Current answer The previous answer assumed that the movement of people to their new positions would be sequential, i.e. people would start moving together within the same sequence. The answer below assumes that the movement within the same sequence is instantaneous. It uses two arrays to map people from their old positions to their new … Read more
You can use cartesian product on sets in Java by the use of com.google.common.collect.Sets. FOR EXAMPLE Set<Integer> s1=new HashSet<Integer>(); s1.add(1);s1.add(4);s1.add(5); Set<Integer> s2=new HashSet<Integer>(); s2.add(2);s2.add(3);s2.add(6); Set<Integer> s3=new HashSet<Integer>(); s3.add(7);s3.add(8);s3.add(8); Set<List<Integer>> set=Sets.cartesianProduct(s1,s2,s3); //Give type safety warning for(List<Integer> l:set){ System.out.println(l); } OUTPUT [1, 2, 7] [1, 2, 8] [1, 3, 7] [1, 3, 8] …. NOTE If you … Read more
Don’t use string logic for this. Parse the string into a BitSet, before entering your loops, i.e. as you read them. Then use methods or(BitSet set), and cardinality(). I just completed challenge doing that. No timeouts. 2 solved I am facing timeout for my solution for a HackerRank
Write a program that takes an array/arraylist of integer and finds longest sub array/arraylist whose entries are equal [closed] solved Write a program that takes an array/arraylist of integer and finds longest sub array/arraylist whose entries are equal [closed]
I believe the quantity you are looking for is called variance. This describes consistency, if you were to say, use the time each day that a student works. 2 solved Algorithm to check consistency? [closed]
include <time.h> include <stdio.h> int main(){ clock_t start = clock(); // Execuatable code clock_t stop = clock(); double elapsed = (double)(stop – start) * 1000.0 / CLOCKS_PER_SEC; printf(“Time elapsed in ms: %f”, elapsed); } 3 solved how to show time performance for merge sort? [closed]
It is 2^n because for each job it is either take it or not take it (0,1). In other words, there are two possibilities for each job. For n jobs it is 2^n. n! Is total of permutations for n jobs, it is used when you have taken n jobs but you want to know … Read more
There’s nothing that makes recursion inherently more efficient. Moreover, the same algorithm implemented with recursion is likely to be less efficient due to function call overhead. Worse than that, and this is exactly your case, using recursions of large depth is likely to cause stack overflow (no pun intended). Unless tail-recursion optimization is in use, … Read more
In your second code block you are first checking if the first character is the same as the last. If it isn’t, return false. Else, return true. So this isn’t going to be a valid palindrome check. It only checks if the first and last letters are the same or not and ignores the letters … Read more
You could use reduce for this kind of thing, here is an example: var calendar = {Q1 : {P1 : {WK1 : {start: ‘1/1/2018’,end: ‘1/7/2018’},WK2 : {start: ‘1/8/2018’,end: ‘1/14/2018’}},P2 : {WK3 : {start: ‘1/15/2018’,end: ‘1/21/2018’}}},Q2 : {P3 : {WK5 : {start: ‘2/1/2018’,end: ‘2/7/2018’},WK6 : {start: ‘2/8/2018’,end: ‘2/14/2018’}},P4 : {WK7 : {start: ‘2/15/2018’,end: ‘2/21/2018’}}}}; var result … Read more
Any number can be represented as a difference of squares if it is odd, or a multiple of 4. The problem arises when a product has only a single 2 in the prime factorisation. So we can mark them. (i.e all the 2) Given this we can easily deduce that only what is not a … Read more
Enter the first string in the variable n. Enter the second string in the variable m. Compare the two string lengths : *if they’re different , return false. Scan the two vector , character per character : *if there’s one difference, return false. Return success. solved Given two strings of length, test if the two … Read more
As everyone is refferring – weight is the sum of the numbers, not digits. And the function is simple: var w = 0; for (var i=0; i<arr.length; i++) { w = w + arr[i]; } console.log(w); But if you want to add together all the digits, you were almost correct – number to string, split … Read more
I got requested result with below codes, another approach than what Marcos provided and I was able to find as well. Only additional thing is you need is to pad arrays to maximum length + 1 with numpy pad. def elementsum2array(arr1, arr2): ”’ 1. Make the arrays of same length padding at the beginning/left side … Read more
In the first example the value of p is reset to 1 on each iteration for i. In the second example the value of p is set just once and then is never reset, so it’s accumulating in all of the iterations of the loop over i. 1 solved I need Help about Loop [closed]