Lowest Common Ancestor

Problem Statment

Given a root of the binary tree and nodes n1 and n2. Find the lowest common ancestor of the nodes.

Example:
The lowest common ancestor for D and E in the above tree is C. The least common ancestor for D and B is A.

Lowest common ancestor with parent pointer
This problem is easily solved if parent pointer is given. All you need is to count distance to the root for both the nodes. If the distance for node n1 is l1 and node n2 is l2, find the larger of l1 and l2 and move the corresponding node up the tree by that distance. Once that is done, then we just need to move both the nodes up the tree one step at a time and check if the parent node is same. If it is then we have the common ancestor.

Here is the code:

Node *lca_parent(Node *n1, Node *n2) {

        Node *t1 = n1, *t2 = n2;

        int c1 = 0, c2 =0;

        // Find the distance to the root for n1
        while(t1 != NULL) {
            t1 = t1->parent;
            c1++;
        }

        // Find the distance to the root for n2
        while(t2 != NULL) {
            t2 = t2->parent;
            c2++;
        }

        // Move up the tree for the node that is
        // furthest from the tree
        if (c1 < c2) {
            while(c1 < c2) {
                n2 = n2->parent;
                c1++;
            }
        } else {
            while (c2 parent;
                c2++;
            }
        }

        // Keep pn moving up the tree  for both nodes
        // n1 and n2 until we have the same parent.
        while (n1 != n2) {
            n1 = n1->parent;
            n2 = n2->parent;
        }

        return n1;
}

Lowest common ancestor without parent pointer
If parent pointer is not given we can find lowest common ancestor by searching the nodes in the tree. We can use post order traversal and keep track if we have seen the node in the current subtree. Once we have seen the node in both subtrees or the current node is one of the node and the other node has been found in one of this node's subtree then we are done. The code is given below:

/*
 * Find lowest common ancestor
 * @param root - root of the tree
 * @param n1 - one of the node whose ancestor is to be determined 
 * @param n2 - the other node whose ancestor is to be determined
 * @param result - the lowest common ancestor
 */
bool lca(Node* root, Node *n1, Node *n2, Node **result) {

    if (root == NULL)
        return false;

    bool l = lca(root->left, n1, n2, result);
    bool r = lca(root->right, n1, n2, result);

    bool res = true;
    if (l && r) {
        *result = root;
    } else if((root == n1 || root == n2) && (l || r)) {
        *result = root;
    } else if (root != n1 && root != n2) {
        res = false;
    }
    
    return res || l || r;
}

Find subset of two or three numbers in a sorted array that sum upto a given value

Problem Statement

Given a number and a sorted array print all possible subsets containing two numbers that sum unto the given number.

Example 1:  array: [-3, -2, -1, 0, 1, 2,  3, 4,  5, 6, 7],  sum: 0
The 2-subsets are [-3, 3], [-2, 2] and [-1 , 1]

Example 2:  array: [1, 3, 4, 5, 6, 7],  sum: 6
The 2-subsets is [1, 5]

This is an instance of well known subset sum problem. But it's not as hard as we just need to produce subsets of 2 numbers. The idea here is to maintain two pointers, one starting from the left (lb) and the other starting from the right (ub). If the sum of x[lb] and x[ub] equals to the sum, we have identified the first subset. If it  is smaller than the sum, then move lb to the right otherwise move ub to the left. Keep on doing this until lb meets ub.

This problem is pretty straightforward and a working solution is given below:

void sum2(int x[], int n, int sum) {

    int lb = 0;
    int ub = n - 1;

    while(lb < ub) {

        if (x[lb] + x[ub] > sum) {
            ub--;
        } else if(x[lb] + x[ub] < sum) {
            lb++;
        } else  {
            std::cout << x[lb] << ", " << x[ub] << std::endl;
            lb++;
            ub--;
        }
    }
}

This solution can easily be extended to 3-subset. All we need is to find subset of two elements for each element in the array. The solution is given below:

void sum3(int x[], int n, int sum) {

    int lb = 0;
    int ub = n - 1;

    for (int i = 0; i < n; i++) {

        int lb = i + 1;
        int ub = n - 1;
        int num = sum - x[i];

        while(lb < ub) {

            if (x[lb] + x[ub] > num) {
                ub--;
            } else if(x[lb] + x[ub] < num) {
                lb++;
            } else  {
                std::cout << x[i] << ", " << x[lb] << ", " << x[ub] << std::endl;
                lb++;
                ub--;
            }
        }
    }
}

Find first missing number in an array containing consecutive numbers

Given a sorted array with no duplicates find the first number that is missing in the sequence.
For an input array where no number is missing, return 1 + largest element in the array.

