Greedy

Intuition My thought was first going left to right checking both neighbours. For example, for [1,2,3,4] would work perfectly. However it’s not possible to solve this in one pass checking both neighbours. We need to use a two pass approach, LTR, RTL. To visualize better the problem, using [1,2,3,4] let’s see how many candies LTR and RTL would give for each children: LTR [1,2,3,4] -> [1,2,3,4] since it's always increase each children get's one more candy than the previous one. RTL [1,2,3,4] -> [1,1,1,1] From RTL perspective, it's always decreasing, so all children get's one. In this case the correct solution is [1,2,3,4] ...

Intuition Find the first station where we have a net positive tank. However that solution is not optimized. Approach After reading the editorial, it’s possible to complete the solution with optimal time complexity using the greedy approach (one pass). The idea is to go iterate over the elements and whenever we find a net negative level we restart the segment and keep the next array index as starting index. Also, while going over the elements we need to keep a the final tank level. If the final tank level is negative, then there’s no solution. ...