Problem Statement

Sherlock Holmes suspects his archenemy, Professor Moriarty, is once again plotting something diabolical. Sherlock’s companion, Dr. Watson, suggests Moriarty may be responsible for MI6’s recent issues with their supercomputer, The Beast.

Shortly after resolving to investigate, Sherlock receives a note from Moriarty boasting about infecting The Beastwith a virus; however, he also gives him a clue—a number, NN. Sherlock determines the key to removing the virus is to find the largest Decent Number having NN digits.

A Decent Number has the following properties:

1. Its digits can only be 3‘s and/or 5‘s.
2. The number of 3‘s it contains is divisible by 5.
3. The number of 5‘s it contains is divisible by 3.
4. If there are more than one such number, we pick the largest one.

Moriarty’s virus shows a clock counting down to The Beast‘s destruction, and time is running out fast. Your task is to help Sherlock find the key before The Beast is destroyed!

Constraints
1T201≤T≤20
1N1000001≤N≤100000

Input Format

The first line is an integer, TT, denoting the number of test cases.

The TT subsequent lines each contain an integer, NN, detailing the number of digits in the number.

Output Format

Print the largest Decent Number having NN digits; if no such number exists, tell Sherlock by printing -1.

Sample Input

4
1
3
5
11


Sample Output

-1
555
33333
55555533333


Explanation

For N=1, there is no decent number having 1 digit (so we print 1−1).
For N=3, 555 is the only possible number. The number 5 appears three times in this number, so our count of 5‘s is evenly divisible by 3 (Decent Number Property 3).
For N=5, 33333 is the only possible number. The number 3 appears five times in this number, so our count of 3‘s is evenly divisible by 5 (Decent Number Property 2).
For N=11, 55555533333 and all permutations of these digits are valid numbers; among them, the given number is the largest one.

Problem Statement

There are N sequences. All of them are initially empty, and you are given a variable lastans = 0. You are given Q queries of two different types:

• 1 x y” – Insert y at the end of the ((x XOR lastans) mod N)th sequence.
• 2 x y” – Print the value of the (y mod size)th element of the ((x XOR lastans) mod N)th sequence. Here, $size$ denotes the size of the related sequence. Then, assign this integer to lastans.

Note: You may assume that, for the second type of query, the related sequence will not be an empty sequence. Sequences and the elements of each sequence are indexed by zero-based numbering.

You can get more information about XOR from Wikipedia. It is defined as ^ in most of the modern programming languages.

Input Format

The first line consists of $N$, number of sequences, and $Q$, number of queries, separated by a space. The following $Q$ lines contains one of the query types described above.

Constraints
1 < N,Q < 10^5
0 < x < 10^9
0 < y < 10^9

Output Format

For each query of type two, print the answer on a new line.

Sample Input

2 5
1 0 5
1 1 7
1 0 3
2 1 0
2 1 1


Sample Output

7
3


Explanation

The first sequence is 5, 3 and the second sequence is 7.

Problem Statement:

Dexter and Debra are playing a game. They have N containers each having one or more chocolates. Containers are numbered from 1 to N, where ith container has A[i] number of chocolates.

The game goes like this. First player will choose a container and take one or more chocolates from it. Then, second player will choose a non-empty container and take one or more chocolates from it. And then they alternate turns. This process will continue, until one of the players is not able to take any chocolates (because no chocolates are left). One who is not able to take any chocolates loses the game. Note that player can choose only non-empty container.

The game between Dexter and Debra has just started, and Dexter has got the first Chance. He wants to know the number of ways to make a first move such that under optimal play, the first player always wins.

Input Format
The first line contains an integer N, i.e., number of containers.
The second line contains N integers, i.e., number of chocolates in each of the containers separated by a single space.

Output Format
Print the number of ways to make the first move such that under optimal play, the first player always wins. If the first player cannot win under optimal play, print 0.

Constraints
1 ≤ N ≤ 106
1 ≤ A[i] ≤ 109

Sample Input

2
2 3


Sample Output

1


Explanation

Only 1 set of moves helps player 1 win.

Player:      1      2      1      2      1
Chocolates: 2 3 -> 2 2 -> 1 2 -> 1 1 -> 0 1

Sample Data

Output:

321143

Theory Data for the solution and algorithm

Problem Statement

There are N strings. Each string’s length is no more than 2020 characters. There are also Q queries. For each query, you are given a string, and you need to find out how many times this string occurred previously.

Input Format

The first line contains N, the number of strings.
The next N lines each contain a string.
The N+2nd line contains Q, the number of queries.
The following Q lines each contain a query string.

Constraints

1N1000
1Q1000
1lengtof anstring20

Sample Input

4
aba
baba
aba
xzxb
3
aba
xzxb
ab


Sample Output

2
1
0


Explanation

Here, “aba” occurs twice, in the first and third string. The string “xzxb” occurs once in the fourth string, and “ab” does not occur at all.

