Interesting problem. If the array is of signed integers, I believe it is possible in O(n) time and O(1) space, with no overflows, assuming the length is small enough to permit such a thing to happen. The basic idea is as follows: We have n numbers. Now on dividing those numbers by n+1, we get … Read more
Create a post value and sort by it: postValue = likes + (isVetted ? 10 : 0) This makes “vetting” worth 10 likes. In other words, if 10 people like something not vetted, it will be ranked equivalently to something vetted with 0 likes. If you want to sort by date instead, then just sort … Read more
The first statement compares the variable ‘a’ to ‘1’, and if the variable ‘a’ is equal to ‘1’, then you will execute the block of code enclosed in brackets. Likewise, the first statement also compares the variable ‘b’ to ‘1’ and if equal will execute a block of code. The second expression evaluates two conditions, … Read more
OK, here’s my answer, which should be O(N*N) worst case. (With smaller constant, since even worstcase I’m testing N against — on average — 1/2 N, but this is computer science rather than software engineering and a mere 2X speedup isn’t significant. Thanks to @Alexandru for pointing that out.) 1) Split cursor (input and output … Read more
I propose a mathematical solution. Number of numbers divided by a is [n/a]. ([] is a floor function.) And about b, c, lcm(a, b), lcm(a, c), lcm(b, c), lcm(a, b, c) so is it. (lcm means least common multiple.) So the answer should be ([n/a]+[n/b]+[n/c])-([n/lcm(a,b)]+[n/lcm(a,c)]+[n/lcm(b,c)])+[n/lcm(a,b,c)] 7 solved Fastest way to solve the divisibility of numbers … Read more
You have your main problem here. double *tabT1; double *tabT2; tabT1[0]=1; tabT1[1]=tabX[j]*(-1); tabT2[0]=1; tabT2[1]=tabX[i]*(-1); You haven’t allocated the memory, instead, you have just declared the double ptrs tabT1 and tabT2 and accessing them by pretending you have allocated. double *tabT1 = new double[2]; double *tabT2 = new double[2]; will fix this, however, I strongly suggest … Read more
Try using OPTICS algorithm, you won’t need to estimate eps in that. Also, I would suggest recursive regression, where you use the python’s best curve fit scipy.optimize.curve_fit to get best curve, and then find the rms error of all the points wrt the curve. Then remove ‘n’ percent of points, and recursively repeat this untill … Read more
Make that 3*a, you left out the actual multiplication. 2 solved Finding the longest collatz sequence for starting numbers < 100
Introduction The DBSCAN algorithm is a popular clustering algorithm used for data mining and machine learning. It is a density-based clustering algorithm that is used to identify clusters of points in a dataset. However, the original research paper on DBSCAN suggested an algorithm for estimating the epsilon parameter, which is an important parameter for the … Read more
Introduction The Collatz Conjecture is an unsolved mathematical problem that has been around since 1937. It states that if you take any positive integer and apply a certain set of rules, you will eventually reach 1. The rules are simple: if the number is even, divide it by two; if the number is odd, multiply … Read more
The key is item 7: For example, if Config.MIN_ACTION is 1 and Config.MAX_ACTION is 3: If the action rankings are {9,90,1} the total is 100. Since actionRanking[0] is 9, then an action of picking up 1 should be chosen about 9/100 times. 2 should be chosen about 90/100 times and 1 should be chosen about … Read more
First, let’s look at two simple nested loops first: for i (1..N) { for j (1..N) { f(); } } for i (1..N) { for j (i..N) { g(); } } f() is called N*N = N2 = O(N2) times. g() is called N+(N-1)+…+5+4+3+2+1 = N(N+1)/2 = N2/2 + N/2 = O(N2) times. As you … Read more
try to separate the coefficients from polynomials and calculate those in separate methods. Example: public class NewClass { public static void main(String[] args) { int greatestExponent = 9; double x = 0.5; System.out.println(calculate(x,greatestExponent)); } public static double getCoefficient(int n){ double coff = 1; for(int i=1; i<n; i++){ if(i%2==0){ coff = coff/i; //if even put it … Read more
Your calculation is alright (not that complicated after all!), I’d just suggest to change your condition statement to reduce a bit your code: double hours, overtimepay, wage; printf(“Enter number of hours: “) scanf(“%f”,&hours); wage=9.73*hours; wage = 9.73 * hours; printf(“Your wage is: %f\n”,wage); if(hours > 40) { overtimepay = (9.73*(hours-40))*1.5; printf(“Your overtime pay is: %f\n”, … Read more
Let’s say the domain is as following String domain[] = { a, b, .., z, A, B, .. Z, 0, 1, 2, .. 9 }; Let’s say the password size is 8 ArrayList allCombinations = getAllPossibleStrings2(domain,8); This is going to generate SIZE(domain) * LENGTH number of combinations, which is in this example (26+26+10)^8 = 62^8 … Read more