Example 1: [1, 2, 3, 4, 7] 
The first missing number is 5.

Example 2: [-1, 0, 1, 2, 3, 4, 5, 6, 7, 9] 
The first missing number is 8.

Example 3: [1, 2, 3, 4, 5, 6] 
There is no missing number. So return 7.

We are aware of binary search technique to find a number in sorted array. But is there a way to extend it to find the missing number? The idea here is to locate the first missing index and we can do that using binary search. A number is missing in the left for an array x, if x[0] + index < x[index], otherwise it's missing in the right. The binary search will eventually lead to to consecutive indices where the number is missing or to the end of the array. In the latter case the result is the largest number in the array + 1.

int find_first_missing_number(int x[], int n) {

    int a = x[0];
    int lb = 0;
    int ub = n - 1;

    // Narrow down the range where the number may be located
    while(ub - lb > 1) {
        int mid = lb +(ub-lb)/2;

        if (x[mid] == a + mid) {
            lb = mid;
        } else if (x[mid] > a + mid) {
            ub = mid;
        }
    }

    // At this point lb + 1 should be equal to ub and
    // if the number NOT missing then x[lb] + 1 should be 
    // equal to x[ub]
    // if the number is missing then x[lb] + 1 is the first
    // missing number
    if (x[lb] + 1 == x[ub])
        return x[ub] + 1;

    return x[lb] + 1;
}

Find first index of a number in the sorted array containing duplicates

This can be done easily in linear time by making a pass through the array. In coding interviews it is not likely the solution the interviewer is looking for. So we have to do better. We know that binary search provides a efficient solution for finding an element in sorted array. Lets look at the problem to see if we can modify the binary search algorithm to solve this problem. We can start with binary search to find the index. Once we find the index we have to look to the left to check if the number is present. If we look to the left linearly then the solution will be O(n). To do better than that we can use another binary search in the left and continue until we find the fist index where the number occurs.

/**
 * Find first index of the number k in the sorted
 * array x of size n
 */
int first_index(int x[], int n, int k) {
    int lo = 0;
    int hi = n - 1;

    while (lo <= hi) {
        int mid = lo + (hi - lo)/2;

        if (x[mid] == k) {
            if (mid > 0 && x[mid - 1] == k) {
                hi = mid - 1;
            } else {
                return mid;
            }

        } else if(x[mid] < k) {
            lo = mid + 1;
        } else {
            hi = mid - 1;
        }
    }

    return -1;
}

Iterative inorder traversal of a binary tree using constant space.

In order traversal of a binary tree can be performed using O(1) storage if the tree stores parent pointer. The trick is to look at possible directions we need to move in the tree.  As shown in the diagram below there are 3 directions that you can move in a binary tree. They are to the left, right or up. Given this all we need is to establish conditions that decides the movement. Consider last visited node as the node we were in before coming to the current node. Here are the cases:

1. If the last visited node is the parent node and the current node's left child exists move left.

2. If the last visited node is the left child of the current node and the current node has right child move right.

3. If the last visited node is on the left or right move up.

For the node to be the next successor for inorder traversal one of the following conditions should hold ( We will print the node value in our sample code below whenever this condition holds):

1. The last visited node was the left child.

2.  The left child of the current node doesn't exist.

It's now easy to translate these conditions into code. All you need is these conditions and some additional null checks as shown in the code below.

void inorder_constant_space(Node *root) {

    if (!root)
        return;

    Node *last = NULL;
    Node *node = root;

    while (node) {

        if (last == node->left || 
                (last == node->parent && node->left == NULL)) {
            std::cout << node->name;
        }

        if (last == node->left && node->right != NULL) {
            last = node;
            node = node->right;
        } else if(last == node->left) {
            last = node;
            node = node->parent;
        } else if (last == node->parent && node->left == NULL
                         && node->right != NULL) {
            last = node;
            node = node->right;
        } else if (last == node->parent && node->left == NULL) {
            last = node;
            node = node->parent;
        } else if (last == node->parent) {
            last = node;
            node = node->left;
        } else {
            last = node;
            node = node->parent;
        }
       
    }
}

Fill in the sibling pointer in the binary tree

You are given a binary tree which has a sibling pointer in addition to left and right pointer. The tree has left and right child set correctly but sibling pointer is not yet set. Your task is to traverse the tree and fill in the sibling pointer. A sibling is defined as a node next to the given node at the same level.

In the tree above B and C, and D and E are siblings. So sibling pointer of B should point to C and sibling pointer of D should point to E. Sibling pointer of C and E should be null because they don't have any siblings to the right.

