Miracle space

Implement atoi

2012-01-15T11:28:00.000-05:00

Implement atoi to convert a string to an integer.

Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.

Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.

Requirements for atoi:
The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.

The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.

If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.

If no valid conversion could be performed, a zero value is returned. If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.

Used Java because it`s cleaner

public int atoi(String str) {
// Start typing your Java solution below
// DO NOT write main() function
String expression = str.trim();
if(expression==null||expression.equals("")){
return 0;
}
int sign = 1;
int index = 0;
if(expression.charAt(0) == '-'){
sign = -1;
}
if(expression.charAt(0) == '-' || expression.charAt(0) == '+'){
index++;
if(expression.length()==1){
return 0;
}
}
int ret = 0;
while(index < expression.length()){
char c = expression.charAt(index);
if(c < '0' || c > '9') break;
int value = c - '0';
if(sign > 0){
if(ret > Integer.MAX_VALUE / 10) return Integer.MAX_VALUE;
}else{
if(ret < Integer.MIN_VALUE / 10) return Integer.MIN_VALUE;
}
ret = ret*10;
if(sign > 0){
if(ret > Integer.MAX_VALUE - value) return Integer.MAX_VALUE;
}else{
if(ret < Integer.MIN_VALUE + value ) return Integer.MIN_VALUE;
}
ret += value * sign;
index++;
}
return ret;
}

Palindrome Number

2012-01-14T17:39:00.000-05:00

Determine whether an integer is a palindrome. Do this without extra space.
Some hints:Could negative integers be palindromes? (ie, -1)

If you are thinking of converting the integer to string, note the restriction of using extra space.

You could also try reversing an integer. However, if you have solved the problem "Reverse Integer", you know that the reversed integer might overflow. How would you handle such case?

There is a more generic way of solving this problem.

public boolean isPalindrome(int x) {
if(x < 0) return false;
int numOfDigits = 1;
int base = 1;
while(x / base > 9) {
base *= 10;
++numOfDigits;
}

int head, tail;
for(int i = 0; i < numOfDigits / 2; ++i) {
head = x / base;
tail = x % 10;
if(head != tail) return false;
x -= head * base;
x /= 10;
base /= 100;
}
return true;
}

In-place array rotation

2012-01-12T23:32:00.000-05:00

The idea that is commonly seen in skill testing interview questions is using three reversals to perform the rotations: rev(1,k) rev(k+1,n-k) rev(1,n). This leads us to the following code:

int reverse(struct baseType * a, int n) {
int i, j;
j = n / 2;
n--;
for (i=0; i < j; i++) {
struct baseType tmp = a[i];
a[i] = a[n-i];
a[n-i] = tmp;
}
return 0;
}

int rotate(struct baseType a[], int n, int k) {
if(a == NULL || n <= 0)
return -__LINE__;
if(k < 0 || k >= n){
k %= n;
if (k < 0) k += n;
}
if (k == 0) return 0;
reverse (a, k);
reverse (a + k, n - k);
reverse (a, n);
return 0;
}

This solution reads and writes each element twice, however, some quick performance testing on random arrays shows that this code is faster for very small sized baseTypes. The reason being that the reverse() is very good for memory access locality. However, for any larger sized baseTypes, the first code sequence given will win by the simple virtue of performing fewer operations.

Rotating a 2D array of integers (matrix) by a given angle (+90, -90, +180, -180)

2012-01-12T21:08:00.001-05:00

This is posted by Rajendra Kumar Uppal

So, I came across this problem on a forum and before looking up, I tried to solve it. After 15 minutes of trying, I figured out that the solution is pretty easy and elegant which was messy otherwise if you are going to do it in an insane way. So, here we go:

Problem definition: You are given a 2D square matrix, or 2D array of integers of size n (n rows and n columns), your output should be n by n 2D matrix rotated by a given angle, which could be +90, -90, +180, -180.

Example:

Input Output

1 2 3 7 4 1

4 5 6 Rotate by +90 8 5 2

7 8 9 9 6 3

If you go by solving it something like manually, swapping elements, its going to be messy, tedious, and error prone in that there will be much more edge cases, test cases to handle. So, after trying on paper, I figured out following elegant solution to solve this in an easy manner.

