https://www.hackerrank.com/challenges/compare-the-triplets/problem function solve(a0, a1, a2, b0, b1, b2) { var solution = [] if (a0 > b0 || a1 > b1 || a2 > b2) { solution += 1; } else if (a0 < b0 || a1 < b1 || a2 < b2 ) { solution += 1; } return solution....
The median of M numbers is defined as the middle number after sorting them in order if M is ...
publicclassSolution{privatestaticBigDecimalone=newBigDecimal(1);privatestaticbooleanisPrime(String n){BigIntegerinteger=newBigInteger(n);returninteger.isProbablePrime(1); }publicstaticvoidmain(String[] args)throwsIOException {BufferedReaderbufferedReader=newBufferedReader(newInputStreamReader(System.in));Stringn...
Solution in JS: function getNumberSwaps(arr, arrMap, sortedArr) { let numberOfSwaps = 0; for (let i = 0; i < sortedArr.length; i++) { if (sortedArr[i] !== arr[i]) { const tmp = arr[i]; const idxSource = arrMap[sortedArr[i]]; arr[i] = arr[idxSource]; arr[idxSour...
* Complete the 'travelAroundTheWorld' function below. * * The function is expected to return an INTEGER. * The function accepts following parameters: * 1. INTEGER_ARRAY a * 2. INTEGER_ARRAY b * 3. LONG_INTEGER c */ inttravelAroundTheWorld(vector<int> a, vector<int> b,longc){ ...
但是复杂度为O(n^3),显然无法满足题目要求。必须优化到O(n)的复杂度。think about it, how to optimize the solution...??? Accepted Code: 1#include <iostream>2#include <cstring>3#include <cstdlib>4#include <vector>5usingnamespacestd;67constintMOD =1000000000+7;8constintMAX_N =100005;9typede...
for (int c: ch) { if (c == 0) continue; count += Math.abs(c); } return count/2; } } Count Duplicates 数组中重复元素的个数 题目 找到一个数组中重复元素的个数。 Complete the countDuplicates function in the editor below. It has 1 parameter: an array of integers, numbers. It must...
Hello Codeforces community, I would like to invite you to join HackerRank's101 Hack 50on June 20, 2017. There will be five tasks and three hours for you to solve them. The contest will be rated and the top ten contestants will receive HackerRank T-shirts!
After that, I remove the boundary values ( points that lie on the axis ) and multiply with number of quadrant, and finally add back the boundary values appropriately. Got only 4 test cases correct. Saw in the editorial that they used generating function. I am curious why my solution does...
function acmTeam(topic) { // Write your code here let possibleTeams = []; for ( let i = 0; i < topic.length; i++ ) { for ( let j = 0; j < topic.length; j++ ) { if ( topic[i] !== topic[j] && !possibleTeams[possibleTeams.length - 1].includes([topic[i]...