This problem is easily solved using level order traversal as shown below.

class Node {
public:
    Node *left;
    Node *right;

    // Sibling pointer that needs to be set
    Node *sibling;

    // Value stored in the node. For simplicity

    // we assume name to be the value
    std::string name;
};

// Fill in the sibling pointer of a binary tree
void fill_node_with_next_sbiling(Node *node) {
    if (!node)
        return;

    // Initialize the queue with the root node.
    std::queue<node> queue;
    queue.push(node);

    // Set count of current level in the queue to 1.
    int cur_count = 1;
    int child_count = 0;

    Node *prev = NULL;

    while (!queue.empty()) {
        Node *n = queue.front();
        queue.pop();

        // Set the sibling pointer and set previous as current node
        if (prev) {
            prev->sibling = n;
            prev = n;
        } else {
            prev = n;
        }

        cur_count--;

        if (n->left != NULL) {
            queue.push(n->left);
            child_count++;
        }

        if (n->right != NULL) {
            queue.push(n->right);
            child_count++;
        }

        // All nodes in the current level are done.
        // Set the cur_count to next level count 
        // for processing next level
        if (cur_count == 0) {
            cur_count = child_count;
            child_count = 0;
            prev = NULL;
        }

    }
}

Convert a binary search tree into doubly linked list

Converting a binary search tree into doubly linked list is a common interview question. It's a simple question as long as we know it can be solved by extending inorder tree traversal. If you are not familiar with inorder tree traversal, I have described it here. In this problem you are given a binary search tree with left and right child and you are supposed to convert left pointer to predecessor node and right pointer to the successor node. Here is the solution to the problem:

// output list
Node *list = NULL;

void convert_to_doubly_ll(Node *root) {
    if (root == NULL)
        return;

    convert_to_doubly_ll(root->left);

    if (list) {
        list->right = root;
        root->left = list;
    } 
    
    list = root;
    
    Node *right = root->right;
    root->right = NULL;

    convert_to_doubly_ll(right);
}

If you want to do this in place, then all you have to do is pass in the previous node in each recursive call. The code is given below:

void convert_to_doubly_ll_in_place(Node *root, Node **prev) {
    if (root == NULL)
        return;

    convert_to_doubly_ll_in_place(root->left, prev);

    Node *prev_node = *prev;
    Node *right = NULL;

    if (prev_node) {
        prev_node->right = root;
        root->left = prev_node;
    } 
    
    *prev = root;
    
    right = root->right;
    root->right = NULL;

    convert_to_doubly_ll_in_place(right, prev);
}

You can call the above method like this:

Node *root = ....a tree...;
Node *prev=NULL;
convert_to_doubly_ll_in_place(root, &prev);

Tree Problems

Many of the tree problems asked in the interviews can be solved by extending traversal algorithms. There are 4 popular ways a tree can be traversed. They are preorder traversal, post order traversal, inorder and level order traversal. We will look at each of these traversals and some sample interview questions that can be answered using the given traversal.

Preorder Traversal

In preorder traversal you start from root, visit left child first and then right child. The processing for that node is done before processing it's children.

void pre_order(Node *root) {
    if (!root) {
        return;
    }
    // Process the node
    std::cout << root->name << std::endl;

    // Traverse left and right children
    pre_order(root->left);
    pre_order(root->right);
}

Sample Interview Problems:

Given a binary tree, serialize and deserialize it.

Given a root and a node in a binary tree, find the path from the root to the node.

Post Order Traversal
In post order traversal you start from root, visit left child first and then right child. The processing for the node is done after processing it's children.

void post_order(Node *root) {
    if (!root) {
        return;
    }

    // Traverse left and right children
    post_order(root->left);
    post_order(root->right);

    // Process the node.
    std::cout << root->name << std::endl;
}

Sample Interview Problems:
Given a binary tree create an in place mirror image of the tree.
Expression evaluation.

In Order Traversal
In in-order traversal you start from root, visit left subtree first and then right subtree. The processing for the node is done before you start traversing right subtree.

void in_order(Node *root) {
    if (!root) {
        return;
    }

    // Traverse left child
    post_order(root->left);

    // Process the node
    std::cout << root->name << std::end;

    // Traverse right child
    post_order(root->right);
}

Sample Interview Problems:
Convert a binary search tree into doubly linked list.

Level Order Traversal
In level-order traversal you start from root, process the root and then move to next level. In the next level you process all the nodes from left to right and then move to next level and so on.