Rotate by +90 (clockwise once):

Input: n by n matrix M, where n >= 2

Algorithm:

Step 1: Transpose M

Step 2: Reverse each row

Dry run:

Step 1: Transpose

M M'

1 2 3 1 4 7

4 5 6 -- 2 5 8

7 8 9 3 6 9

Step 2: Reverse each row

M' M''

1 4 7 7 4 1

2 5 8 -- 8 5 2

3 6 9 9 6 3

M'' is rotated form of M by +90 degree.

Pseudocode: is here which is self explanatory and easily convertible to source code in a language of your choice.

Transpose

for i in [0, n)

for j in [0, n)

if ( i < j )

swap( M[i][j], M[j][i] )

Reverse a row (rowidx)

start = 0

end = cols - 1

while ( start < end ) {

swap( M[rowidx][start], M[rowidx][end] )

++start

--end

}

Rotate

Transpose

for i in [0, rows)

Reverse( i )

That was fair enough until only asked to rotate by +90 degree. If problem is further extended to be solved for any given angle, of course the rotation should make sense, for example, 47 degree is not a choice ;-). So, here are some elegant techniques (I'll be brief now for other angles, because for +90 degree I have elaborated the solution):

Rotation by -90 degree (anticlockwise once):

Step 1: Transpose

Step 2: Reverse each column

Rotation by +180 degree (clockwise twice): Two methods follows

First:

Rotate input matrix +90 degree twice, if routine for which is available to you

Second: (You'll be amazed!)

Step 1: Reverse each row

Step 2: Reverse each column

Rotation by -180 degree (anticlockwise twice): Three(!!!) methods follows

First:

Rotate input matrix -90 degree twice, if routine for which is available to you

Second: (You'll be amazed again!)

Step 1: Reverse each column

Step 2: Reverse each row

Third: (Aha!)

Because rotating a matrix +180 degree or -180 should produce same result. So you can rotate it by +180 degree using one of above methods.

Concluding note: Above techniques are elegant and very simple and straightforward to implement. Try them for some input of your choice and rotate it in all angles. These techniques are tested with a working C program.

Learn Vim Progressively — 1st Level : Survive

2011-12-25T19:34:00.000-05:00

I`m now learning vim, and I found a very interesting article which talks about how to learn vim progressively in four steps. I just paste the whole article here to share with others.

Want to learn vim (the best text editor known to human kind) the fastest way possible. I suggest you a way. Start by learning the minimal to survive, then integrate slowly all tricks.

Vim the Six Billion Dollar editor

Better, Stronger, Faster.

Learn vim and it will be your last text editor.There isn’t any better text editor I know.Hard to learn, but incredible to use.
I suggest you to learn it in 4 steps:

Survive
Feel comfortable
Feel Better, Stronger, Faster
Use vim superpowers

By the end of this journey, you’ll become a vim superstar.
But before we start, just a warning.Learning vim will be painful at first.It will take time.It will be a lot like playing a music instrument.Don’t expect to be more efficient with vim than with another editor in less than 3 days.In fact it will certainly take 2 weeks instead of 3 days.

1^st Level – Survive

Install vim
Launch vim
DO NOTHING! Read.

In a standard editor, typing on the keyboard is enough to write something and see it on the screen.Not this time.Vim is in Normal mode.Let’s get in Insert mode.Type on the letter i.
You should feel a bit better.You can type letters like in a standard notepad.To get back in Normal mode just tap the ESC key.
You know how to switch between Insert and Normal mode.And now, the list of command you can use in Normal mode to survive:

i → Insert mode. Type ESC to return to Normal mode.
x → Delete the char under the cursor
:wq → Save and Quit (:w save, :q quit)
dd → Delete (and copy) current line
p → Paste
Recommended:
hjkl (highly recommended but not mandatory) → basic cursor move (←↓↑→). Hint: j look like a down arrow.
:help <command> → Show help about <command>, you can start using :help without anything else.

