We can see this problem as an instance XOR-SAT problem, though it is more general than the problem posed here, that’s one way to go. Just to gain some intuition I provide a very simple example. Suppose you have systems of three switches and three bulbs like this: S B 1 1, 3 // toggles … Read more
Here is a function that works fine for all cases and return the list of all first candidates encountered if no choice can be made to separate them. def find_best(list_of_lists): i = 0 while len(list_of_lists[i]) == 0: i+=1 list_containing_candidates = list_of_lists[i][:] if len(list_containing_candidates) == 1 : return list_containing_candidates[0] else: if i+1 < len(list_of_lists): for next_list … Read more
Yes, you can. Your algorithm is running in O(n*m) time, and is basically saying: Are the two lists denoted as A and B have any common items? This can be done more efficiently if you first store one list into a set, which is implemented with more efficient data structure (Usually self balancing binary tree … Read more
The way I see it, your code would be O(n) if it wasn’t for the a.pop(0) part. Since lists are implemented as arrays in memory, removing an element at the top means that all elements in the array need to moved as well. So removing from a list is O(n). You do that in a … Read more
The question itself is interesting imho As a starting point I would create all possible combinations of the quadruples (?) that might occur and store them as (List Of) INTEGER (because that will allow you to work with bitwise compare). The number of combinations should be (20*19*18*17)/4! => 4845. Then you convert your input “lines” … Read more
At the end of the for loop s and k are equal. Before the next iteration k != n is checked. This is equivalent to s != n. So the loop runs until s == n holds and then n is returned. So the function get the input n, runs for some time and returns … Read more
You should keep a reference of the position of each letter in the string. You can use the toCharArray method. String s = “ciao”; String underscore = s.replaceAll(“.”, “_ “).trim(); char[] sC = s.toCharArray(); if(s.contains(“a”){ StringBuilder myUnderscore = new StringBuilder(underscore); for(int i = 0; i < sC.length; i++){ if(sc[i] == ‘a’){ myUnderscore.setCharAt(i, ‘a’); } } … Read more
The solution is : for n=1: = 1 for n=2: = 1+2 => 1+2+3 => 1-2+3 for n=3: = 1+2 for n=4: = 1+2+3 => -1+2+3 for n=5: = 1+2+3 => 1+2+3+4-5 for n=6: = 1+2+3 for n=7: = 1+2+3+4=>1+2+3+4+5=>1+2+3-4+5 for any n, first calculate the S(k)=1+2+3+…+k where S(k)>n and x=S(k)-n is an even number. … Read more
Not sure I get your question right but my bet is you’re asking how to convert between ASCII and number representation of characters. Encoding/Decoding has nothing to do with RSA. You just use dynamic range shift. As the alphabet is in increasing order in ASCII so the only thing left is to offset the a … Read more
You want all permutations of each member of the powerset of the input. permSub(“abc”, “”) func permSub(input, perm) print perm if input = “” return for i = 0 to input.length-1 permSub(input[0..i]+input[i+1..input.length), perm+input[i] end end Where input[i..j] represents the sub string of input from i(inclusive) to j(exclusive), and + is string concatenation. Note that this … Read more
Master theorem is just a shorthand for another, more cumbersome method: approximating manually the result (i.e. thinking of the size of the recursion trees and how much time is spent in each node) and checking the result. 0 solved Master method does not solve all questions [closed]
It is a variation of Subset-Sum problem – with an additional constraint on the size of the subset (larger then 1). The problem is NP-Complete, but for relatively small integers can be solved using Dynamic Programming in pseudo-polynomial time. A possibly simpler alternative which is feasible for small arrays is brute-force – just search all … Read more
For 1 to n the formula is sum =n(n+1)/2 , so from n1 to n2 you have to calculate sum of n1-1 terms snd sum of n2 terms and then subtract. sum_of_n1_to_n2 = n2(n2+1)/2 – (n1-1)((n1-1) +1)/2; 9 solved algorithm for the sum of integer peer from n to n [closed]
You made a mistake in the formulae. The way you’re looking at it, you should write: (a + sqrt(b^2 + c^2)) (b + sqrt(a^2 + c^2)) (c + sqrt(a^2 + b^2)). And even then, you wouldn’t get the shortest distance. To give an example, suppose the cube is of 1x1x1 units with sides along the … Read more
The fastest way that the rumour will travel is when the roumor starts in the middle of the matrix. Since it travels only horizontally or vertically, if it starts from the middle then it will take m/2 + n/2 days to reach a corner in the worst case. Just try this printf(“%d\n”,m/2+n/2); it should work. … Read more