#include <queue>
#include <iostream>
void level_order(Node *node) {
    if (!node)
        return;

    std::queue queue;

    // Count of the nodes in current level
    int cur_count = 0; 

    // Count of the nodes in the next level
    int child_count = 0; 

    queue.push(node);

    // Start with 1 as we have root in the queue
    // to begin with
    cur_count += 1;

    while (!queue.empty()) {
        Node *n = queue.front();
        queue.pop();

        // Process the node. We will just print to 
        // indicate processing.
        std::cout << n->name << " ";
        
        // Decrement of the count of nodes remaining in the
        // queue  for current level.
        cur_count--;

        if (n->left != NULL) {
            queue.push(n->left);
            child_count++;
        }

        if (n->right != NULL) {
            queue.push(n->right);
            child_count++;
        }

        // When we hit a count of 0, we start at next level
        if (cur_count == 0) {
            cur_count = child_count;
            child_count = 0;
            std::cout << std::endl;
        }
    }
}

Sample Interview Problems:
Pretty print a binary tree.
Given a binary tree, return each level as a list.
You have a binary tree.  Lets say you add an extra sibling pointer to it. Can you fill all the sibling pointers to point to it's sibling starting  from the left to the right.


I will add more sample problems as well as provide solutions to as many I can.

Median and Order statistics

Finding kth median in linear time

The idea is to partition the array we do in quick sort.  Assuming a 0 based index the item that you are looking for is positioned at k-1  if the array is ordered. Here is how we can find that element. First partition the array using a pivot (the pivot partitions the array into two parts; the elements on the left are smaller than the pivot element and elements on the right are greater than the pivot element).  Lets say the pivot index is i. if i equals k-1 then it is your median. Otherwise check if k-1 is less than  i. If it is  then repeat the same process for elements from 0 to i - 1 else repeat the process for elements from i + 1 to the end of partition that you are working on. As we chose the pivot randomly this algorithm would take expected linear time.

Here is the code that implements the above algorithm.

int find_kth_median(int x[],int start, int end, int k) {

 int pivot =  start + ((float)rand() - 1)/INT_MAX * (end - start +1);
 int a = x[pivot];

 int up = end, down = start;
 while(down < up) {

  while(x[down] <= a && down <= end)
   down++;
  while(x[up] > a)
   up--;
  if(down < up) {
   std::swap(x[up], x[down]);
  }
 }

 x[start] = x[up];
 x[up] = a;

 if (up == k-1)
  return x[up];
 else if (up < k-1) 
  return find_kth_median(x,up+1, end, k);
 else
  return find_kth_median(x, start, up -1, k);
}

Here are some of the questions that are similar to the above question:

Given an algorithm that finds median in linear time. How would you find kth element ?

Use the technique above. The idea is simple: find the median of the whole array ( the array will be partitioned around the media). Say the median index is i. Then if k is greater than i repeat the process on the right side of the array else repeat on the left side. Continue until you find the median.

Given a set of elements divide the set into k equal parts

The idea is to find the median of the whole set and then find median of the set at the left and right of the median and then continue so on. This requires O (n log k).

Java Language Concepts

What's the difference between abstract classes and interface?

interface can only have method signatures on it. It can't have implementations(body of methods) but abstract classes can have implementations. A class can be labeled as abstract even if no methods in the class are abstract. Abstract classes cannot be instantiated. If all the methods in a class are abstract it behaves like interface. Is it? Then why do we need interface? Only limitation of abstract classes is that they are classes which means we cannot extend from more than one abstract class.

What is the difference between static and not static methods?

Static methods can be accessed using class name while non static method requires instance of the class to access it. Both static and non static methods have single copy in the application. So under the hood what is the difference? One important distinction is that static methods cannot be virtual. Other is that static methods don't have access to this pointer. Non static methods take the object (on which it is called as argument). Static and non static methods look like this:

class X {

public static void staticMethod() {

//do sth
}

public void nonStaticMethod() {
//do sth
}
}

Under the hood the compiler generates method like this:

public void staticMethod() {
//do sth
}

public void nonStaticMethod(X this) {
//do sth
}

What is the difference between Comparable and Comparator ?

Comparable can be implemented by any class that requires ordering. compareTo method of the Comparable interface takes another object to which the current object is compared. For example you might have Person object and you have implemented Comparable to compare those objects using the first name and last name. But later your need changes and you have to order person objects based on their address. In this case, with Comparable, you have to change the implementation. You might not want to do that because you might need sorting using first name and last name in other places.So in that case you can always use Comparator interface. The sort method of Collections class can also take Comparator as argument. This helps you sort objects based on your needs. With comparable you are stuck to one implementation.

When do you have to override hashcode method?

