

return;

}

else

{

           //go on adding ")" if open > close

if(open>close)

{

                                //appending ")"
                temp.append(")");

backtracking(open, close+1, a, temp, res);

temp.setLength(temp.length()-1);

                                //removing ")"

}

           //go on adding "(" if open <n

if(open<a)

{

                                //appending "("

                                temp.append("(");

backtracking(open+1, close, a, temp, res);

temp.setLength(temp.length()-1);

                                //removing "("

}

}

}

public static void main(String[] args) {

// TODO Auto-generated method stub

      Solution1 s1 = new Solution1();

      System.out.println(s1.generateParenthesis(3));

}

}

GeeksForGeeks Print Martix in Spiral Order

Print a given matrix in spiral form

Given a 2D array, print it in spiral form. See the following examples.

Input:
        1    2   3   4
        5    6   7   8
        9   10  11  12
        13  14  15  16
Output: 
1 2 3 4 8 12 16 15 14 13 9 5 6 7 11 10 


Input:
        1   2   3   4  5   6
        7   8   9  10  11  12
        13  14  15 16  17  18
Output: 
1 2 3 4 5 6 12 18 17 16 15 14 13 7 8 9 10 11

Below is the clear link for explanation and pseudo code implementation


https://www.youtube.com/watch?v=siKFOI8PNKM&feature=youtu.be

Solution in Java

public class Solution {
// DO NOT MODIFY THE LIST
public ArrayList<Integer> spiralOrder(final List<ArrayList<Integer>> a) {

                //Initialization of 4 pointers top bottom left and right
int top = 0,bottom = a.size()-1,left = 0,right = a.get(0).size()-1;
                // below variable will control the direction
int direction = 0;
ArrayList<Integer> res = new ArrayList<Integer>();
               // the while loop will execute until the top meets bottom and left meets right
while(bottom>=top && right>=left)
{
   if(direction==0)
   {
       for(int i =left;i<=right;i++)
       {
         
           res.add(a.get(top).get(i));
       }
     
       top++;
   }
   else if(direction==1)
   {
       for(int i =top;i<=bottom;i++)
       {
           res.add(a.get(i).get(right));
       }
     
       right--;
   }
   else if(direction==2)
   {
       for(int i=right;i>=left;i--)
       {
           res.add(a.get(bottom).get(i));
       }
     
       bottom--;
   }
   else if(direction==3)
   {
       for(int i =bottom;i>=top;i--)
       {
           res.add(a.get(i).get(left));
       }
     
       left++;
   }
                   //this helps to put back the direction to 0 so that we are again pointed to the top row of the
                   //matrix
   direction = (direction+1)%4;
}
   

return res;
}
}