Deleting a middle node from a single linked list when pointer to the previous node is not available

This updated information has been further expanded upon on my new website. You can find the updated details here: https://k5kc.com/cs/algorithms/delete-non-tail-node-in-linked-list-given-only-access-to-that-node/.

Problem

Implement an algorithm to delete a node in the middle of a single linked list, given only access to that node.

EXAMPLE
Input: the node ‘c’ from the linked list a->b->c->d->e
Result: nothing is returned, but the new linked list looks like a->b->d->e

Solution
If we are allowed to make some assumption, it can be solved in O(1) time. To do it, the strictures the list points to must be copyable. The algorithm is as the following:
We have a list looking like:

......... -> Node(i-1) -> Node(i) -> Node(i+1) -> ... 

and we need to delete Node(i).

1. Copy data (not pointer, the data itself) from Node(i+1) to Node(i), the list will look like: ... -> Node(i-1) -> Node(i+1) -> Node(i+1) -> ...
2. delete the second Node(i+1), it doesn't require pointer to the previous node.

Pseudocode:
As the solution given here:

void delete_node(Node* pNode)
{
    pNode->Data = pNode->Next->Data;  // Assume that SData::operator=(SData&) exists.
    Node* pTemp = pNode->Next->Next;
    delete(pNode->Next);
    pNode->Next = pTemp;
}

Here the problem is we are not actually deleting the node, but just replacing the value from the next node. Hence it is sort of quiz. Moreover, if some node is already referring to pointer(i+1) (not data), then it may be problem.

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Array implementation of stack

Here will discuss the array implementation of stack.

Problems with array implementation
Underflow - Array may be empty but people may try to pop the element
Overflow - Array is full. To over come we have to use re-szing of array. We can see how we can solve this problem here - http://k2code.blogspot.com/2013/09/resizing-array-implementation-of-stack.html
Null items - Can nulls be added - Yes in this case, nulls can be added in stack
Loitering (java specific) - Holding a reference to an object wen it is no longer needed. To over come this we should explicitly set array index to null. For eg.

public string pop() //stack of string
{
  String item = stackArray[--top];
  stackArray[top] = null;
  return item;
}

Doing so, makes your program less of memory consumer.

Stack in java
Following is the implementation of stack in java:

public class MyStack {
   private int maxSize;
   private long[] stackArray;
   private int top;
   public MyStack(int s) {
      maxSize = s;
      stackArray = new long[maxSize];
      top = -1;
   }
   public void push(long j) {
      stackArray[++top] = j;
   }
   public long pop() {
      return stackArray[top--];
   }
   public long peek() {
      return stackArray[top];
   }
   public boolean isEmpty() {
      return (top == -1);
   }
   public boolean isFull() {
      return (top == maxSize - 1);
   }
   public static void main(String[] args) {
      MyStack theStack = new MyStack(10); 
      theStack.push(10);
      theStack.push(20);
      theStack.push(30);
      theStack.push(40);
      theStack.push(50);
      while (!theStack.isEmpty()) {
         long value = theStack.pop();
         System.out.print(value);
         System.out.print(" ");
      }
      System.out.println("");
   }
}

CPP implementation (without classes)
Following is the implementation in cpp:

#include <stdio .h>  
 #include <conio .h>  
 #define MAX 20  
 int top=-1,ch;  
 int a[MAX];  
 int main(){  
   void menu();  
   void push();  
   void pop();  
   void display();  
   menu();  
   while(ch!=3){  
      switch(ch){  
           case 1:  
               push();  
               display();  
               break;  
           case 2:  
               pop();  
               display();  
               break;  
           }  
      menu();       
      }  
 }  
 void push()  
 {  
    if(top==MAX-1)  
     printf("Stack is full\nStack overflow\n");  
    else  
    {  
     printf("Enter element:");  
     scanf("%d",&a[++top]);  
    }  
 }  
 void pop()  
 {  
    if(top==-1)  
    printf("Stack is emplty\nSTACK UNDERFLOW\n");  
    else  
    {  
      top--;  
      printf("Deleted %d\n",a[top+1]);  
    }  
 }  
 void display()  
 {  
    int i=0;  
    printf("\nStack is :\n");  
    for(i=top;i>=0;i--)  
    printf("%d\n",a[i]);  
 }  
 void menu()  
 {  
    printf("\n1.Push\n2.Pop\n3.Exit\n");  
    scanf("%d",&ch);  
    while(ch < 1 ch=>3)  
    {  
     printf("Enter valid choice 1,2 or 3\n");  
     scanf("%d",&ch);  
    }  
 }  