Only 5 commands. This is very few to start.Once these command start to become natural (may be after a complete day), you should go on level 2.
But before, just a little remark on Normal mode.In standard editors, to copy you have to use the Ctrl key (Ctrl-c generally).In fact, when you press Ctrl, it is a bit like if all your key change meaning.With vim in Normal mode, it is a bit like if your Ctrl key is always pushed down.
A last word about notations:

instead of writing Ctrl-λ, I’ll write <C-λ>.
command staring by : will must end by <enter>. For example, when I write :q it means :q<enter>.

Some Linux commands useful for manipulating text files

2011-12-25T19:31:00.000-05:00

I have learned some Linux utilities for manipulating text files these days. They are extremely handy which is far beyond my expectation. I should have learned them before I used Java to write those unresuable, verbose and error-prone codes to process texts a week ago.

1. cat

This is the simplest tool in this category. The cat command copies its input to output unchanged (identity filter). When supplied a list of file names, it concatenates them onto stdout.

2. head

Display the first few lines of a specified file. Particularly useful when the file is big and you wants to see only the header. You don`t have to open the whole file (which makes the system slow) using this command.

3. tail

Displays the last part of a file, similar to head.

4. cut

The cut command prints selected parts of input lines. It can select columns (assumes tab-separated input); can select a range of character positions; can also specify delimiter characters.

5. sort

Sort each line of the file using either lexicographic or arithmetic order. It can use a key for comparison to sort the lines in delimiter seperated files.

6. uniq

Remove or report adjacent duplicate lines. Useful for finding duplicate and non-duplicate lines.

7. wc

The word count utility, wc, counts the number of lines, characters or words. The line counting feature is my best favorite because it does not have to open the file.

8. tr

Copies standard input to standard output with substitution or deletion of selected characters. It can be used to filter a range of characters in order to make certain conversions.

9. grep

Search substrings that match the given regular expression, and print the line they reside in. Often used to search for the results we are interested in.

10. sed

More powerful tool compared to the above. It Looks for patterns one line at a time, like grep , but changes lines of the file. It`s a non-interactive text editor. Editing commands come in as a script. There is an interactive editor ed which accepts the same commands. It`s a A Unix filter which is the superset of previously mentioned tools, and it`s syntax is also used in VIM.

Above are the most common used Linux text manipulating utilities. Although they are already very convenient individually, combining them using pipes can make them even more powerful and flexible. If you are a linux admin or academic researcher, these are the commands you must know.

Maximum subsequence sum problem

2011-12-10T11:37:00.001-05:00

The problem we’d like to solve can be stated as:

Given a sequence of numbers, find the maximum sum of a contiguous subsequence of those numbers.

One obvious solution is that we can simply enumerate every possible subsequence and calculate their sum. This will take O(n^2) time.

There is a linear solution called Kadane`s algorithm. Following is the code:

int max_sub_sum(int arr[], int len) {
    int max_local = 0;
    int max_global = 0;
    for(int i = 0; i < len; i++) {
       max_local += arr[i];
       if(max_local < 0)
           max_local = 0;
       if(max_local > max_global)
           max_global = max_local;
    }
    return max_global;
}

Why does it work? If max_local becomes negative, it will not contribute to the next subsequence by adding the next element. Actually the next element itself would be larger than the sum of it and max_local, so we can simply restart summing from here.

Note that this algorithm only works when there is at least one positive integer in the array.

Some interviewers may ask this question with slight modification that the input array is a circular array. In this case, we can concatenate the input array with itself, and apply Kadane`s algorithm.

Ellipse fitting and parameter conversion

2011-12-09T16:20:00.001-05:00

The third assignment of Computer Vision course involves an interesting problem. Given a large number of points in Cartesian coordinate system, how to find an ellipse that best fit these points. In this paper the illustration and solution of this problem is presented.

The theory and proofs are not trivial and I can`t understand all of them. However, it`s implementation is very simple. Below is a six-line Matlab implementation of the algorithm.

function a = fit_ellipse(x, y)
    D = [x.*x x.*y y.*y x y ones(size(x))];
    S = D' * D;
    C(6, 6) = 0; C(1, 3) = -2; C(2, 2) = 1; C(3, 1) = -2;
    [gevec, geval] = eig(S, C);
    [NegR, NegC] = find(geval < 0 & ~isinf(geval));
    a = gevec(:, NegC);
end

The input of this function is two vectors containing x and y coordinates of the points, respectively. The output is the 6 parameters of implicit equation of the ellipse.

