[ad_1] Is there a general method for calculating the end of an animation and switching between animations when they are finished? [ad_2] solved Is there a general method for calculating the end of an animation and switching between animations when they are finished?
[ad_1] If there are any obstacles, use breadth-first-search algorithm. If there aren’t, the task is even easier. Just get to the right column and then continue to the right row. [ad_2] solved How to find the shortest way to destination [duplicate]
[ad_1] “Hello World” is a constant so it is O(1) whereas string str is variable of length n so the complexity is O(n) 2 [ad_2] solved Time complexity difference in System.out.println() [closed]
[ad_1] 121 = 11 * 11 would be the first non-prime listed as prime by your code. So your code is only a solution up to n = 120. I.e.: No. 3 [ad_2] solved Is this a good approach for finding number of prime numbers within a certain range? [closed]
[ad_1] Finally, I discovered that the problem was in the code. Just small mistake using substr func. So, the correct code is: $buffer = file_get_contents(“image.png”); $length = filesize(“image.png”); if (!$buffer || !$length) { die(“Reading error\n”); } $_buffer=””; for ($i = 0; $i < $length; $i++) { $_buffer .= sprintf(“%08b”, ord($buffer[$i])); } echo $_buffer.”<br>”; $nb = … Read more
[ad_1] You could take the remainfer of the index with three and address the dame group for every third element. If necessary take destructuring assignment, like [array0, array1, array2] = result; const array = [{ id: 0 }, { id: 1 }, { id: 2 }, { id: 3 }, { id: 4 }, { … Read more
[ad_1] You can cast it on input like this: hrs = int(input(“Enter Hours:”)) However, there is no need to do this as the compiler can tell which data type has been input. Example: >>> hrs = input(“Enter Hours:”) Enter Hours:30 >>> type(hrs) <type ‘int’> You can see that python already knows what type has been … Read more
[ad_1] find() seems to be heavily used and they are very different in D and C++ implementations: int find(int x) { return ds[x] = (x == ds[x] ? x: find(ds[x])); } vs: long long find(long long node) { if(parents[node] == node)return node; else return find(parents[node]); } find() in D modifies array (looks like some kind … Read more
[ad_1] A. The board is being shuffled, so 1000 means ‘a lot, but not that much that you really have to start waiting for the board being shuffled’ B. getCandidates() returns the candidates by filling them in the vector, v.clear() resets the vector for new candidates. C. Here 1,2,4,8 just means up,down,left,right (not in particular … Read more
[ad_1] Solution: change palindrome to int palindrome ( int n,int base) { int isTrue = 1; int digitCount = digitCountBase(n,base); int power = intPow(base,digitCount-1); int original = n; while (n>0&& digitCount >0) { if (n%base != (original/power) % base &&digitCount!=1) { isTrue =0; return 0; } n=n/base; power = power /base; digitCount=digitCount-2; } return isTrue; … Read more
[ad_1] Since your files are sorted, you don’t have to store them or even parse them outside of lines. You can keep advancing in the sequence with the lesser current element: def count_equal(a, b): “”” Counts the number of values that are equal in two sorted iterables. >>> odds = [1, 3, 5, 7, 9, … Read more
[ad_1] For this to work, you would have to sort your coins in descending order, or use reversed(coinList) instead of coinList in the first for loop and vice versa in the second if you would like to sort them forwards: total = 250 value = 0 coins = [] coinList = [100, 100, 20, 20, … Read more
[ad_1] int countUniVals(node* head, bool* unival) { if (!node) { *unival = true; return 0; } bool uniL,uniR; int sum = countUniVals(head->l, &uniL) + countUniVals(head->r, &uniR); if (uniL && uniR && (!head->l || head->l->val == head->val) && (!head->r || head->r->val == head->val)) { sum++; *unival = true; } return sum; } 0 [ad_2] solved Counting … Read more
[ad_1] You can try swaping first and last char value like below. =========================================== String s = “asd321tre2”; char[] charr = s.toCharArray(); int end = 0; for (int i = s.length(); i <= 0; i–) { if (Character.isLetter(charr[i])) { end = i; } } String output = charReplace(end, charr); } private static String charReplace(int end, char[] … Read more
Introduction [ad_1] A unival subtree is a tree where all the nodes have the same value. Counting the number of unival subtrees in a binary tree is a common problem in computer science. This problem can be solved using a recursive approach, which involves traversing the tree and checking each node to see if it … Read more