Let p1, p2, p3, … pk be the prime factors of some positive integer n. By the fundamental theorem of arithmetic, n can be represented as p1e1•p2e2•p3e3• … pkek for some positive integers e1, e2, e3, … ek. Furthermore, any such set of such positive integers represents one positive integer in this way. Every factor … Read more
To find the average in the given array and find the number which is closest to the average, public static void main(String[] args) { int array[] = new int[] { 1, 2, 3, 4, 5, 6, 7, 8 }; int sum=0; double avg; for (int i = 0; i < array.length; i++) { sum += … Read more
Does this return what you want? [[1,2] , [3,7] , [2,9] , [10,19]].sorted(by: {$0[0]<$1[0]}) 3 solved Is there a way to sort an array by a subarray’s index in Swift?
[EDIT] So based on you’re edit i thought this solution. Since you don’t specify any programming language i’ll use some c-like functional programming. I’ll leave you the work of transforming it to object oriented if you need it. Input: Global maxtrix that starts in M[0,0] and ends in M[100’000, 100’000] (Note that, to make it … Read more
Using the algorithm you propose in your question (i.e. to first find all unique values and then find the lowest one), a possible implementation looks like this: #include <stdio.h> #include <stdlib.h> // NOTE: The second parameter array_size specifies the number of // int elements in the array, NOT the size of the array in bytes. … Read more
The for-loop solution should be the easiest, as you can use the Math.floor function to get the start of your desired list: for (double d = Math.floor(start); d <= Math.ceil(end); d += 1.0) list.add(d); 13 solved Filter element in array of doubles
The first mistake is pushing all the characters on the stack outside of the if statement. Also you should check if stack is empty before removing items from it. Otherwise EmptyStackException is thrown. // stack1.push(S.charAt(i)); <– remove this line if (S.charAt(i)!=’#’) { stack1.push(S.charAt(i)); }else if (!stack1.isEmpty()) { // <– add this check stack1.pop(); } The … Read more
You need to choose a standard unit of measure. The question suggests meters, so use that (you can use float or double for this, depending on what precision you need). The problem is then simple, create some variables for the sum, smallest seen, and largest seen, the for each new input, convert to the standard … Read more
You can slightly improve your code by removing queue, you don’t need this. The rest of the code looks algorithmically optimal. #include <map> #include <iostream> using namespace std; int main() { int a,b; map<int, int> AB; map<int, int>::iterator it; std::ios::sync_with_stdio(false); while (1) { cin >> a; if (a == -1)break; cin >> b; AB.insert(pair<int, int>(a, … Read more
This is basically a link only answer but it gives you the information you need to formulate a good question. Bucket Sort Wikipedia’s step 1, where you “Set up an array of initially empty buckets”, will need to include buckets for negative numbers. Counting Sort “Compared to counting sort, bucket sort requires linked lists, dynamic … Read more
So you have power of 2 operations which can be easily converted to bit operations… no bigint lib is needed for such triviality. Let assume you got number const int n=1000000000; // number size in bytes BYTE x[n]; // number bytes let num[0] be LSW (least signifficant Word) So the operations involved are: mod: y … Read more
I believe your error is in your input file: you read i and j and use it to access your arrays, without verifying that these value are within expected boundaries. For instance, if you improve your questioned code with an error message, like: for (k=0; k<M; k++){ input>>i>>j>>V; if (i>=N || j>=N) { cerr << … Read more
Since you didn’t really give more details, we can only give you broad pointers. First Check out: Stable Marriage Problem. And also search the web for Bipartite matching (or in cases of weighted edges: Assignment Problem, which can be solved using: Hungaring Algorithm). Note that a solution to the stable marriage problem might also solve … Read more
Change your code into: int main() { int N,count=0,i,j; scanf(“%d”,&N); int A[N]; … 0 solved Find the number of pairs (A[i], A[j]) such that i < j and A[i]**A[j] > A[j]**A[i]
IsWordValid(S) = for word in dict: if S.startsWith(word) and IsWordValid(S[word.length:]) return true return false IsWordValid(null) = true 4 solved Recurrence relation for DP?