Stack Implementation in CPP (using classes)

#include <iostream>
using namespace std;
#define STACKSIZE 10
class stack
{
        private:
                int arr[STACKSIZE+1];
                int tos;
        public:
                stack();
                void push(int x);
                int  pop();
                bool is_empty();
                bool is_full();
                int  size();
                void display();
};
stack::stack()
{
        tos = 0;
}
void stack::push(int x)
{
        if(!is_full())
                arr[tos++] = x;
        else
                cout << "Stack is full, Can not push " << x << endl;
}
int stack::pop()
{
        if(!is_empty())
                return arr[--tos];
        else
                cout << "Stack is empty, cannot pop" << endl;
        return -1;
}
bool stack::is_empty()
{
        if(tos == 0)
                return true;
        else
                return false;
}
bool stack::is_full()
{
        if(tos == STACKSIZE)
                return true;
        else
                return false;
}
int stack::size()
{
        return tos;
}
void stack::display()
{
        if(tos == 0)
        {
                cout << "No elements to display" << endl;
                return;
        }
        for(int i=0;i
                cout << arr[i] << " ";
        cout << endl;
}
int main()
{
        stack mystack;
        cout << mystack.size() << endl;
        mystack.push(1);
        if(mystack.is_full())
                cout << "stack is full" << endl;
        mystack.pop();
        if(mystack.is_empty())
                cout << "stack is empty" << endl;
}

Thanks.

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5 Pirates Problem

Problem :

Five pirates have 100 gold coins. they have to divide up the loot. in order of seniority (suppose pirate 5 is most senior, pirate 1 is least senior), the most senior pirate proposes a distribution of the loot. they vote and if at least 50% accept the proposal, the loot is divided as proposed. otherwise the most senior pirate is executed, and they start over again with the next senior pirate. what solution does the most senior pirate propose? assume they are very intelligent and extremely greedy (and that they would prefer not to die).

Solution :

(to be clear on what 50% means, 3 pirates must vote for the proposal when there are 5 for it to pass. 2 if there are 4. 2 if there are 3. etc… )

Most of the time i get people who give answers like “the most senior pirate takes half and divides the rest up among the least senior pirates.” um, you missed the whole point to begin with. sorry.

any answer without a specific logic behind it is invalid. if i ask you why pirate 5 gave x coins to pirate 1, please don’t say “because he’s nice”.

now for the real solution. pirate 5 being the most senior knows that he needs to get 2 other people to vote for his solution in order for him not to be executed. so who can he get to vote for him, and why would they choose to vote for him? if you start thinking that pirate 4 will never vote for him, because he would rather have 5 die and then be in charge and take it all for himself, you are on the right track. but it gets more complicated.

lets consider if there were only 1 pirate. obviously he would take it all for himself and no one would complain.

if there were 2 pirates, pirate 2 being the most senior, he would just vote for himself and that would be 50% of the vote, so he’s obviously going to keep all the money for himself.

if there were 3 pirates, pirate 3 has to convince at least one other person to join in his plan. so who can he convince and how? here is the leap that needs to be made to solve this problem. pirate 3 realizes that if his plan is not adopted he will be executed and they will be left with 2 pirates. he already knows what happens when there are 2 pirates as we just figured out. pirate 2 takes all the money himself and gives nothing to pirate 1. so pirate 3 proposes that he will take 99 gold coins and give 1 coin to pirate 1. pirate 1 says, well, 1 is better than none, and since i know if i don’t vote for pirate 3, i get nothing, i should vote for this plan.

now we know what happens when there are 3 pirates. so what happens with 4? well pirate 4 has to convince 1 other person to join in his plan. he knows if he walks the plank then pirate 3 will get 99 coins and pirate 1 will get 1 coin. pirate 4 could propose giving pirate 1 two coins, and surely pirate 1 would vote for him, since 2 is better than 1. but as greedy as he is, pirate 4 would rather not part with 2 whole coins. he realizes that if he gets executed, then pirate 3’s scenario happens and pirate 2 gets the shaft in that scenario (he gets zero coins). so pirate 4 proposes that he will give 1 coin to pirate 2, and pirate 2 seeing that 1 is better than 0 will obviously vote for this plan.

