Question : What is the optimal algorithm to find the minimum element given a rotated sorted array of integers? A rotated sorted array of integers is the output array of a rotation operation performed on a sorted array. Eg:  3 5 6 1 2

Answer : One way to solve the above problem in linear time is to get the minima from the array using difference array and then finding the sign change from positive to negative.This will solve the problem in O(n) time.

/*
* Solution for O(n) time complexity
*/
int getRotationPoint(int *arr,int length)
{
    int *diff=(int *)malloc(sizeof(int)*length-1);
    if(length==1)
    return 0;
    assert(length>=2);
    for(int i=1;i0)?PS:NG;
    for(int i=0;i0)?PS:NG;
        if((sg==NG)&&(sign==PS))
        {
            index=i;
            break;
        }
        sign=sg;
    }
    return index+1;
}

However for an even tighter bound we modify the binary search algorithm and use it.We also could use the fact that for a sorted array, if the array were circular the index of the smallest would be the index after the largest element.Hence this solution is a O(log n) solution.

/*
* Solution for O(nlogn) time
*/
int getRotationPointBi(int *arr,int length)
{
    int first=0;
    int last=length-1;
    int middle=(last+first)/2;
    while(last>first)
    {
        middle=(last+first)/2;
        if(arr[first]>=arr[middle])
        last=middle;
        else if(arr[middle]>=arr[last])
        first=middle;
    }
    return middle+1;
}

Question : In a BST to every element add sum of elements greater than it.

Answer: The first thing is to notice that on doing this the resulting tree would be the reverse of the binary search tree. Implementing this  you should notice that the inorder travesal prints the elements in increasing order.Hence we need to modify the inorder traversal a bit to do this.Here is a sample C++/C code for insert and inorder for a BST.

#include<iostream>
#include<cstdlib>

using namespace std;

struct node
{
    int data;
    struct node *parent;
    struct node *left;
    struct node *right;
}*root;

typedef struct node node;

/**
*Makes a new node from the integer
*value and returns a pointer to the
*new created node
*/
node* makeNode(int num)
{
    node *temp=(node*)malloc(sizeof(node));
    temp->data=num;
    temp->right=NULL;
    temp->left=NULL;
    return temp;
}

/**
 * Inserting a node in a BST
 */
void insert(int num,node *T)
{
    if(T==NULL)
    root=makeNode(num);
    else
    {
        if(T->data>num)
        {
            if(T->left==NULL)
            T->left=makeNode(num);
            else
            insert(num,T->left);
        }
        else
        {
            if(T->right==NULL)
            T->right=makeNode(num);
            else
            insert(num,T->right);
        }
    }
}

/**
*Inorder traversal of the tree
*/
void inorder(node* T)
{
    if(T!=NULL)
    {
        inorder(T->left);
        cout<<T->data<<" ";
        inorder(T->right);
    }
}

Now to add the elements which are greater than a particular node we need to reverse the inorder by accessing the bigger elements i.e. right first and then the left. On accessing the right add the data to the data already collected and then change the node value accordingly.Hope you better understand by the code.

void chorder(node *T,int *data)
{
    if(T!=NULL)
    {
        chorder(T->right,data);
        *data+=T->data;
        T->data=*data;
        chorder(T->left,data);
    }
}

Notice that that data needs to be reference so that after being updated it is assigned to the other nodes.

PROBLEM : The problem is to rotate a matrix by 90. Fairly easy , just focus on double pointers if you are using C/C++ and make it for Clock Wise and anti clockwise.The code is as follows:

#include<iostream>
#include<cstdlib>
#define CL 0
#define ACL 1

using namespace std;

void printMatrix(int **matrix,int width,int height)
{
  for(int i=0;i<width;i++)
    {
      for(int j=0;j<height;j++)
 cout<<*(*(matrix+i)+j)<<" ";
      cout<<endl;
    }
}

void matrixRot90(int **matrix,int width,int height,int dir)
{
  //dir 0 for CL and 1 for ACL
  if(dir==CL)
    {
      for(int j=0;j<height;j++)
 {
   for(int i=width-1;i>=0;i--)
     cout<<*(*(matrix+i)+j)<<" ";
   cout<<endl;
 }
    }
  else
    {
      for(int j=height-1;j>=0;j--)
 {
   for(int i=0;i<width;i++)
     cout<<*(*(matrix+i)+j)<<" ";
   cout<<endl;
 }
    }
}

int main(void)
{
  int width,height;
  cin>>width>>height;
  int **matrix=(int**)malloc(sizeof(int**)*width);
  for(int i=0;i<width;i++)
    *(matrix+i)=(int*)malloc(sizeof(int)*height);
  for(int i=0;i<width;i++)
    for(int j=0;j<height;j++)
      cin>>*(*(matrix+i)+j);
  cout<<"Original Matrix::";
  printMatrix(matrix,width,height);
  cout<<"After clockwise rotation ::\n\n";
  matrixRot90(matrix,width,height,CL);
  cout<<"After anticlockwise rotation ::\n\n";
  matrixRot90(matrix,width,height,ACL);
  return 0;
}

PROBLEM : Given a directed graph check whether the graph contains a cycle or not.