Our TA is very smart and he used a trick to simplify the problem for us to implement. So we can avoid solving the general eigenvector problem, which is not provided by OpenCV library. However, the eigen()function provided by OpenCV can only solve symmetric matrix, which is not our case. Fortunately, there is another linear algebra C++ library called Eigen, and OpenCV provided functions to convert data structures needed by both libraries back and forth. With Eigen, we can get things done, but it leaves me rather messy code. This is why I don`t post my C++ implementation here.

Having only the implicit equation is not sufficient, because we cannot get any useful information from it directly. To visualize the ellipse, we have to get semi-major axis, semi-minor axis, rotation and center from the 6 parameters. Below is the C++ code:

    // Input parameters: a, b, c, d, e, f
    phi = atan(b / (a - c)) * 0.5; // rotation
    double cos_phi = cos(phi);
    double sin_phi = sin(phi);
    double u = (2 * c * d - b * e) / (b * b - 4 * a * c);
    double v = (2 * a * e - b * d) / (b * b - 4 * a * c);
    center = cv::Vec2d(u, v);        // center

    // eliminate rotation and recalculate 6 parameters
    double aa = a * cos_phi * cos_phi - b * cos_phi * sin_phi + c * sin_phi * sin_phi;
    double bb = 0;
    double cc = a * sin_phi * sin_phi + b * cos_phi * sin_phi + c * cos_phi * cos_phi;
    double dd = d * cos_phi - e * sin_phi;
    double ee = d * sin_phi + e * cos_phi;
    double ff = 1 + (d * d) / (4 * a) + (e * e) / (4 * c);

    sx = sqrt(ff / aa);              // semi-major axis
    sy = sqrt(ff / cc);              // semi-minor axis

    if(sx < sy) {
        double temp = sx;
        sx = sy;
        sy = temp;
    }

Now we can use information to visualize the ellipse. Here is an example of my result:

The red dots are cross section points of human body and we want to fit. The ellipse shown is the result. We can see that waist, left and right biceps are quite good, but chest, hips, left and right thigh are not satisfactory. One possible explaination is that there may be many noisy points outside this visual range that affect the result.

Find the sum of all prime numbers below N

2011-12-09T14:14:00.001-05:00

This is another classic question. Everyone would immediately come up with an idea that we can iterate from 2 to N and check if it is a prime number. If so, we add it to the sum.

bool isprime(int num) {
    int j = sqrt((float)num);
    for(int i = 2; i <= j; i++) {
       if(num % i == 0)
           return false;
    }
    return true;
}

int primesum(int range) {
    int sum = 2;
    for(int i = 3; i <= range; i += 2) {
       if(isprime(i))
           sum += i;
    }
    return sum;
}

This solution uses Trail Division to test if a number is prime. The trial factors need go no further than sqrt(n) because, if n is divisible by some number p, then n = p * q and if q were smaller than p, n would have earlier been detected as being divisible by q or a prime factor of q.

When calculating the sum, we only need to check even numbers. This would save half of the time.

The time complexity of this method is O(n * sqrt(n)), and it requires O(1) space. The algorithm can be further optimized if we have a smaller pre-generated prime number table and test only prime factors.

If extra space is allowed, there is a better solution which is called Seive of Eratosthenes. The basic idea is to directly generate composite numbers rather than testing for prime numbers. The Seive of Eratosthenes algorithm is described as follows:

Create a list of consecutive integers from 2 to n: (2, 3, 4, ..., n).
Initially, let p equal 2, the first prime number.
Starting from p, count up in increments of p and mark each of these numbers greater than p itself in the list. These numbers will be 2p, 3p, 4p, etc.; note that some of them may have already been marked.
Find the first number greater than p in the list that is not marked; let p now equal this number (which is the next prime).
If there were no more unmarked numbers in the list, stop. Otherwise, repeat from step 3.

