There are basically two mistakes in your logic First is when you are finding smallest number if(a[k]<small) { small=a[k]; } This is wrong as there will some condition arise when a[k] will be 0. So each time when the loop runs small will always be zero. See for example So it should be replace with … Read more
You could do something like this in javascript var data = [ [1, 2], [3, 4], [5, 6] ]; var getPos = function(elem, data) { var result=””; data.forEach(function(value, index) { value.forEach(function(v, i) { if (v == elem) result += (index + 1) + ‘-‘ + (i + 1); }); }); return result; } console.log(getPos(5, data)); … Read more
This sequence is Stern’s diatomic sequence, and the function you’ve been given is called fusc(n). One way to compute it in O(log n) time and O(1) space complexity is to find the n’th value in the Calkin-Wilf tree. That value’s numerator will be fusc(n). You can read some background on the wikipedia page: https://en.wikipedia.org/wiki/Calkin%E2%80%93Wilf_tree Here’s … Read more
It is hard (if not impossible) to label an heuristic as “best”. One idea I have in mind is evaluate the heuristic for the current state as the maximum value of all the tiles at this state. And then, that with the higher value is supposed to be better (“closer”) to the goal. And it … Read more
The number of iterations of the while-loop is exactly floor(x/y). Each iteration takes n operations, because that is the complexity of the subtraction r – y. Hence the complexity of the algorithm is n * floor(x/y). However, we want to express the complexity as a function of n, not as a function of x and … Read more
Here is the greedy algorithm – always the best place to start. allocate all teams IF not all sections covered output -1 stop mark all teams non-critical flag_improved = true WHILE( flag_improved == true ) flag_improved = false find most expensive section find most expensive non-critical team on most expensive section IF team found that … Read more
Well, there is a very easy and naive recursive solution to retrieving all subsets. You take away one element from your set, then find all subsets for this new, smaller set. Then you copy the result and add the element you previously removed to the copy. Add the results together and you’re done. For example: … Read more
Is this a class assignment? I have a feeling you’re asking us to do your homework for you. The short answer is that the algorithm is O(n^3). The value of n is read from the console. The outer loop is i = 1 .. n (i is only ever incremented in the loop, i.e. no … Read more
You can solve this using dynamic programming #include <iostream> using namespace std; #define INF 1000000007 int main() { // your code goes here int dp[101] = {0}; int a,b,i,j; scanf(“%d%d”,&a, &b); for(i=a;i<=b;i++) dp[i] = INF; dp[a] = 0; for(i=a; i<=b; i++){ for(j=2; j*j<=i; j++){ if(i%j == 0){ dp[i+j] = min(dp[i+j], dp[i]+1); dp[i+i/j] = min(dp[i+i/j], dp[i]+1); … Read more
Basically it stores deltas rather than the final list; that means each operation only takes 2 reads and writes rather than (b – a + 1). Then the final max scan adds the deltas as it goes along, which is still an O(n) operation which you would have had to do anyway. n, inputs = … Read more
Can someone tell me the Complexity of the Addition & Subtraction for the Divide & Conquer Matrix Multiplication algorithm? solved Can someone tell me the Complexity of the Addition & Subtraction for the Divide & Conquer Matrix Multiplication algorithm?
try pythons high level file operations library ( import shutil ) Also consider using pythons excellent directory walking function ( from os import walk ) You shouldn’t need to copy the files but just rename the directories and filenames. solved How to copy all files with a particular prefix to new directories with python?
I don’t understand how this change modifies the head too. I thought curr was just a variable. If this assignment were an assignment to curr you’d be right, and it wouldn’t mutate the list, but the assignment is not to curr, but to curr.next, and that is not a variable, but an attribute of a … Read more
For positive ? the number of executions of count = count + 1 is representative of the complexity, as the other lines of code don’t execute significantly more times. The recurrence relation can use the first iteration of the outer loop: that represents ? iterations of the inner loop. The loop continues for ?/2, and … Read more
Try this: matrix = [[0, 0, 0, 1, 0], [0, 0, 1, 1, 1], [0, 1, 0, 1, 0], [1, 1, 1, 0, 1], [0, 1, 0, 1, 0]] matrix.sort(key=sum, reverse=True) print(matrix) Update: This is in your expected output form: matrix = [[0, 0, 0, 1, 0], [0, 0, 1, 1, 1], [0, 1, 0, … Read more