Microsoft Software Engineer Interview Questions

I would first think about splitting the sentence into words, but that would use extra space if I create a new string. To optimize it, I would reverse the whole string first and then reverse each word one by one so the words end up in the correct order. The reason I choose this approach is that it is still O(n) time and can be done in O(1) extra space if I have a mutable buffer like a character array. I would also mention edge cases like multiple spaces, leading and trailing spaces, and an empty string. If the interviewer wants a simpler version, I would explain the same logic step by step before coding it.

I would use the slow and fast pointer technique. The slow pointer moves one step at a time, and the fast pointer moves two steps at a time, so when the fast pointer reaches the end, the slow pointer is at the middle. If the list has an even number of nodes, I would clarify whether the interviewer wants the first middle or the second middle, because that affects the stopping condition. The reason I choose this approach is that it gives O(n) time and O(1) space, which is very efficient. I would also mention edge cases like a single-node list and an empty list.

I would first think about sorting both strings and comparing them, which is easy to explain but not the most efficient. To optimize it, I would count the frequency of each character using a hash map or a fixed-size array if the input is only lowercase English letters. If the character counts match, then the strings are anagrams. The reason I choose counting over sorting is that it gives O(n) time instead of O(n log n). I would also mention edge cases like different lengths, uppercase versus lowercase handling, and Unicode if the problem allows it.

I would use the BST property to walk down the tree instead of searching both sides. If both target values are smaller than the current node, I move left; if both are larger, I move right; otherwise, the current node is the split point and therefore the LCA. The reason I choose this approach is that it is simple and runs in O(h) time, where h is the height of the tree. The space complexity is O(1) if I do it iteratively. I would also mention that for a general binary tree, I would need a recursive approach instead.

I would first think about binary search over the answer range, especially if the interviewer wants an integer square root. If I need better convergence for floating-point precision, I would also consider Newton's method. The reason I choose binary search is that it is easier to explain and works well for bounded precision, while Newton's method is faster once I am comfortable with the math. I would mention edge cases like 0, 1, and negative input if the problem allows it. The time complexity is O(log n) for binary search over the value range, and the space complexity is O(1).