a common objection is that pirate 2 is not guaranteed to vote for this plan since he might hope for the case when there are only 2 pirates and then he gets all the booty. but that is why i said that the pirates are extremely intelligent. pirate 2 realizes that pirate 3 is smart enough to make the optimal proposal, so he realizes that there will never be 2 pirates left, because 3 doesn’t want to die and we just showed that 3 has a winning proposal.

so lets sum up at this point

Pirate 1  2  3  4  5

    5. ?  ?  ?  ?  ?

    4. 0  1  0 99  -

    3. 1  0 99  -  -

    2. 0 100 -  -  -

    1.100

once you see the pattern it becomes very clear. you have to realize that when a pirate’s plan does not succeed then that means you are in the same situation with one less pirate.

1. pirate 1 needs 0 other people to vote for him. so he votes for himself and takes all the money. 2. pirate 2 needs 0 other people to vote for him. so he votes for himself and takes all the money. pirate 1 gets 0. 3. pirate 3 needs 1 other person to vote for him. he gives 1 coin to pirate 1 for his vote - if we are reduced to 2 pirates, pirate 1 gets 0 so pirate 1 knows 1 is better than none. pirate 3 takes 99. pirate 2 gets 0. 4. pirate 4 needs 1 other person to vote for him. he gives 1 coin to pirate 2 - if we reduce to 3 pirates, pirate 2 gets 0 so pirate 2 knows 1 is better than none. pirate 4 takes 99. pirate 3 gets 0. pirate 1 gets 0. 5. pirate 5 needs 2 other people to vote for him. its clear now that the 2 people he needs to convince are the 2 who get shafted in the 4 pirate scenario - pirate 3 and pirate 1. so he can give them each 1 coin (which is better than 0 - what they would get otherwise) and keep 98 for himself.

Pirate 1  2  3  4  5

    5. 1  0  1  0 98

what happens if there are 15 pirates? pirate 15 needs 7 other people to vote for him, so he recruits pirates 13,11,9,7,5,3, and 1 with 1 coin each and keeps 93 coins himself. those pirates will all vote for him because they know that they get 0 coins if he dies and pirate 14 is in charge.

hope you enjoyed this one. its my favorite interview question of all. it really allows the candidate to ask a lot of interesting questions and its really amazing when they reach the solution all by themselves. 

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100 Programmers Problem

Problem :

100 programmers are lined up in a row by an assassin. the assassin puts red and blue hats on them. they can’t see their own hats, but they can see the hats of the people in front of them. the assassin starts in the back and says “what color is your hat?” the fogcreek programmer can only answer “red” or “blue.” the programmer is killed if he gives the wrong answer; then the assassin moves on to the next programmer. the programmers in front get to hear the answers of the programmers behind them, but not whether they live or die. they can consult and agree on a strategy before being lined up, but after being lined up and having the hats put on, they can’t communicate in any way other than those already specified. what strategy should they choose to maximize the number of programmers who are guaranteed to be saved?

Solution :

this is a very difficult problem to solve during an interview (especially if you’ve already taxed the candidate’s brain). look for obvious solutions first, and the reasoning behind them and then try to lead them to the ultimate solution.

a logical answer could be all the programmers would just say “red” and that way about half of them would survive on average, assuming the hats were distributed randomly.

this is a good start and should naturally lead to having every other programmer say the color of the hat in front of them. the first programmer would say the color of the hat in front of him, then the next programmer would just say that color that was just said. so we can guarantee that half survive - the even numbered programmers (since the person behind them told them the answer). and potentially if the hats were distributed randomly some of the programmers would get lucky and the hat in front of them would be the same color as their own. so this strategy should save more than half, and on average 75% of them would live.

at this point, if the solution is not clear, the candidate may give answers like, “they could agree that if they said their hat color in a soft voice, it means the hat in front of them is the same color, and if they say it in a loud voice, it means the hat in front is a different color”. this is definitely good and on the correct track. another option is they could say “reeeeeeeeeeed” for x number of seconds, where x represented the distribution of hats where a hat was a bit in a binary number, (red = 1, blue = 0). another interesting answer. there are many others like these that “bend” the rules and come to a solution.

