Delete node in LinkedList

Delete a given node in a singly linked list. You don't have an access to the root node of the list. You have only access to the given node. Let's suppose this is your definition of a node.

class Node {
    Node next;
    int data;
}

Deleting a node in the middle of a singly linked list could happen if you copy the value from the next node over the given node and then to delete the next node.

public static void deleteMiddleNode(Node n){
    if(n == null){
        System.out.println("Node value is incorrect");
    }
    // in case of last element
    if(n.next == null){
        n = null;
    }
  
    Node next = n.next;
    n.data = next.data;
    n.next = next.next;
}

Travel from string to string (Google Interview question)

Google Interview Question (http://www.careercup.com/question?id=14947965):

Given a source string and a destination string write a program to display sequence of strings to travel from source to destination. Rules for traversing:
1. You can only change one character at a time
2. Any resulting word has to be a valid word from dictionary
Example: Given source word CAT and destination word DOG , one of the valid sequence would be
CAT -> COT -> DOT -> DOG
Another valid sequence can be
CAT -> COT - > COG -> DOG
One character can change at one time and every resulting word has be a valid word from dictionary

The solution here includes not only functions but global variables as well.

public class SourceToDestinationString {
    // Container for the valid words
    HashSet dictionary = null;
    // Container for words already passed by a word sequence
    HashSet visited;

    public SourceToDestinationString() {
        super();
        visited = new HashSet();
        dictionary = new HashSet();
        initDictionary();
    }

    private void initDictionary() {
        // Add the wrods in the dictionary
    }

    // Returns all valid words suitable for the next step of the path
    private ArrayList generateWords(String source, int position) {
        if (position < 0 || position >= source.length())
            return null;
        ArrayList result = new ArrayList();
        for (char ch = 'A'; ch <= 'Z'; ch++) {
            StringBuffer tmpWord = new StringBuffer(source);
            tmpWord.setCharAt(position, ch);
            String word = tmpWord.toString();
            if (dictionary.contains(word)) {
                result.add(word);
            }
        }

        return result;
    }

    // The sequence of words is written in the stack 'path' 
    public void sourceToDestinationString(Stack path, String destination) {
        String source = path.peek();
        if (source.equals(destination)) {
            System.out.print("Solution is ");
            System.out.println(path);
            return;
        }

        visited.add(source);

        for (int i = 0; i < source.length(); i++) {
            ArrayList words = generateWords(source, i);
            for (String nextWord : words) {
                if (visited.contains(nextWord))
                    continue;
                path.push(nextWord);
                sourceToDestinationString(path, destination);
                path.pop();
            }
        }
  
        visited.remove(source);
    }
}

Fibonacci numbers - 4 solutions

Find the n-th Fibonacci number. The Fibonacci numbers are integers 0, 1, 1, 2, 3, 5, 8, 13, 21, ...
Wikipedia says:

By definition, the first two numbers in the Fibonacci sequence are 0 and 1 (alternatively, 1 and 1), and each subsequent number is the sum of the previous two. 

For example the 7-th Fibonacci number in the sequence is 8. Here are a few solutions. The first one is a trivial one and follows the definition - i.e. it uses recursion.

long fibonacciRecursive(int n) {
    if (n == 0 || n == 1) {
        return n;
    }
    return fibonacciRecursive(n - 1) + fibonacciRecursive(n - 2);
}

The second solution is iterative. A drawback here is that the size of the array which is used to store the Fibonacci numbers needs to be defined statically.

long fibonacciIterative(int n) {
    int size = 100;
    long[] fibonacciNumbers = new long[size + 1];
    fibonacciNumbers[0] = 0;
    fibonacciNumbers[1] = 1;
  
    for (int i = 2; i <= n; i++) {
        fibonacciNumbers[i] = fibonacciNumbers[i - 1]
                            + fibonacciNumbers[i - 2];
    }
    return fibonacciNumbers[n];
}

The third solution is again iterative. The difference here is that it uses an array with size 3. This is a very elegant and optimized solution.

long fibonacciIterativeLowMem(int n) {
    int[] fibonacciNumbers = new int[3];
    fibonacciNumbers[0] = 0;
    fibonacciNumbers[1] = 1;

    for (int i = 2; i <= n; i++) {
        fibonacciNumbers[i % 3] = fibonacciNumbers[(i - 1) % 3]
                                + fibonacciNumbers[(i - 2) % 3];
    }
    return fibonacciNumbers[n % 3];
}

Here is one more solution that uses an optimization called memoization. The idea here is that we avoid the calculation of the tasks we already had calculated.

long fibonacciMemoization(int n) {
    int size = 100;
    long[] fibonacciNumbers = new long[size+1];
    fibonacciNumbers[0] = 0;
    fibonacciNumbers[1] = 1;
    for (int i = 2; i < size; i++) {
        fibonacciNumbers[i] = -1;
    }
  
    if (fibonacciNumbers[n] != -1) {
        return fibonacciNumbers[n];
    } else {
        fibonacciNumbers[n] = fibonacciMemoization(n - 1)
                            + fibonacciMemoization(n - 2);
        return fibonacciNumbers[n];
    }
}

Reverse a LinkedList

Given a singly linked list. You have to revers it.

