You can do this by building a set of integers and adding larger seen in the set, and holding the minimum not seen in as a counter. Once there is a number that is equal to the latter, go through the set removing elements until there is a missing integer. Please see below for implementation. … Read more
static void Main() { var cnk = comb(new [] {1,2,3},2); foreach ( var c in cnk) { } } public static IEnumerable<int[]> comb(int[] a, int k) { if (a == null || a.Length == 0 || k < 1 || k > a.Length) yield break; int n = a.Length; // 1 if ( k == … Read more
use this Method to determine your string public static string ConvertExp(int Value) { int exp = Math.Abs(Value).ToString().Length -1; return string.Format(“{0} * 10^{1}”,( Value / Math.Pow(10 , exp)), exp); } and to get 2 values public static double[] ConvertExp(int Value) { int exp = Math.Abs(Value).ToString().Length -1; return new double[] { Value / Math.Pow(10, exp), Math.Pow(10, exp) … Read more
You can use DP to solve it. Here is a good website for learning DP: https://www.geeksforgeeks.org/dynamic-programming/ # For example X=3, Y=5, D=24. If we know solution for D=21 (24-3) and D=19 (24-5), then we know for D=24 the solution is min(D=21, D=19) +1. # And for D=19, we know it’s min(D=16, D=14) +1. So all … Read more
You miss the initialization of _N, so it will never be equal to N and you will rest into the while loop forever 4 solved Why this C code makes infinite loop?
My friend your approach is absolutely right , Only check your datatype to print answer . Your answer may come >10000000000 for any test case . check it, Its open forum , So I won’t mention by pointing. 3 solved Codechef : Correct approach or not?
The average case complexity for both cases is O(n). if k is the number of times the first if fails, then the number of comparisons is 2*n – 2 – k. maxmin(a,n,max,min){ max=min=a[1]; for i=2 to n do{ // goes through the loop n-1 times if a[i]>max then max:=a[i]; // out of n-1 times succeeds … Read more
Alright I’ve done the hard part hoping you can finish it up: tmp = [] def recal(_list): n = [] if ‘-‘ in _list: for i in 0,1: t = _list[:] t[t.index(‘-‘)] = i n.append(recall(t)) else: tmp.append(l) return l recall([‘-‘,’-‘,’0′,’1’]) for l in tmp: print int(”.join(l),2) 0 solved combinations program help in python [closed]
Are you sure that you are checking power of 2? Rather, being checked divisible with 2. If you are serious, use the piece of cake snippet return n > 0 && ((n & -n) == n); The second way, private static boolean powerOf2(int num) { if(num <= 0){ return false; } while(num > 1) { … Read more
Here ya go, use std::string::find: // Notice the double quotes std::string::size_type position = my_string.find(“34”); bool found = ((position != std::string::npos) && (position > 0) && (my_string[position – 1] == ‘6’)); The find method returns the position of substring, or std::string::npos if it is not found. The position must be greater than 0, because of the … Read more
def consecutiveLetters(word): currentMaxCount = 1 maxChar=”” tempCount = 1 for i in range(0, len(word) – 2): if(word[i] == word[i+1]): tempCount += 1 if tempCount > currentMaxCount: currentMaxCount = tempCount maxChar = word[i] else: currentMaxCount = 1 return [maxChar,tempCount] result = consecutiveLetters(“TATAAAAAAATGACA”) print(“The max consecutive letter is “, result[0] , ” that appears “, result[1], ” … Read more
It’s probably wrong when there is some segments having same start time. I run this test case with your code: 10 1 1 1 2 3 3 1 4 3 5 3 6 3 7 8 8 3 9 3 10 Your code output 10 but the answer is actually 9 I have several WA … Read more
There is a solution with a time complexity of O(T): Use the formula for sum of integers 1+2+3+…+n = n*(n+1)/2. Also note that 3+6+9+…+(3*n) = 3*(1+2+3+…+n) = 3*n*(n+1)/2. Figure out how many numbers divisible by 3 there are. Calculate their sum. Figure out how many numbers divisible by 5 there are. Calculate their sum. Figure … Read more
“Better” is a subjective term. Some points to consider: Yes, if you had a Map or an object keyed by id for patients, beds, and departments, in general looking up patients, beds, and departments by id would be faster using that Map/object instead of a linear searches of those arrays. But, presuming that you need … Read more
Compare X ≡ s1 * m2 with Y ≡ s2 * m1. If X > Y, then s1 / m1 > s2 / m2. No division is required to do the comparison. The caveat to this solution is that s1, s2, m1, and m2 should all have the same sign, and m1 and m2 should … Read more