Find and Replace in Terminal

To Find and Replace in Terminal use this

for first occurrence:
sed -i 's/FIND/REPLACE/' <file_name>

for global occurrence: 
sed -i 's/FIND/REPLACE/g' <file_name>

In Java Create Static Instance of a class

public class MyClass {

private static MyClass instance = null;
public synchronized static MyClass getInstance() {
if (instance == null) {
instance = new MyClass();
}
return instance;
}
public MyClass() {
super();
}

.
.
.

}

from external classes, this class can be accessed through

MyClass.getInstance()

Find and Replace in vi Editor

How do we find and replace a string in vi Editor? 
Open the file in vi and use this: 

:%s/FIND_PATTERN/REPLACE_STRING/ 

This will replace only the first occurrence of that string in each line. If you want to replace all occurrence, use the below: 

:%s/FIND_PATTERN/REPLACE_STRING/g 

How to take one string pattern from find pattern and use it in replace string? 
In such case, you have to add that string pattern inside \( and \) in the find pattern and use \1 in replace string in the position where you want to replace it. 
For Example: 
Let's say you wan't to take all abc_NAME.test (NAME can be any name) and put it differently as lmn_NAME.test. 

:%s/abc_\(.*.test\).*/lmn_\1/ 

You can also do in multiple times. Only the order in which you open \( matters. It can even be nested. In replace you can use \1 \2 ]3 ... etc 

Replace String in Multiple Files Linux Bash

We can copy all file paths into a file. Let's call the file name as `list`.
There are multiple ways of creating `list`. Here are some.

ls > list
grep -l "FIND_STRING" file_path > list

In file_path you can mention something like *.java for example to search all java files alone.

Below is the bash script which you can put in an executable file and run.

#!/usr/bin/ksh
for var in `cat list`
do
sed -i "s/FIND_STRING/REPLACE_STRING/g" $var
done

Level Order Traversal in a Binary Tree from Bottom to Top

    public void LevelOrderReverse()    {
        System.out.println(LevelOrderReverse(this));
    }
    private String LevelOrderReverse(BinarySearchTree T)    {
        StringBuffer sb = new StringBuffer();
        if(T!=null)    {
            Queue<BinarySearchTree> Q = new LinkedList<BinarySearchTree>();
            Stack<BinarySearchTree> S = new Stack<BinarySearchTree>();
            Q.add(T);
            while(Q.size() > 0)    {
                BinarySearchTree C = Q.remove();
                if (C.right != null)
                    Q.add(C.right);
                if (C.left != null)
                    Q.add(C.left);
                S.push(C);
            }
            while(S.size() > 0)
                sb.append(S.pop().data).append(", ");
        }
        return sb.toString();
    }

Amazon Question: +1 or -1 Array Searching

Given an array, next element is either +1 or -1 of previous element then find any number k ?

    public int getIndex(int element, int[] input) { //element = k
        int size = input.length;
        for(int i=0; i < size;) {
            if(input[i]==element)
                return i;
            i=i+getPositive(element - input[i]);
        }
        return -9999; // element not found
    }
    private int getPositive(int i)    {
        if (i>0) return i;
        return (i * -1);
    }

Example 1:
Input: 4 5 4 5 6 7 8 9 8 9 10 11
k = 8
Output = 6

Example 2:
Input: 11 10 9 8 7 6 5 4 5 6 7 8 9 10 11
k = 4
Output = 7

Bucket Sort

Amazon Question: Given an array with 3 distinct elements, sort the elements in O(n) complexity
Input: 1,3,1,2,3,1,2,2,3
Output: 1, 1, 1, 2, 2, 2, 3, 3, 3


import java.util.ArrayList;

public class BucketSort {
    ArrayList> bucket;
    private int count;
    public BucketSort(int unique_int_count) {
        this.count = unique_int_count;
        this.bucket = new ArrayList>();
        for(int i=0; i < this.count; i++) {
            bucket.add(new ArrayList());
    }
    public ArrayList Sort(ArrayList input)    {
        for(int c:input)    {
            if(c==1)    {
                bucket.get(0).add(c);
            }    else if(c==2)    {
                bucket.get(1).add(c);
            }    else    {
                bucket.get(2).add(c);
            }
        }
        ArrayList sorted_output = new ArrayList<>();
        for(int i=0; i < count; i++) {
            sorted_output.addAll(bucket.get(i));
        return sorted_output;
    }
    @Override
    protected void finalize() throws Throwable {
        bucket=null;
        super.finalize();
    }
}
//----------------------Main Method------------------------
    public static void main(String[] args) {
        BucketSort b = new BucketSort(3);
        ArrayList input = new ArrayList();
        input.add(1);
        input.add(3);
        input.add(1);
        input.add(2);
        input.add(3);
        input.add(1);
        input.add(2);
        input.add(2);
        input.add(3);
        System.out.println(b.Sort(input));
    }

Amazon Question: Print Last N nodes of Linked List in reverse

Given a singly linked list's head and an integer N, print last N nodes of the list in reverse order.
Example:
List = 1->2->3->4->5->6->7->8
N = 3
Output = 8, 7, 6,

    public void printRev(int count)    { // count = N
        temp = count;
// this here is the head node
        System.out.println(printRev(this, new StringBuffer()));
    }
    private int temp;
    private String printRev(LinkedList L, StringBuffer sb)    {
        if(L.next!=null)
            printRev(L.next, sb);
        if(temp>0)    {
            sb.append(L.data).append(", ");
            temp--;
        }
        return sb.toString();
    }

Tamil Kumari FM Talk about Birds

This is a talk about Birds which I gave 3 years ago in Kumari FM about Origin of Birds, Living Birds, Conservation of Birds.

Find if a Tree is a Binary Search Tree

public boolean isBinarySearchTree() {
return isBinarySearchTree(this, -99999, 99999);
}
public boolean isBinarySearchTree(BinarySearchTree T, int min, int max) {
if(T==null)return true;
if(T.data>min && T.data<max 
&& isBinarySearchTree(T.left, min, T.data) && isBinarySearchTree(T.right, T.data, max))
return true;
return false;
}