When you will be using a hashed container to put your objects in, you need to override hashcode. If you override equals it's always good to override hashcode too. The hashcode provided by Object class in Java Language takes memory address of the object as hashcode. This is not what we want with our objects. For example consider following example:

class Person {

private String name;
private String dob;

public Person(String name, String dob) {
   this.name = name;
   this.dob = dob;
}
public boolean equals(Object obj) {
//implementation
}

//..


public static void main(String[] args) {

   Set<Person> persons = new HashSet<Person>();
  Person p1 = new Person("Person1", "12/12/2008");
  
   
  persons.add(p1);

 Person p2 = new Person("Person1", "12/12/2008");
  if(persons.contains(p2)) {
     //do sth
   }
    
}   
}

I am assuming here that person with same name and Date of Birth are equal. Do you think the condition will be true? Not necessarily. It's because the two person objects might get mapped into two different buckets in the hashset. Unless we return same hashcode for two equal objects there is no guarantee that the objects can be located in the hashed containers. So keep in mind that whenever you override equals you should override hashcode and two equal objects should always return same hashcode.

finally in C/C++

If you have programmed in C/ C++ you know there is no such thing as finally in C/C++. If you have used Java you know how convenient it is with finally to handle resources in code that is exception prone. There is a pretty nice and clean way to simulate finally in C++ (using destructors) but C doesn't have nice and clean way to do it. I am not sure if it is the recommended approach as it utilizes goto keyword. Since I started programming, I have been told not to use goto. Even while I read books i find authors ranting about use of goto. But here is a technique in C that simulates finally using goto.

void do_sth() {

   int *arr1 = 0,*arr2 = 0, * arr3 = 0;

   //calling function on will return error code, 0 being success
   // and other values a failure
   arr1 = (int *)malloc(100 * sizeof(int));
   if(!arr1)
      goto finally; // u wud have free(arr1) without finally
  
   arr2 =(int*) malloc(100 * sizeof(int));
 
   if(!arr2)
     goto finally; //u wud have free(arr1) and free(arr2) without finally

   arr3 =(int*) malloc(100 * sizeof(int));

   if(!arr3)
      goto finally; //u wud have free(arr1) and free(arr2) and free(arr3) without finally

 //do sth here 

   finally:
     free(arr1);
     free(arr2);
     free(arr3);
}

This is a simple example showing how you can handle exceptions in memory allocation. This makes sure that everything is freed when function exists and makes code lot cleaner. The resource allocation used here was simple. For example if you were opening 3 files instead. How would the code ? You would probably check if the file handle is valid in finally and if it is valid you would close them. The main advantage of this approach is that your resource handling code is localized. it's not spread over places and also you would write less code with this approach.

Now how would you do same thing in C++? Well C++ has facility for destructor which can do the job for you. When the object goes out of scope, the resource is cleaned up. There are lots of examples in C++ standard library. For example, instead of using dynamic arrays as shown above in C code, you could use std::vector in C++ which takes care of freeing the resources itself. Also there is std::auto_ptr which can be used for exception prone code. Of course auto_ptr doesn't support arrays.For supporting arrays you can use equivalent of boost's scoped_array.

Producer Consumer Problem

Well this is one of the frequently asked questions. It's a good way to see how much you understand multithreading. The solution is pretty simple but it's always good to understand the subtleties. The code would look something like this:

 
public void produce() {

  synchronized(queue) {

      while(queue.isFull()) {
      queue.wait();
      }

  //produce
  queue.enqueue(someObj);

  queue.notifyAll();
  }
}




public void consume() {

  synchronized(queue) {
 
     while(queue.isEmpty()) {
     queue.wait();
     }

  //consume
  someObj = queue.dequeue();
  //do something with the object

  queue.notifyAll();
  }

}

The code is pretty simple but even there are things where people get confused easily. First of all both the producer and consumer should synchronize on same object. Beginners tend to forget that. Also notice the use of while loop, when required conditions haven't been met threads are made to wait. A while loop is needed because we are using notifyAll here.notifyAll wakes up all the threads waiting on an object but only one thread will be able to obtain the lock. So in this case only one thread is allowed to proceed and others go back to wait. Another question arising in the context is the placement of notifyAll. Does it have to be the last statement inside synchronize block? The answer is no. It doesn't matter where you keep the notifyAll till it is inside synchronize block. The lock will only be released when synchronize block is escaped.

Recommendation

Never try to do produce consumer problem this way in production code. Always try to use the concurrent collections or facilities provided by the Java Language. For example, you could have used BlockingQueue rather than hand coding the program above.

