Jump Game Problem: How Do You Solve It?
Learn the greedy O(n) approach to the jump game problem, why it beats O(n^2) DP, and how to answer it in interviews.
Expected Interview Answer
The jump game problem — determining if you can reach the last index of an array where each element is the maximum jump length from that position — is best solved with a greedy O(n) single pass that tracks the farthest index reachable so far, rather than the O(n²) dynamic programming approach of checking every reachable index explicitly.
Walking left to right, maintain a variable "farthest" representing the furthest index reachable using any position visited up to now; at each index i, if i exceeds farthest, the array is unreachable there and the answer is false, otherwise update farthest to max(farthest, i + nums[i]). If farthest ever reaches or exceeds the last index, the answer is true immediately. This greedy approach works because reachability is monotonic — if you can reach index i, you never need to "go back" and reconsider it, so tracking the single best frontier is sufficient and avoids the O(n²) DP of checking canReach[i] for every prior index. A common follow-up, "jump game II", asks for the minimum number of jumps to reach the end, which extends this greedy frontier-tracking idea by counting how many times the frontier needs to be advanced.
- O(n) time with a single greedy pass
- O(1) space, no DP array needed
- Reachability is monotonic, which justifies the greedy correctness
- Extends naturally to the minimum-jumps variant (Jump Game II)
AI Mentor Explanation
Determining whether a team can reach the final over of an innings, where each ball’s outcome tells you the maximum number of overs you could theoretically skip ahead from there, is the jump game problem. Walking over by over, you track the farthest over you could theoretically reach given everything seen so far, and if the current over ever falls beyond that farthest reachable point, the innings stalls out right there. If the farthest reachable point ever covers the final over, you know reaching it is possible without simulating every possible sequence of over-skips. This single running frontier, updated once per over, replaces checking every possible combination of skips explicitly.
Step-by-Step Explanation
Step 1
Track the farthest reachable index
Initialize farthest = 0; this represents the best frontier reached using positions visited so far.
Step 2
Scan left to right
At each index i, if i > farthest, the array is unreachable from here — return false immediately.
Step 3
Update the frontier
Otherwise set farthest = max(farthest, i + nums[i]) using the jump length at position i.
Step 4
Check completion
If farthest >= last index at any point, return true — no need to keep scanning.
What Interviewer Expects
- Propose the greedy O(n) "farthest reachable" approach, not just O(n²) DP
- Justify why greedy works (monotonic reachability — no benefit to revisiting)
- Correctly handle the edge case of a single-element array (trivially true)
- Connect to the Jump Game II variant (minimum jumps) when asked as a follow-up
Common Mistakes
- Defaulting to O(n²) DP (canReach[i] checked against every prior index) without mentioning the greedy optimization
- Forgetting to check i > farthest as an early-exit unreachable condition
- Confusing "jump game" (can you reach the end) with "jump game II" (minimum jumps to the end)
- Not short-circuiting once farthest already covers the last index
Best Answer (HR Friendly)
“The jump game asks whether you can reach the last index of an array, where each number tells you the farthest you can jump from that position. I solve it greedily by walking through the array once and tracking the farthest index reachable so far — if I ever land on an index beyond that frontier the answer is no, and if the frontier ever reaches the end the answer is yes, all in linear time.”
Code Example
def can_jump(nums: list[int]) -> bool:
farthest = 0
for i, jump in enumerate(nums):
if i > farthest:
return False
farthest = max(farthest, i + jump)
if farthest >= len(nums) - 1:
return True
return True
# can_jump([2, 3, 1, 1, 4]) -> True
# can_jump([3, 2, 1, 0, 4]) -> FalseFollow-up Questions
- How would you solve Jump Game II, which asks for the minimum number of jumps to reach the end?
- How would you modify this to also return one valid sequence of jump indices?
- What is the time complexity of a brute-force DP approach, and why is greedy better here?
- How would you handle an array where some values are zero, creating potential dead ends?
MCQ Practice
1. What does the "farthest" variable represent in the greedy jump game solution?
It tracks the best frontier achievable from any position scanned up to the current one, which is the key greedy invariant.
2. What condition immediately proves the array is unreachable from the current index?
If the current index exceeds the farthest reachable point established so far, no prior jump can get you here, so the end is unreachable.
3. Why does the greedy approach give the correct answer for the jump game?
Once an index is reachable, it never needs to be reconsidered, so maintaining just the farthest frontier captures all necessary information.
Flash Cards
What is the optimal time complexity for the jump game problem? — O(n), using a single greedy pass tracking the farthest reachable index.
What triggers an early "false" return in the greedy jump game solution? — When the current index i exceeds the tracked farthest reachable index.
What is Jump Game II? — A follow-up variant asking for the minimum number of jumps needed to reach the last index.
Why is greedy correct for the jump game instead of full DP? — Reachability is monotonic — once reachable, an index need never be reconsidered, so tracking one frontier suffices.