but the real solution acknowledges that the programmers can only say “red” or “blue” and cannot alter their voice in such a convincing way as to signal any information other than the word they said. a good way to get this point across, is simply to change the problem slightly by saying “the assassin gets to hear their plan before she puts the hats on, and so will try to thwart the plan however she can.”

so if they decide to all say “red”, she’ll put blue hats on all of them. if they decide to all say the color of the hat in front of them, she’ll alternate the hats on every head, guaranteeing half will die. even with the assassin hearing their plan, there is still a way to save almost everyone.

we know that the first person is never going to have any information about the color of their hat, so they cannot be guaranteed to survive. but, i’ll give you a hint to the solution: i can save every other person for sure.

solution: they agree that if the number of red hats that the back person can see is even, that programmer will say “red”. if they add up to an odd number, they will say “blue”. this way number 99 can look ahead and count the red hats. if they add up to an even number and number 100 said “red”, then 99 must be wearing a blue hat. if they add up to an even number and number 100 said “blue”, signalling an odd number of red hats, number 99 must also be wearing a red hat. number 98 knows that 99 said the correct hat, and so uses that information along with the 97 hats in front to figure out what color hat is on 98’s head.

sample:

100  99  98  97  96  95  94 ... facing ->

 R    B   B   R   B   R   B ... -> 45 R and 48 B

this shows #100 wearing a red hat, 99 a blue, 98 a blue, 97 a red, 96 a blue, 95 a red, 94 a blue and 45 red hats - 48 blue hats on the people in front of them.

100 counts up the red hats: 47 total. so 100 says “blue”. the assassin kills 100. 99 counts up the red hats in front: 47. 100 said blue, so 100 saw an odd number. 99 sees an odd number, so 99 says “blue” and lives. 98 had counted 47 red hats, and 99 didn’t say “red” so thats still the total. 98 says “blue”. 97 counts up and finds 46 red hats. 99 and 98 didn’t say “red”, so his count is missing a red hat (its on his head, he realizes). he says “red”. 96 heard the “red” and now knows that there are an even number of “red” hats in front of 95. 96 sees 46, so he knows he has a “blue” hat. etc…

even if the assassin knows the plan, she can’t thwart it. she hears the plan, but she still has to put the hats on their heads. the plan doesn’t rely on any ordering of the hats, so the worst the assassin can do is to make sure #100 gets killed and thats the worst damage she can do.

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Sum it Up

Problem: 

you are given a sequence of numbers from 1 to n-1 with one of the numbers repeating only once. (example: 1 2 3 3 4 5). how can you find the repeating number? what if i give you the constraint that you can’t use a dynamic amount of memory (i.e. the amount of memory you use can’t be related to n)?
what if there are two repeating numbers (and the same memory constraint?)

Solution:

as a programmer, my first answer to this problem would be make a bit vector of size n, and every time you see the number, set its correspond index bit to 1. if the bit is already set, then that’s the repeater. since there were no constraints in the question, this is an ok answer. its good because it makes sense if you draw it for someone, whether they are a programmer, mathemetician, or just your grandpa. its not the most efficient answer though.

now, if i add the constraint that you can only use a fixed amount of memory (i.e. not determined by n) and it must run in O(n) time… how do we solve it. adding all the numbers up from 1 to n-1 would give us a distinct sum. subtracting the total sum of all the numbers from the sum of n to n-1 ( which is (n)(n-1)/2 ) would give us the secret extra number.

what if you can only use a fixed amount of memory, and two of the numbers are repeated? we know that the numbers have a distinct sum, and the difference would be equal to the sum of our unknowns

c = a + b

where c is the sum and a and b are the unknowns - c is a constant

if we had another similar formula we could solve the two unknown equations. my first thought was that the numbers would have a distinct product - (n-1)!

if we divide the total product by the (n-1)! product, we would get another equation

c2 = ab

we could then solve the two equations to get them into quadratic formula notation

0 = ax^2 + bx + c

and solve for the two values of x. this answer is correct but factorial grows really fast.

some sort of sum would be better. the sum of the squares from n-1 to 1 would work. that would yield a function of the form

c2 = a^2 + b^2

which could also be solved by using the quadratic equation.

i think its fine to remind someone of the quadratic equation… (maybe only because i myself had to look it up to solve the problem) i mean really though, the last time i used it was probably in 10th grade. as long as they get the idea that given two unknowns and two equations you can solve for the unknowns - thats the point.

