Leetcode 55 - Jump Game

30-Jun-2026

Things are changing too fast.

Couple years back, compsci majors weren’t expected to have junior-level experience to get a job, but now we’re expected to have much more. What’s more, we have AI evolving every day, heck even every hour, which makes it harder to envision what my career path will look like in the coming future. With that said, it’s never bad to keep practicing and honing my programming and problem solving skills.

Jumping to conclusions

Jumping from the first step all the way to the last one requies checking for a SINGLE valid run through. That means, we just need to check if we can reach the final step without worrying about the minimum number of jumps or anything. To do this, the simple way would be to go from the top and see if we can reach the current position from any of the previous steps, closer to the bottom step the better.

That was the idea behind my thought process here, but kinda wishy-washy.

After I finished writing the solution above, I figured things could be done with less variables. Instead of having two index variables, we can just have one and use a for-loop that’ll iterate from the last stair all the way to the first to determine the validity of our array.

The edit came out a lot cleaner than before, which made me feel kinda warm inside!

Performance evaluation

Simply put, my solution is a single pass through the array from the last to the first index, so the time complexity is simply $O(n)$ . This solution only uses a sinlge additional variable curr_idx, so additional space complexity is $O(1)$ .

All and all, my solution was satisfactory for a medium difficulty question!