DSA interview question

Jump Game

Easy · Algorithms · Topics: greedy, array

The problem

A platformer level is a row of tiles, and tile i lets the player jump up to nums[i] tiles forward. Starting on the first tile, can the player reach the last one? Exploring every jump is exponential. But reachability is monotone: if tile j is reachable, every tile before it is too. So you only need the farthest reachable index so far. Walk forward while you're inside that reach, extending it with i + nums[i]. If you ever stand beyond it, you're stuck.

Example: [2, 3, 1, 1, 4] → true; [3, 2, 1, 0, 4] → false.

Function signature

bool canJump(const vector<int>& nums);

Test cases

Input	Output	Explanation
`[2, 3, 1, 1, 4]`	`true`	0 → 1 → 4
`[3, 2, 1, 0, 4]`	`false`	Every path lands on the 0 at index 3
`[0]`	`true`	Already at the end
`[1, 0, 1]`	`false`	Stuck at index 1
`[2, 0, 0]`	`true`	Jump straight to the end

(Confirm: jumps can be any length from 1 to nums[i]; values are non-negative.)

How interviewers phrase it

"Each element in the array represents your maximum jump length at that position. Determine if you can reach the last index" (LeetCode 55).

Clarifying questions to ask

Can I jump shorter than nums[i]? (Yes, any length up to it.)
Is a single-element array already at the goal? (Yes.)
Can values be 0? (Yes, and zeros create traps.)

Follow-ups to expect

"Why is the farthest reach enough?"
"When do you fail?"
"Backward greedy?"
"DP alternative?"
"Complexity?"

Practise questions like this with a real engineer

On Practick another engineer interviews you live for 60 minutes on a question from this bank. They have the full interviewer guide beside the call: the expected answer to every follow-up, hints to give if you stall, and a scoring rubric. You get the model solution once the session settles. It is free: interview someone for an hour and you have earned your own.

Book a free mock interview