When the algorithm terminates, all the numbers in the list that are not marked are prime.

int primesum(int range) {
    int sum = 2;
    bool* prime_table = new bool[range];
    memset(prime_table, true, sizeof(bool));

    for(int i = 2; i <= range / 2; i++) {
        if(prime_table[i]) {
            for(int j = 2 * i; j <= range; j += i) {
                prime_table[j] = false;
            }
        }
    }
    for(int i = 3; i <= range; i += 2) {
        if(prime_table[i])
            sum += i;
    }
    return sum;
}

The time complexity is O(nloglogn), and space complexity if O(n).

Above is only the implementation of original version. There are some refinements of this algorithm. For further informaton, please go to this link http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes.

Reverse a singly linked list

2011-12-09T00:20:00.001-05:00

This is a very classic interview question. It demonstrates the ability to work with basic data structure and pointers. The seemingly easy question has some pitfalls. Nearly every candidate starts with two temporary pointers and eventually finds out that you need three. You need to point to the current node (the one you’re handling), the previous node (so you can point back to it), and the next node (so you can prevent lost of the next node). Also, you should not forget to make the first node point to NULL.

There are two solutions to this question. One is iterative and the other is recursive.

Node* reverse_iterative(Node* head) {
    Node* prev = head;
    Node* curr = head->next;
    Node* next = head->next->next;
    prev->next = NULL;

    while(curr->next) {
        curr->next = prev;
        prev = curr;
        curr = next;
        next = next->next;
    }
    curr->next = prev;
    return curr;
}

The recursive version is shown as follow:

Node* reverse(Node* prev, Node* curr) {
   // If reached the end of the list, return the head
   if(curr->next == NULL) {
       curr->next = prev;
       return curr;
   }
   // Remember the next node
   Node* next = curr->next;
   // Change current pointer to previous node
   curr->next = prev;
   return reverse(curr, next);
}

Node *reverse_recursive(Node* curr)
{
   return reverse(NULL, curr);
}

Although recursive version is more straightforward, in general we prefer the iterative version because if the list is very large, this version would fail because of stack overflow.

Note that there are some variations of this question like "How to print a singly linked list in reverse order". In this case, if we are allowed to alter the data structure, we can reverse linked list first and print it from the beginning.

Find projection of a vector on a plane

2011-12-08T17:16:00.001-05:00

Given a vector V in 3D space and a normal vector N of a plane, how to calculate the projection of V on the plane?

The following is my C++ and OpenCV implementation of the algorithm.

Vec3d findProjection(Vec3d V, Vec3d N) {
    Vec3d v, n;
    // We convert both vectors to unit vectors
    normalize(V, v);
    normalize(N, n);
// Then calculate the cos angle between n and v
    double cosVN = v.dot(n);
// Make sure the angle is in range of -90 to 90 degrees
    if(cosVN < 0) {
        cosVN = -cosVN;
        n = -n;
    }
    double V_len = sqrt(V.dot(V));
    double _N_len = V_len * cosVN;
// _N is the projection of V on N
    Vec3d _N = _N_len * n;
    return V - _N;
}

Miracle space

Implement atoi

Palindrome Number

In-place array rotation

Rotating a 2D array of integers (matrix) by a given angle (+90, -90, +180, -180)

Learn Vim Progressively — 1st Level : Survive

1^st Level – Survive

Some Linux commands useful for manipulating text files

More array problems

More linked list problems

Maximum subsequence sum problem

Ellipse fitting and parameter conversion

Find the sum of all prime numbers below N

Reverse a singly linked list

Find projection of a vector on a plane

Miracle space

Implement atoi

Palindrome Number

In-place array rotation

Rotating a 2D array of integers (matrix) by a given angle (+90, -90, +180, -180)

Learn Vim Progressively — 1st Level : Survive

1st Level – Survive

Some Linux commands useful for manipulating text files

More array problems

More linked list problems

Maximum subsequence sum problem

Ellipse fitting and parameter conversion

Find the sum of all prime numbers below N

Reverse a singly linked list

Find projection of a vector on a plane

1^st Level – Survive