Fullstack developer Intern

Step 1: I started by thinking about how to approach this problem using two pointers. Initially, I set both the left and right pointers at the start of the string, and thought about expanding the window by moving the right pointer to the right.

Step 2: I realized that as the window expanded, I needed to track how many distinct characters were in the window. So, I used a hashmap (or dictionary) to count the frequency of each character within the window.

Step 3: As I moved the right pointer, I kept updating the hashmap with the character counts. If the number of distinct characters (i.e., the size of the hashmap) ever exceeded k, I knew I needed to shrink the window. So, I moved the left pointer to the right until the window had at most k distinct characters again.

Step 4: While doing this, I kept track of the maximum length of the window that satisfied the condition (at most k distinct characters).

Step 5: This approach only required one pass through the string, making it much more efficient (O(n)) compared to using nested loops or sorting

Step 1: I started by thinking about how to handle the intervals. The first thing that came to mind was sorting the intervals based on their start times. This seemed like a natural approach because if intervals are sorted, any overlapping intervals will be next to each other.

Step 2: Once the intervals were sorted, I knew I had to check each interval against the last merged interval. If the current interval started before the last merged interval ended, I would merge them. This would mean updating the end of the merged interval to the maximum of the two ends.

Step 3: If the current interval didn’t overlap with the last merged interval, I would just add the current interval to the list of merged intervals.

Step 4: After completing the entire list of intervals, I would have the final list with all overlaps merged.

Step 5: I realized this could be done in O(n log n) time complexity because sorting the intervals takes O(n log n), and then merging the intervals in a single pass is O(n). This seemed to be the most efficient approach.

Step 1: I started by iterating through the array and comparing each element with the current highest and lowest values. Initially, I set the first element as both the highest and lowest. This seemed like the simplest approach.

Step 2: The interviewer asked me to optimize the solution since this could be done in a more efficient way, and it didn’t require sorting or multiple passes through the array.

Step 3: I then realized that I could achieve this in a single traversal by updating the highest and lowest values as I went through the array, comparing each element with the current maximum and minimum values. This way, I reduced the complexity to O(n), which was much better.

Step 1: At first, I considered using recursion to traverse the tree, but I quickly realized that it wasn’t the best approach for visiting nodes level by level. 

Step 2: So, I switched to using a queue. I added the root node to the queue and started processing its children one at a time. 

Step 3: As I moved through each level, I kept adding the child nodes of the current level into the queue. This helped me traverse the tree level by level, and the interviewer seemed happy with how I approached the problem.

Step 1: I started by iterating through the array and comparing each element with the current highest and lowest values. Initially, I set the first element as both the highest and lowest. This seemed like the simplest approach.

Step 2: The interviewer asked me to optimize the solution since this could be done in a more efficient way, and it didn’t require sorting or multiple passes through the array.

Step 3: I then realized that I could achieve this in a single traversal by updating the highest and lowest values as I went through the array, comparing each element with the current maximum and minimum values. This way, I reduced the complexity to O(n), which was much better.