Bubble Sort Algorithm

Below is the code to do Bubble Sorting in Java for integer values

public class BubbleSort {

// logic to sort the elements
public static void srtbubble(int array[]) {
int n = array.length;
int k;
for (int m = n; m >= 0; m--) {
for (int i = 0; i < n - 1; i++) { k = i + 1; if (array[i] > array[k]) {
swap(i, k, array);
}
}
print(array);
}
}

private static void swap(int i, int j, int[] array) {

int temp;
temp = array[i];
array[i] = array[j];
array[j] = temp;
}

private static void print(int[] input) {

for (int i = 0; i < input.length; i++) {
System.out.print(input[i] + ", ");
}
System.out.println("\n");
}

public static void main(String[] args) {
int[] input = { 14, 21, 19, 62, 233, 142, 134, 10, 41 };
srtbubble(input);
}
}




Output:

14, 19, 21, 62, 142, 134, 10, 41, 233,

14, 19, 21, 62, 134, 10, 41, 142, 233,

14, 19, 21, 62, 10, 41, 134, 142, 233,

14, 19, 21, 10, 41, 62, 134, 142, 233,

14, 19, 10, 21, 41, 62, 134, 142, 233,

14, 10, 19, 21, 41, 62, 134, 142, 233,

10, 14, 19, 21, 41, 62, 134, 142, 233,

10, 14, 19, 21, 41, 62, 134, 142, 233,

10, 14, 19, 21, 41, 62, 134, 142, 233,

10, 14, 19, 21, 41, 62, 134, 142, 233,

Finding loop in a singly linked-list

You can detect it by simply running two pointers through the list.
Start the first pointer p1 on the first node and the second pointer p2 on the second node.

Advance the first pointer by one every time through the loop, advance the second pointer by two. If there is a loop, they will eventually point to the same node. If there’s no loop, you’ll eventually hit the end with the advance-by-two pointer.

Consider the following loop:

head -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8
^                        |
|                        |
+------------------------+

Starting A at 1 and B at 2, they take on the following values:

p1   p2
=    =
1    2
2    4
3    6
4    8
5    4
6    6


Because they’re equal, and P2 should always be beyond p1 in a non-looping list (because it’s advancing by two as opposed to the advance-by-one behavior of p1), it means you’ve discovered a loop.

The pseudo-code will go something like this:

• If hn==null; return false; //empty list
• if hn.next!=null
• p1=hn; p2=hn;
• While p2!=null loop
• p1=p1.next//Advancing p1 by 1
• if p2.next!=null
• p2=p2.next.next//Advancing p2 by 2
• else return false
• if p1==p2
• return true// Loop was found
• return false// till now no loop was found

Once you know a node within the loop, there’s an O(n) guaranteed method to find the start of the loop.

Let’s return to the original position after you’ve found an element somewhere in the loop but you’re not sure where the start of the loop is.

                                 p1,p2 (this is where p1 and p2
|               first met).
v
head -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8
^                        |
|                        |
+------------------------+


This is the process to follow:

• First, advance p2 and set the loopsize to 1.
• Second, while p1 and p2 are not equal, continue to advance p2, increasing the loopsize each time. That gives the size of the loop, six in this case.
If the loopsize ends up as 1, you know that you must already be at the start of the loop, so simply return A as the start, and skip the rest of the steps below.
• Third, simply set both p1 and p2 to the first element then advance p2 exactly loopsize times (to the 7 in this case). This gives two pointers that are different by the size of the loop.
• Lastly, while p1 and p2 are not equal, you advance them together. Since they remain exactly loopsize elements apart from each other at all times, p1 will enter the loop at exactly the same time as p2 returns to the start of the loop. You can see that with the following walk through:
• loopsize is evaluated as 6
• set both p1 and p2 to 1
• advance p2 by loopsize elements to 7
• 1 and 7 aren’t equal so advance both
• 2 and 8 aren’t equal so advance both
• 3 and 3 are equal so that is your loop start

Now, since each those operations are O(n) and performed sequentially, the whole thing is O(n).

Source

Algorithm Name: Floyd–Warshall algorithm

Problem Statement:

There is a building of 100 floors  If an egg drops from the Nth floor or above it will break. If it’s dropped from any floor below, it will not break. You’re given 2 eggs.

Give an Algorithm to find N, while minimizing the number of drops for the worst case.

Solution:

The Problem comes with a general equation as X(Max Number of floors) = n*n;

So in this case for example, X= 100 Floors( => n=10)

now use the value n to traverse. This means that drop the first egg from 10th floor, then 20th floor, then 30th and so on up to 90.

The floor from which it breaks, say 60th, then go back 10 floors and try each floor to find N.

After Floor 90, Start moving 1 by 1.

Worst case scenario:

eggs break at 99th floor.

