Another Casual Coder

Problem is here: https://leetcode.com/problems/peak-index-in-a-mountain-array/description/ Let's call an array  A  a  mountain  if the following properties hold: A.length >= 3 There exists some  0 < i < A.length - 1  such that  A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1] Given an array that is definitely a mountain, return any  i  such that  A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1] . Example 1: Input: [0,1,0] Output: 1 Example 2: Input: [0,2,1,0] Output: 1 If you play close attention to the problem, it actually ends up being much simpler than the whole problem statement suggests. Since the input is confirmed to be a mountain (hence you don’t need to check whether it is a mountain or not), then all you really need is to find the max element in the array! Now even if the question was to deter...

Problem is labeled as easy by LeetCode, here it is:  https://leetcode.com/problems/backspace-string-compare/description/ : Given two strings  S  and  T , return if they are equal when both are typed into empty text editors.  #  means a backspace character. Example 1: Input: S = "ab#c" , T = "ad#c" Output: true Explanation : Both S and T become "ac". Example 2: Input: S = "ab##" , T = "c#d#" Output: true Explanation : Both S and T become "". Example 3: Input: S = "a##c" , T = "#a#c" Output: true Explanation : Both S and T become "c". Example 4: Input: S = "a#c" , T = "b" Output: false Explanation : S becomes "c" while T becomes "b". My solution won't be the most efficient - it will run in O(n)-time and O(n)-space. There is a way to do in O(1)-space but this is a straightforward problem to use a stack that I decided to use i...