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Palindromes

Problem: 

This year on October 2, 2001, the date in MM/DD/YYYY format will be a palindrome (same forwards as backwards).
10/02/2001
when was the last date that this occurred on? (see if you can do it in your head!)

Solution:

we know the year has to be less than 2001 since we already have the palindrome for 10/02. it can’t be any year in 1900 because that would result in a day of 91. same for 1800 down to 1400. it could be a year in 1300 because that would be the 31st day. so whats the latest year in 1300 that would make a month? at first i thought it would be 1321, since that would give us the 12th month, but we have to remember that we want the maximum year in the 1300 century with a valid month, which would actually be 1390, since 09/31 is a valid date.

but of course, a question like this wouldn’t be complete without an aha factor. and of course, there are not 31 days in september, only 30. so we have to go back to august 08 which means the correct date would be 08/31/1380.

palindromes also offer another great string question.

write a function that tests for palindromes

bool isPalindrome( char* pStr )

if you start a pointer at the beginning and the end of the string and keep comparing characters while moving the pointers closer together, you can test if the string is the same forwards and backwards. notice that the pointers only have to travel to the middle, not all the way to the other end (to reduce redundancy).

bool isPalindrome( char* pStr )

{

  if ( pStr == NULL )

   return false;

  char* pEnd = pStr;

  while ( *pEnd != '\0' )

    pEnd++;

  pEnd--;

  while(pEnd > pStr)

  {

    if ( *pEnd != *pStr )

      return false;

    pEnd--;

    pStr++;

  }

  return true;

}

thanks to tom for sending me this one! congrats on the wedding…

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Daughters’ Ages

Problem :

Two MIT math grads bump into each other at Fairway on the upper west side. They haven’t seen each other in over 20 years.

the first grad says to the second: “how have you been?”
second: “great! i got married and i have three daughters now”
first: “really? how old are they?”
second: “well, the product of their ages is 72, and the sum of their ages is the same as the number on that building over there..”
first: “right, ok.. oh wait.. hmm, i still don’t know”
second: “oh sorry, the oldest one just started to play the piano”
first: “wonderful! my oldest is the same age!”

How old are the daughters?

Solution :

Start with what you know. you know there are 3 daughters whose ages multiply to 72. let’s look at the possibilities…

Ages:            Sum of ages:

1 1 72            74

1 2 36            39

1 3 24            28

1 4 18            23

1 6 12            19

1 8 9             18

2 2 18            22

2 3 12            17

2 4 9             15

2 6 6             14

3 3 8             14

3 4 6             13

after looking at the building number the man still can’t figure out what their ages are (we’re assuming since he’s an MIT math grad, he can factor 72 and add up the sums), so the building number must be 14, since that is the only sum that has more than one possibility.

finally the man discovers that there is an oldest daughter. that rules out the “2 6 6” possibility since the two oldest would be twins. therefore, the daughters ages must be “3 3 8”.

(caveat: an astute reader pointed out that it is possible for two siblings to have the same age but not be twins, for instance one is born in january, and the next is conceived right away and delivered in october. next october both siblings will be one year old. if a candidate points this out, extra credit points to him/her.)

this question is pretty neat, although there is certainly a bit of an aha factor to it. the clues are given in such a way that you think you are missing information (the building number), but whats important isn’t the building number, but the fact that the first man thought that it was enough information, but actually wasn’t.

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int atoi( char* pStr )

Problem: 

Write the definition for this function without using any built-in functions. if pStr is null, return 0. if pStr contains non-numeric characters, either return 0 (ok) or return the number derived so far (better) (e.g. if its “123A”, then return 123). assume all numbers are positive. plus or minus signs can be considered non-numeric characters. in order to solve this program, the programmer must understand the difference between the integer 0 and the character ‘0’, and how converting ‘0’ to an int, will not result in 0. in other words, they have to understand what ascii is all about.

