More linked list problems

There are a lot of commonly asked linked list questions. They can range from simple to much more challenging. The following are some questions that can all be solved using two pointers ptr1 and ptr2 with different moving speed. They are relatively simple so I won`t post code here, but just the method:

Find the middle element of a linked list.

Let ptr1 moves 1 step and ptr2 moves 2 steps in each loop. When ptr2 reaches the end, ptr1 is pointing the middle element. 

Find the nth last element of a linked list.

Let ptr2 move n steps ahead first, and then let two pointers move together. When ptr2 reaches the end, ptr1 is pointing the nth last element.

Two linked lists of different length intersect at one node. Find this node.

We should first find the lengths of two linked lists by iterating through the whole list. Suppose the length of two lists are len1 and len2, len1 < len2, the difference d = len2 - len1. Now reset two pointers and let the pointer of longer list move d steps first. Then move two pointers simultaneously and compare the data of them. When data is equal, we find the node.

Given a circular linked list, find node at the beginning of the loop.

Let N be the node we want to find. In each loop, ptr1 moves 1 step and ptr2 moves 2 steps, until they meet at one node. The distance from this node to N is equal to the distance from head node to N. Now we reset ptr1 to head node and begin to move two pointers simultaneously. When they meet again, N is found there.

Maximum subsequence sum problem

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

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

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:

1. Create a list of consecutive integers from 2 to n: (2, 3, 4, ..., n).
2. Initially, let p equal 2, the first prime number.
3. 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.
4. 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).
5. 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

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

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;
}