Conversion from infix to postfix

Converting a given infix expression to postfix expression is used when numerical expressions are evaluated. Given an expression like A+B*C, the problem is to convert it into expression ABC*+. This is quite a simple problem and involves use of a stack. The idea is to add every digit encountered into postfix string. A stack is used to store operators. Whenever an operator is encountered, if it is of lower precedence than the operator on top of stack then the stack is popped and the operator is added to the postfix string. The popping is continued until the stack is empty. This is a pretty simple problem and the code is given below:

//returns true if stk precedes infix
bool precedes(char stk, char infix) {

 //declare an array of operators in precedence order
 //considers only 4 operators now
 static char prec[] ={'/','*','+','-'};

 //check for precedence
 for(int i = 0; i <4; i++) {
  if(prec[i] == stk)
   return true;
  if(prec[i] == infix)
   return false;

 }
}

//convert to postfix
std::string postfix(const std::string &infix) {

 int i = 0;

 //operator stack
 std::stack<char> op_stack;
 std::string post_fix;
 while(i < infix.length()) {

  //if it is a digit add it to postfix 
  if(isdigit(infix[i]))
   post_fix.append(1,infix[i]);
  else {

   //if operator at the top of stack 
   //precededs current operator
   //pop the operator and add it to 
   //the post fix string
   //repeat this until stack is empty
   //or operator at top of stack 
   //doesn't have higher precedence
   while(!op_stack.empty() && 
    precedes(op_stack.top(),infix[i])) {
    post_fix.append(1, op_stack.top());
    op_stack.pop();
   }
   
   //push the current operator
   op_stack.push(infix[i]);
  }
  i++;
 }

 //add everything to the post fix string
 while(!op_stack.empty())
 { 
  post_fix.append(1, op_stack.top()); 
  op_stack.pop();
 }
 return post_fix;
}

Number base conversion and arithmetic

Perhaps most of us came across base conversions when we were in high school. It's pretty easy to convert numbers from and to decimal if we remember 2 simple rules. If we desire to convert from/to base b to/from decimal we can use following rules:

Conversion from decimal to base b

Divide the given decimal number by b and store the remainder in a string. Keep on dividing until we the number can be divided no more.

num be the given decimal number

digit[i] = (num /b ^ (i+1)) % b.

resulting number in base b is digit[0] digit[1] ...digit[n-1].

Conversion from base b to decimal

num be the given number in base b and n be the number of digits

decimal = digit[n - 1] * b ^ (n -1)+...+digit[i] * b ^ (i) + digit[1] * b^1 + digit[0] * b^0


These 2 techniques are sufficient to convert between numbers in any bases.
Lets say you want to convert a number from base x into base y. To solve this you can convert base x into decimal and then convert the decimal into base y. These 2 techniques are easily transformed into programs. But it might not be a good idea to use these techniques when bases are power of 2. If bases are power of 2, bit manipulation techniques can be applied to convert. For example for converting from decimal to hex, it's always lot faster to take 4 bits at a time and convert them to hexadecimal. For example if given number is 35 (00100011, take 4 rightmost bits (0011) which is 3 and again take next 4 bits(0010) which is 2. Thus the Hex number for this is 23. This is likely to be faster than division.

Arithmetic

How would you add/subtract or do some other arithmetic operations on 2 numbers in base b? The simple idea is to convert the numbers into decimal and do the operation and then convert the resulting number back to b. The operation can be implemented for the given base itself but requires significant effort than the conversion technique.

Web Application Performance Improvement

When a web application is accessed by few users the performance may not be a huge problem but as the numbers of users grow it might bring your application to a halt if not designed properly with scalability in mind.It's always good to start out by profiling the application first. The profiling can give you insight on the piece of code that is expensive. A piece of code that is expensive might not necessarily be the main culprit. Always look for the piece of code that is extensively called. For example you might have one method that takes 2 seconds to run and then the other method that takes 200 ms. First intuition might be to make the first method efficient. But it might be the case that your application calls the method 1 or 2 times whereas it calls the second method thousands of time. In this case it's always good to optimize the second method.

Connection Pooling

If your application accesses database a lot make sure you are using connection pooling. And also consider using Stored Procedure if you need to filter results or make multiple requests to the database in the same code path.

Caching

If results can be reused across multiple requests, it's always a good idea to employ cache.

Compression

Make sure server supports compression and is using it to send results. This decrease communication time.


Clustering

Use server cluster for your application.

Resource contention

Minimize resource contention because if lot of requests are competing for same resource they degrade performance and hinder scalability. For example servlet is a shared resource and having synchronized method in servlets may cause scalability problems.