Solution :

string manipulation functions are great programming questions. they test whether the user can understand and translate into code simple algorithms. string functions test pointer arithmetic which usually shows a knowledgeable programmer. also there are usually multiple solutions, some more efficient than others. plus people use them all the time so they should understand how they work. my favorite is atoi and i start the problem like this:

int atoi( char* pStr )

write the definition for this function without using any built-in functions. if pStr is null, return 0. if pStr contains non-numeric characters, either return 0 (ok) or return the number derived so far (better) (e.g. if its “123A”, then return 123). assume all numbers are positive. plus or minus signs can be considered non-numeric characters. in order to solve this program, the programmer must understand the difference between the integer 0 and the character ‘0’, and how converting ‘0’ to an int, will not result in 0. in other words, they have to understand what ascii is all about. if they are stuck solving this problem, just ask them first to write:

charToInt(char c)

if they can’t do that then they basically missed half the problem. any moderately talented programmer who has a CS degree knows how to convert a char to an int. (note i said convert, not cast. charToInt('9') should return 9.)

when they start to solve the problem you will notice that they must make a choice in how they will process the string - from left to right or right to left. i will discuss both methods and the difficulties encountered in each.

“right to left” - this method starts at the right hand letter of the string and converts that character to an int. it then stores this value after promoting it to its correct “tens” place.
int atoi( char* pStr )
{
  int iRetVal = 0;
  int iTens = 1;

  if ( pStr )
  {
    char* pCur = pStr;
    while (*pCur)
      pCur++;

    pCur--;

    while ( pCur >= pStr && *pCur <= '9' && *pCur >= '0' )
    {
      iRetVal += ((*pCur - '0') * iTens);
      pCur--;
      iTens *= 10;
    }
  }
  return iRetVal;
}

“left to right” - this method keeps adding the number and multiplying the result by ten before continuing to the next number. e.g. if you had “6234” and you processed from left to right you’d have 6, then if you kept reading you’d multiply your result by 10 (6*10) to add a zero for where the next number would go. 60, and then you’d slide the 2 into the zero place you just made. 62. do it again, 620, slide the next number in, 623.
int atoi( char* pStr )
{
  int iRetVal = 0;

  if ( pStr )
  {
    while ( *pStr && *pStr <= '9' && *pStr >= '0' )
    {
      iRetVal = (iRetVal * 10) + (*pStr - '0');
      pStr++;
    }
  }
  return iRetVal;
}

i think the “left to right” method is a little bit cleaner, or maybe its just cooler. but both are “correct”.

remember that debugging code on paper is somewhat hard. most programmers aren’t used to studying code that much when you can just hit F-7, compile and see if the compiler barfs or not. if you notice an error, just ask them to step through a sample string drawing out what is happening with all the variables and the pointers in every step. they should find their mistake then and fix it (no points deducted).

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Bumblebee

Problem: 

two trains enter a tunnel 200 miles long (yeah, its a big tunnel) travelling at 100 mph at the same time from opposite directions. as soon as they enter the tunnel a supersonic bee flying at 1000 mph starts from one train and heads toward the other one. as soon as it reaches the other one it turns around and heads back toward the first, going back and forth between the trains until the trains collide in a fiery explosion in the middle of the tunnel (the bee survives). how far did the bee travel?

Solution: 

This puzzle falls pretty high on my aha scale. my first inclination when i heard it was to think “ok, so i just need to sum up the distances that the bee travels…” but then you quickly realize that its a difficult (not impossible) summation which the interviewer could hardly expect you to answer (unless i guess if you are looking for a job as a quant). “there must be a trick” you say. eh, sort of i guess, enough to say that this question is a stupid interview question.

the tunnel is 200 miles long. the trains meet in the middle travelling at 100 mph, so it takes them an hour to reach the middle. the bee is travelling 1000 mph for an hour (since its flying the whole time the trains are racing toward one another) - so basically the bee goes 1000 miles.

there is no process to explain, so this question can’t possibly teach you anything about the person. they either know it or they don’t and if they already knew it before you asked, you’re not going to be able to tell when they give you the answer. so don’t ask this question. and if someone asks you this question, just tell them you’ve already heard it before.

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100 Doors in a Row - Solution

Problem: 

you have 100 doors in a row that are all initially closed. you make 100 passes by the doors starting with the first door every time. the first time through you visit every door and toggle the door (if the door is closed, you open it, if its open, you close it). the second time you only visit every 2nd door (door #2, #4, #6). the third time, every 3rd door (door #3, #6, #9), etc, until you only visit the 100th door.
What state are the doors in after the last pass? which are open which are closed?

