Minimum Operations to Reduce X to Zero - Leetcode 1658 - Python 

yeah man, hard for chumps like me to solve

THANK YOU SO MUCH, but his solution is more efficient since you are using a hash map the memory will be O(n) and the memory in his code is O(1)

"so simple" 🙄

He actually says target < 0 is part of the reason left and right could cross when it is actually the whole reason He has a bias towards unnecessary checks in loops that felt inefficient at first but doesn't seem to affect runtime. I don't know how it plays in interviews.

Great question, can be confusing but note the ordering, since r always points to a valid element, in the real and valid case that l == r == len(nums - 1) it will first subtract the element from l which is the last element and is valid, l (left) will then increase out of bounds but the for loop will terminate at the top because r is at its max and we never access the element at left after we increment it potentially out of bounds. So our code may end with l == len(nums), an invalid index and right and left have crossed which we usually hate, but at that point everything is guaranteed to terminate and we never access those elements at those pointers again and thus never have to check for those conditions.

no this doesn't work. For example, lets say you had x =5 and nums = [2,2,1,2,10,4], your algorithm would subtract 4 first then 2 which will never work

@@impolitebigmac3044 oh thanks for the explanation

That's what I did initially. Greedy approach. Failed after 45 test cases

No, I don't think the array is sorted. Check the example.

Why would it be sorted?

Got it 😅