SOLUTION : In a graph containing a cycle suppose vi,...vk are the vertices that constitute the cycle and suppose that the backedge is from vk to vi then on applying dfs to the graph "when the dfs covers the edge from vk to vi then vi must have already been visited." vi will already be in the recursion stack when the edge will be covered.Hence we know a cycle is present.Thus the given graph will not be cyclic.

Thus for each node visited earlier we need to store the recursion stack to tell whether a edge points to a node already in the recursion stack.Then it would be a back edge.In this way this algorithm can be modified to find in a graph:

• The number of back edges
• The components i.e. the trees in a forest
• The number of cycles = number of back edges .. etc

The cpp code for the following :

#include<iostream>
#include<list>

using namespace std;

class Graph
{
    int V; //number of vertices
    list<int> *adj; //pointer to array containing adjacency lists
    bool isCyclicUtil(int v,bool visited[],bool *rs);

    public:
        Graph(int v);
        void addEdge(int v,int w);
        bool isCyclic();
};

Graph::Graph(int v)
{
    this->V=v;
    adj=new list<int>[V];
}

void Graph::addEdge(int v,int w)
{
    adj[v].push_back(w);
}

bool Graph::isCyclicUtil(int v,bool visited[],bool *rs)
{
    if(visited[v]==false)
    {
        visited[v]=true;
        rs[v]=true;
        list<int>::iterator i;
        for(i=adj[v].begin();i!=adj[v].end();++i)
        {
            if(!visited[*i] && isCyclicUtil(*i,visited,rs))
            return true;
            else if(rs[*i])
            return true;
        }
    }
    rs[v]=false;
    return false;
}

bool Graph::isCyclic()
{
    bool *visited=new bool[V];
    bool *rs=new bool[V];
    for(int i=0;i<V;i++)
    {
        visited[i]=false;
        rs[i]=false;
    }
    for(int i=0;i<V;i++)
    if(isCyclicUtil(i,visited,rs))
        return true;
    return false;
}

int main()
{
    Graph g(4);
    g.addEdge(0,1);
    g.addEdge(0,2);
    g.addEdge(1,2);
    g.addEdge(2,0);
    g.addEdge(2,3);
    g.addEdge(3,3);

    if(g.isCyclic())
    cout<<"Graph contains a cycle"<<endl;
    else
    cout<<"Graph doesnot contain a cycle"<<endl;
    return 0;
}

Given a binary tree, print all the nodes in the Zigzag order.For eg.


 
 For the given tree the answer should be : 1324567

Answer:This question is fairly simple if we keep a flag to decide whether we want to print it in normal order or to print it in reverse order.The pseudo algorithm for the implementation of this is as follows:

def print(Queue Q,int order)
{
    newQueue=new Queue();
    while(iterate over Q)
    {
      node=Q[iter];
      newQueue.enqueue(node.rightchild());
      newQueue.enqueue(node.leftchild());
    }
    if(order==0)
    Q.reverse();
    while(Q.size()>0)
    print(Q.dequeue());
    if(newQueue.size()>0)
    print(newQueue,~order);
}

You should also start with:

Q.enqueue(Tree.root);
print(Q,0);

So you always used sqrt() function to find out the square roots of a number .. So what if i ask you to write a function that will display the square root without using the math library functions.. STUNNED ?
       Well lets try to figure out what we do in finding the square root of a number.So how do we start.The first thing that comes into my mind is that square root can be any real value less than the given number.Hence there are infinite possibilities and we cannot loop through to check if multiplying the number to itself will give us the number.That would be a never ending process and we do not know the increment value.So what do we do now ?
       Ok.What if we take any number as the initial guess and start.When we find the square of the number is greater than the number than we know we have to select a number that is less than the current and if it is smaller than choose a greater number.In this way after a fixed number of iterations we will eventually reach near to the square root.This method is also called as Approximation by Bisection.

def sqrt(x,epsilon):
    assert((x>0)&&(epsilon>0)),"Initial Conditions not met"
    low=0
    high=x
    val=(low+high)/2
    count=1
    while val**2-x>epsilon and crt<100:
        crt+=1
        if(val**2>x)
            high=val
        else:
            low=val
        val=(high+low)/2
   return val

However even this cannot find to much approximation and what if in count number of times it doesn't reach the required value.For this we have another method called the Newtons method where we use the initial guess to make a better guess.It works like taking 1 as the initial guess.Then

1+(2/1)=3  3/2=1.5
1.5+(2/1.5)=2.83  2.83/2=1.416 ~ 1.414 sqrt of 2

Hence using Newtons Method of Approximation we get:

float sqrt(float x)
{
 float ng;
 float g=x/2;
 float d=(g*g)-x;
 d=(d>0)?d:-d;
 while(d>EPS)
 {
  ng=(g+(x/g))*0.5;
  g=ng;
  d=(g*g)-x;
  d=(d>0)?d:-d;
  //cout<<g<<endl;
 }
 return g;
}