The first solution uses recursion.

Node reverseList(Node previous, Node current) {
    Node tmp;
    if (current.next == null) {
        current.next = previous;
        return current;
    }
    tmp = reverseList(current, current.next);
    current.next = previous;
    return tmp;
}

The second solution is non-recursive.

Node reverse(Node current) {
    Node tmp;
    Node previous = null;
    while (current != null) {
        tmp = current.next;
        current.next = previous;
        previous = current;
        current= tmp;
    }
    return previous;
}

Sum of nodes at odd and even levels of Binary Tree

Given a Binary Tree, write a function which calculates the difference between the sum of node values at odd levels and sum of node values at even levels, i.e. function(BinaryTree) = (sum of values of nodes at odd height) - (sum of values of node at even height). Consider the root node is at height 1.

int calculateDifference(BinaryTree root) {
    if (root == null) {
        return 0;
    }
    int result = root.getData() - calculateDifference(root.getLeft())
              - calculateDifference(root.getRight());
    return result;
};

The solution is short and uses recursion. The idea is that you negate all levels under the current one (the level of the current node) and you do that on each step of the recursion.

sum[l1] - (sum[l2] - (sum[l3] - (sum[l4] - ... = sum[l1] - sum[l2] + sum[l3] - sum[l4]...

Deifference between sum of nodes on even height and sum of nodes on odd height

Determine if a number is power of 2

How can you check if a number is power of 2? This is not so hard to solve and in fact there are many ways you can implement a solution. One of the most obvious solutions is probably the one where you repeatedly divide the number by 2 until you end with 1 (in case the number is power of 2) or you reach an odd number higher than 1 (in case the number is not power of 2).

boolean isPowerOf2(int x) {
    while (((x % 2) == 0) && x > 1) {
        x /= 2;
    }
    return (x == 1);
}

A lot nicer solution is the one using bit operation:

boolean isPowerOf2(int x) {
    return x > 0 && (x & (x - 1)) == 0;
}

Here the idea is that every number that is a power of 2 has exactly one bit set to 1 in its binary representation and if you subtract 1 from this number you get a number where all first bits (starting from right) are 1 in the binary representation:

2    =  00000010     2  - 1 = 1  = 00000001
4    =  00000100     4  - 1 = 3  = 00000011
8    =  00001000     8  - 1 = 7  = 00000111
16   =  00010000     16 - 1 = 15 = 00001111
...

And if you do a bitwise AND (operator '&') on the original number and the original number subtracted by 1, we get 0:

2  & 1  = 00000010 & 00000001 = 0
4  & 3  = 00000100 & 00000011 = 0
8  & 7  = 00001000 & 00000111 = 0
16 & 15 = 00010000 & 00001111 = 0
...

Rotate Matrix

Here is another Google interview question (http://www.careercup.com/question?id=14891673):

Given a matrix represented as int[n][n], rotate it 90 degrees clockwise in-place. (In-place means minimal extra memory to be used, i.e. don't make a new array to copy into). Rotate clockwise means top-row becomes right-column, right column becomes bottom-row etc.

In the solution uses one extra integer variable to rotate elements in each iteration. There are two for loops - the outer one deals with one layer of the matrix per iteration, while the inner one deals with rotation of the elements of the layers.

void rotateMatrix(int a[][]) {
    int n = a.length;
    if (n <= 1) {
        return; // nothing to do
    }

    /* layers */
    for (int i = 0; i < n / 2; i++) {
        /* elements */
        for (int j = i; j < n - i - 1; j++) {
            int saved = a[i][j];
            a[i][j] = a[n - j - 1][i];
            a[n - j - 1][i] = a[n - 1 - i][n - 1 - j];
            a[n - 1 - i][n - 1 - j] = a[j][n - 1 - i];
            a[j][n - 1 - i] = saved;
        }
    }
}

In the picture below you can see the different layers in different colours. It shows what happens after each iteration of the outer for loop.