Egg drops: 10 -> 20 -> 30 -> 40 -> 50 -> 60 -> 70 -> 80 -> 90 -> 91 -> 92 -> 93 -> 94 -> 95 -> 96 -> 97 -> 98 -> 99

Max Attempts: 18

The water jug problem

We have three water jugs, and each can hold 3oz., 5oz., and 8oz. of water, respectively.
Without the possibility of water spilling when poured from one jug to another, and given that the jugs have no calibration, how do we divide the 8oz. of water equally among two jugs?

We will define a class named State holding the capacity of A and B jars.
It should be noted that only 2 jars are sufficient to define a state, as water held in third jar can be calculated by subtracting the sum of two from the total.

Define class State like this…

package mystate;import bfs.threejugproblem.NotSupportedException;import java.util.ArrayList;import java.util.List;public class State{    int a=0;//3    int b=0;//5    int c=8;//8    public State(int a, int b)    {        this.a=a;        this.b=b;        this.c=8-a-b;    }    public boolean isGoal()    {        return (b==4 && c==4);    }    public boolean equals(Object xx)    {        State x = (State) xx;        if(this.a==x.a && this.b==x.b && this.c==x.c)        {            return true;        }        else        {            return false;        }    }    public int hashCode()    {        return 8;    }    public List getChildren()     {        List children = new ArrayList();        // a -> b        if(a!=0 && b!=5)// if a is not empty        {            if(a+b<=5)            {                children.add(new State(0, a+b));            }            else            {                children.add(new State(a+b-5,5));            }        }        //a->c        if(a!=0 && c!=8)        {            // We are pouring completely from a to c            // a will be 0            // b will be 8-a-c            // c will be a+c            children.add(new State(0, 8-a-c));        }        //b->a        if(b!=0 && a!=3)        {            if(a+b<=3)            {                children.add(new State(a+b, 0));            }            else            {                children.add(new State(3, a+b-3));            }        }        // b->c        if(b!=0 && c!=8)        {            // We are pouring completely from b to c            // a will be 8-b-c            // b will be 0            // c will be b+c            children.add(new State(8-b-c, 0));                    }        //c->a        if(c!=0 && a!=3)        {            if(c+a<=3)            {                children.add(new State(c+a, 8-c-a));            }            else            {                    // a will be full i.e. 3 liters                    // b will be 8-c-a                    // c will be c+a-3                    children.add(new State(3, 8-c-a));                            }        }        // c->b        if(c!=0 && b!=5)        {            if(c+b<=5)            {                children.add(new State(8-c-b , c+b));            }            else            {                children.add(new State(8-c-b, 5));            }        }        return children;    }    @Override    public String toString()    {        return "{"+a+","+b+","+c+"}";    }}

Depth First Search Algorithm

public class DFSThreeJugProblem {    public static void main(String[] args)    {        State currentState = new State(0,0);        List visitedStates=new ArrayList();          // Check if the current State has a solution        // given a set of visited States.        dfs(currentState, visitedStates);    }    public static void dfs(State currentState, List vStates)    {        // if it is GOAL        if(currentState.isGoal())        {            // That's it we are done.            for(State v : vStates)            {                System.out.println(v);                System.out.println(currentState);            }            System.exit(0);                    }                // if visisted state contains currentState, then just return.        // This is the wrong branch, and we need not traverse it further.        if(vStates.contains(currentState))            return;                // Add current state to visited states.        vStates.add(currentState);                        // Make clone of visited states.        List clonedVStates = new ArrayList(vStates);        // Find the set of possible children of current state.        List children = currentState.getChildren();        for(State c : children)        {            // if a children C is not in the visited states             // again call DFS on current child and visited States.            if(!clonedVStates.contains(c))            {                dfs(c, clonedVStates);            }        }    }}

Breadth First Search algorithm…

public class BFSThreeJugProblem {    private static List visitedStates=new ArrayList();    private static Queue stateQueue = new LinkedList();    public static void main(String[] args) throws NotSupportedException     {        State currentState = new State(0,0);        // Add current state to state Queue.        stateQueue.add(currentState);        do        {            // Get the first Element from Queue.            State firstElementInQueue = stateQueue.peek();            // If the first Element is the Goal            // We are done.            if(firstElementInQueue.isGoal())            {                for(State p : visitedStates)                {                    System.out.println(p.toString());                }                // There is no recursion here, so simple return would do.                return;            }            else            {                // Add firstElement to visited States                visitedStates.add(firstElementInQueue);                    // Get the children of first element                List children = firstElementInQueue.getChildren();                for(State v : children)                {                    // if children has not already been visited.                    if(!visitedStates.contains(v))                    {                        // add the child to state Queue.                        stateQueue.add(v);                    }                }                // Remove the first element from state queue.                stateQueue.remove(firstElementInQueue);            }            // do this till state queue is empty.        }while(!stateQueue.isEmpty());    }}