CareerCup

wrong solution

class Stack{
int top;
int[] stack;
MinimumStack miniStack;
// constructor for stack
Stack(int capacity){
top=0;
stack=new int[capacity];
miniStack = new MinimumStack(capacity);
}
boolean isFull(){
return(top==stack.length);
}
public void push(int val){
if(!isFull()){
stack[top++]=val;
miniStack.push(val);
}else{
System.out.println("The stack is full");
}
}

public Object pop(){
if(top!=0){
miniStack.pop(stack[top-1]);
return stack[top--];
}else{
return null;
}
}

public void displayStack(){
for(int i=0;i<top;i++){
System.out.println(stack[i]);
}
}

public void displayMinimum(){
System.out.println("The current minimum is"+miniStack.getMinimum());
}

}
class MinimumStack{
int topMin;
int[] Ministack;
public MinimumStack(int capacity) {
// TODO Auto-generated constructor stub
topMin=0;
Ministack=new int[capacity];
}
public int getMinimum() {
// TODO Auto-generated method stub
if(topMin!=0){
return Ministack[topMin-1];
}
return (Integer)null;
}
boolean isMinStackFull(){
return(topMin==Ministack.length);
}

public void push(int val){
if(!isMinStackFull()){
if(topMin!=0){
if(val<Ministack[topMin-1])
Ministack[topMin++]=val;
}
else{
if(topMin==0)
Ministack[topMin++]=val;
}

}else{
System.out.println("The stack is full");
}
}
public void pop(int val){
if(Ministack[topMin]==val){
int Minval=Ministack[topMin--];
}
}


}

I have thise as an answer to your question. Only problem in this solution is I am failing to guess where EOF file is encountering.
{{
public void serializeBinaryTree(BTNode root, ObjectOutputStream oos) throws IOException{
//if nde is null then put some #
String s = "#";
if(root==null){
oos.writeObject("#");
}else{
oos.writeObject(" "+root.data);
serializeBinaryTree(root.left,oos);
serializeBinaryTree(root.right,oos);
}
}

public void dserializeBinaryTree() throws IOException, ClassNotFoundException{
//if nde is null then put some #
String node = null;
FileInputStream fin = new FileInputStream("Tree.ser");
ObjectInputStream ois = new ObjectInputStream(fin);
Scanner input = new Scanner("Tree.ser");
while(true){
System.out.println((String) ois.readObject());
}

}
}}

what do you mean by greater nodes ? Please give detail example of it.

This works with +ve as well as -ve Numbers{{
public static double getPower(double a, double b){

if(b==0) return 1;
//if b is negative
if(b<0){
return 1/a*(getPower(a,b+1));
}

return a*getPower(a,b-1);

}
}

}}

public static double getPower(double a, double b){

if(b==0) return 1;
//if b is negative
if(b<0){
return 1/a*(getPower(a,b+1));
}

return a*getPower(a,b-1);

}
}

you need to right customized hash function to decide index of an array where you wants to insert such value. The trick is to write such hash function o that you will not waste space and retrieve value in O(1) time.

{{
public static boolean isStringIsoMorphic(String str1, String str2){
if(str1.length()!=str2.length())
return false;
int len=str1.length();
//declaring two arrays to keep count of position of each char encountered
int[] str1arr = new int[256];
Arrays.fill(str1arr, -1);
int[] str2arr = new int[256];
Arrays.fill(str2arr, -1);
for(int i=0;i<len;i++){
if(str1arr[str1.charAt(i)]==-1){
if(str2arr[str2.charAt(i)]!=-1){
return false;
}else{
str1arr[str1.charAt(i)]=i;
str2arr[str2.charAt(i)]=i;
}

}else{ // if char in first string has already occurred somewhere
// then look for the char present at same index in str2 previous occurance
System.out.println(str1arr[str1.charAt(i)]);
System.out.println(str2arr[str2.charAt(i)]);
if(str1arr[str1.charAt(i)]!=str2arr[str2.charAt(i)])
return false;
}
}

return true;
}
}}