 

Solution :

For example, after the first pass every door is open. on the second pass you only visit the even doors (2,4,6,8…) so now the even doors are closed and the odd ones are opened. the third time through you will close door 3 (opened from the first pass), open door 6 (closed from the second pass), etc..

you can figure out that for any given door, say door #42, you will visit it for every divisor it has. so 42 has 1 & 42, 2 & 21, 3 & 14, 6 & 7. so on pass 1 i will open the door, pass 2 i will close it, pass 3 open, pass 6 close, pass 7 open, pass 14 close, pass 21 open, pass 42 close. for every pair of divisors the door will just end up back in its initial state. so you might think that every door will end up closed? well what about door #9. 9 has the divisors 1 & 9, 3 & 3. but 3 is repeated because 9 is a perfect square, so you will only visit door #9, on pass 1, 3, and 9… leaving it open at the end. only perfect square doors will be open at the end.

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Reverse a String

Problem:

A typical programming interview question is “reverse a string, in place”. if you understand pointers, the solution is simple. even if you don’t, it can be accomplished using array indices. i usually ask candidates this question first, so they get the algorithm in their head. then i play dirty by asking them to reverse the string word by word, in place. for example if our string is “the house is blue”, the return value would be “blue is house the”. the words are reversed, but the letters are still in order (within the word).

 

Solution:

Solving the initial problem of just reversing a string can either be a huge help or a frustrating hinderance. most likely the first attempt will be to solve it the same way, by swapping letters at the front of the string with letters at the back, and then adding some logic to keep the words in order. this attempt will lead to confusion pretty quickly.

for example, if we start by figuring out that “the” is 3 letters long and then try to put the “t” from “the” where the “l” from “blue” is, we encounter a problem. where do we put the “l” from “blue”? hmm… well we could have also figured out how long “blue” was and that would tell us where to put the “l” at… but the “e” from “blue” needs to go into the space after “the”. argh. its getting quite confusing. in fact, i would be delighted to even see a solution to this problem using this attack method. i don’t think its impossible, but i think it is so complex that it’s not worth pursuing.

here’s a hint. remember before when we just reversed “the house is blue”? what happened?
initial: the house is blue
reverse: eulb si esuoh eht

look at the result for a minute. notice anything? if you still don’t see it, try this.
initial: the house is blue
reverse: eulb si esuoh eht
wanted : blue is house the

the solution can be attained by first reversing the string normally, and then just reversing each word.

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The Circular Lake Monster Problem

Problem:

For #1, how about row in a circle a bit smaller 1/4r in size. Since he won't be able to keep up with you, when he is on the opposite shore you can make a brake for it. You have r*3/4 to travel, but 4*3/4 is less then pi, so he won't be able to catch you in time

Solution:

Assume x is the monster's speed. Then to get the circle trick to work, you row a circle a little less than 1/x of the radius, leaving you 1 - 1/x to row when he is opposite of you. If the monster can travel pi times the radius faster than you can travel the radius, you're hosed. In the time you travel 1 - 1/x, he'll travel x times that. Set that equal to pi, and you x * (1 - 1/x) = pi, which solves to x = pi + 1.

pi + 1 would be my guess for the speed that impossible to escape from, but I could be making an easy mistake.

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Some terms and their synonyms in programming languages

Scope/Visibility
Nested/Inner Class
Instance/Member 
Class/Static
Local/Automatic variable

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Arrays tip : Ignoring the zeroth row

This trick can be used in any language but is shown in java right now.
Sometimes you want to use natural input values as an index, the real values that that data has instead of starting with zero. Let's take the case of data that starts with the value 1, like the day of the month. The standard approach is to subtract one from every day value that's used as an index. This is annoying and error prone. Another way to do handle this case is to declare the array with an extra element, eg, 32 if dealing with the days in the month, then ignoring the zeroth element. If you're dealing with a two dimensional array, for example the accidents array from the previous page, you can even deallocate the first row so there won't be any possibility of referencing the zeroth day. For example,

static final int DAYS  = 32;
static final int HOURS = 24;
. . .
int[][] accidents = new int[DAYS][HOURS];
accidents[0] = null;

Because two-dimensional arrays are stored by row, you can do this. You can use the trick of allocating more columns to use the natural data, but you can't deallocate a column.

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