Memory Usage

Watch memory usage of your applications and try to reduce unnecessary memory usage.Use of large amount of memory might trigger garbage collector frequently which uses precious CPU time which could have been used for serving users.

Designing Cache

Cache is one very important concept in computer science. It is one design technique that is likely to help you when your system is thwarted by performance problems. Give this it is highly likely that at some point in your life as a Software Guy you have to design a cache. Cache generally employs multiple data structures for efficiency. Cache has limited space to store data so depending on your application you might have to chose a replacement policy. Ideally you would want to throw the element that would be unused to make room for new elements. But our world is not perfect and we have no way of knowing which element is likely to be unused in the future. The best thing we can do is to make a guess based on our application or may be log the usage statistics and fine tune the cache. Based on the replacement policy we can design caches in different ways.

LRU Cache

LRU cache is the simple cache where the least recently used item is thrown off the cache to make room for new one. As cache requires fast lookup we can always rely on hashtable for fast lookup. But how can we find least recently used item in Hashtable? We can't. So we need to find some other data structure that does it effectively. Why not use a queue? Whenever element is accessed put it on the tail. So front of the queue will be the least recently used one. Cool but we cannot access elements randomly from the queue. Queue is a data structure where you can remove element from the front and put the element from the back. To be able to access element anywhere we can use LinkedList. This solves the problem pretty effectively. SO here is the simple design: Create Node class that holds cache key and data. This node will be node of the linked list and also based on the key, it is inserted into hastable. We can use pointers so that both hashtable and Linked List point to same node. Lookup will require looking up in the hastable, if the data is not there and cache is full then remove the first element from the linked list and fetch new element and put it in back of the list. If cache is not full and data is there simply put the Node to the back of the list. The operations are quite simple.

MRU Cache
The neeed of this kind of cache may not arise that often but it can be implemented similarly as before but you can remove the tail of the list to remove most recently used item.

LFU cache
Least frequently used cache. This is quite difficult to design because we need to keep count of the accesses of each entry in cache. Here is my idea: Use Binary Search Tree with frequency of access as key and Hashtable for fast lookup. For lookup, if entry is in the cache, take the entry Node , delete it from the BST, increase its frequency and reinsert it into BST. While replacing the entry chose the leftmost element.

There are different other schemes. I would also like to hear from readers about these schemes and their solution.

Reversing a Linked List

I heard this is one of the most frequently asked questions in an interview. It seems to be asked quite a lot in phone screens. May be because it's simple, gives a good idea whether you have grasp of pointers and whether you have an idea of basic data structure (Of course linked list). There are 2 ways to approach this problem. The first one is recursive. But you might not want to come up with recursive solution during interview unless interviewer explicitly directs you to do so. Why? Because recursion is slow and takes up lot of memory if linked list is long. The second solution is also obvious. You go through each node in the list and adjust the pointers. Before proceeding it might be wise to ask the interviewer whether it's an singly or doubly linked list. Here we assume it's a singly linked list. The code is given below.

Node* reverse_list(Node *head) {
if(head == NULL)
return;

Node *current = head, *temp, *new_list = 0;

while(current != 0) {
temp = current->next;
current->next = new_list;
new_list = current;
current = temp;
}

//return new head
return new_list;

}

Young Tableau

Young tableau consists of rows and columns of numbers in the form of a matrix. The special property of young tableau is that its rows and columns are sorted. Young tableau can also function as a min heap. There are different interesting algorithms relating to young tableau.

Example of young tableau.

1 2 3
4 6 7
5 7 12

It's clear from the example that each row and column is sorted.

Here is one interesting problem on Young Tableau: How to find an element in an Young Tableau?

Naive search on the table would require O(mn) time. Where m is the number of rows and n is the number of columns. Can we do bit better? Turns out we can. The trick is to start from the top right corner> if the element we are searching for is greater we move down in the column. If the element we are searching for is smaller we move left in the row. We repeat the procedure until we find the element or we end up in a row or column beyond the table. This would translate to a simple code below:

The example assumes 3 X 3 matrix for simplicity. row is the number of rows and col is the number of columns in the matrix and n is the element being looked for.

int find_n(int x[][3],int row, int col, int n) {

 int r =0 , c = col -1;
 while( r < row && col >=0) {
  if(x[r][c] == n)
   return n;
  else if ( x[r][c] > n)
   c--;
  else
   r++;
 }
 return -1;
}

Another problem in Young Tableau is extracting minimum element. The minimum element is the element in (0,0) position. The main problem is replacing the spot (0,0) after minimum element has been removed. The code is given below (sorry that i have it in Java because I did these two things separately but still i believe it is simple enough so i thought it was not necessary to put both of these in same language)

/*
a is young tableau, i and j are current position in the tableau that we are filling
m and n are number of rows and column
*/
void youngify(int[][] a, int i, int j, int m, int n) {
  if (i > m || j > n)
   return;

  int x = -1, y = -1;

  if (i + 1 < m && a[i + 1][j] < a[i][j]) {
   a[i][j] = a[i + 1][j];
   x = i + 1;
   y = j;

  }
  if (j + 1 < n && a[i][j + 1] < a[i][j]) {
   a[i][j] = a[i][j + 1];
   x = i;
   y = j + 1;
  }

  if (x != -1) {
   //the empty space is filled with marked
   //by maximum value a integer can hold
   a[x][y] = Integer.MAX_VALUE;
   youngify(a, x, y, m, n);
  }

 }

Inserting an element into non full young tableau follows the similar process as extracting minimum element. The idea is to put the element into last row and column and move it upto the position where is should be.

Sorting can also be implemented for n X n young tableau in O(n ^3) time by using the operation by which we extract minimum element. The idea is to extract minimum element which takes O(n+n) = O(n) time, IF we repeat this for n^2 element it will take O (n ^3)time.

Finding overflows during arithmetic operations

We know that data in computers have limited size. For example, integer may be limited to 4 bytes. So, while writing programs sometimes we might need to check for overflows to make sure that the resulting value fits the size. But finding overflows are easy.

Finding overflow during addition

Lets say you are adding number a and b. Lets say maximum value that the types of a and b can hold be MAX. Then we can check for overflow using following condition:

if ( (MAX - a) < b then the result overflows else not.

Finding overflow during multiplication

One example of its use is while computing factorial. Lets say you want to compute n!. But you also want to make sure that appropriate error is thrown when it overflows. This is also quite simple. Lets say again that the maximum value a data type can hold be MAX and you are multiplying a and b. The result overflows if MAX/a < b.

Finding median or Kth element in Binary Search Tree (BST)

Finding kth element in BST is quite simple. We know that all the elements in left subtree is smaller than all the elements in right subtree. If we count the number of elements in left and right subtree then we know in which subtree the required element exists and then we move to this subtree and repeat the same steps again. For example starting from root we are looking for 11th element. Lets say left subtree has 7 elements and right subtree has 6 elements.Then as the number 11 is greater than 7 we know that the 11th median exists in the right subtree. Now in the right subtree we look for (11 - 7 -1 = 3)rd element (Why?? because we have discarded left subtree). We proceed similarly in the new subtree until we get the kth median as the root.

The example described above can very well be described in recursive form. The code below is a complete program in recursive form.

  
 //find kth element in a BST recursively
Node * find_kth_element(Node *tree, int &k) {

 if(tree == 0)
  return 0;

 int count = 0;

 //count the nodes in left subtree
 count_nodes(tree->left_, count);

 //check if the root is median
 if( k == count + 1)
  return tree;

 //check if median falls on left subtree
 else if (count >= k) {

  return find_kth_element(tree->left_,k);
 }
 //the median falls on right subtree
 else {
  k = k - (count + 1);
  return find_kth_element(tree->right_, k);
 }
}

Iterative Solution

We know that recursive solutions are not that efficient and it's always good to avoid it if it's possible. The problem however is equally simple in iterative form as well. The code below translates recursive form into iterative form.

  
 

//find kth element iteratively
Node * find_kth_element_iterative(Node *tree, int &k) {
 if(tree == 0)
  return 0;

 Node *node = tree;
 int count = 0;
 int pos = k;


 while(node != 0) {
  count = 0;

  //count nodes on the left subtree
  count_nodes(node->left_, count);

  //check if root is the median
  if( pos == count + 1)
   return node;

  //check if median falls on left subtree
  else if (count >= pos)
   node = node->left_;

  //median falls on right subtree
  else {
   pos -= (count + 1);
   node = node->right_;
  }

 }
 return 0;
}

Complexity

What's the complexity of above algorithms? While finding median You are moving along a path in the tree, so in the worst case it can be height of the tree (h) and also you are counting the number of nodes in left subtree at each step. Which requires O(n) time in the worst case. Thus the overall algorithm is O(nh).

Is there a way to improve it? Yes, of course ( at least one technique I know of). If you need to keep on computing the kth median on a single tree many times, then you can construct a tree with a count of elements in its subtree. If you do this the counting of number of nodes in left subtree takes O(1) time. Effectively the algorithm will reduce to O(h) algorithm. In case tree is balanced this is just O(log n). Not bad :).