Approach Summary
Track max reachable index. If current index exceeds max reachable, return false. Update max at each step.
Practice Mode
2 Progressive Approaches
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Solution Progression (Brute Force → Most Optimized)
Understand the architectural evolution from naive baseline to optimal complexity.
| Approach | Time | Space | Core Strategy / Trade-off |
|---|---|---|---|
| 1 Approach 1: Backward Greedy Target Shift | O(n) | O(1) | Work backwards from destination n - 1, shifting target to i whenever i + nums[i] >= target. |
| 2 Approach 2: Forward Greedy Maximum Reach (Optimal) | O(n) | O(1) | Track max reachable boundary as you iterate forward, failing early if current index > maxReachable. |
Step 1: Baseline
Approach 1: Backward Greedy Target Shift
Time: O(n)
Space: O(1)
Set destination target = n - 1. Iterate backwards from n - 2 down to 0. If index i can jump to or beyond target (i + nums[i] >= target), update target = i. If target successfully reaches index 0 by the end of the loop, a valid path exists.
Approach 1: Backward Greedy Target Shift Implementation
Approach 1: Backward Greedy Target Shift
class Solution {
public boolean canJump(int[] nums) {
int target = nums.length - 1;
for (int i = nums.length - 2; i >= 0; i--) {
if (i + nums[i] >= target) {
target = i;
}
}
return target == 0;
}
}class Solution {
public:
bool canJump(vector<int>& nums) {
int target = nums.size() - 1;
for (int i = nums.size() - 2; i >= 0; i--) {
if (i + nums[i] >= target) {
target = i;
}
}
return target == 0;
}
};class Solution:
def canJump(self, nums: list[int]) -> bool:
target = len(nums) - 1
for i in range(len(nums) - 2, -1, -1):
if i + nums[i] >= target:
target = i
return target == 0function canJump(nums) {
let target = nums.length - 1;
for (let i = nums.length - 2; i >= 0; i--) {
if (i + nums[i] >= target) {
target = i;
}
}
return target === 0;
} Step 2: Most Optimized
Approach 2: Forward Greedy Maximum Reach (Optimal)
Time: O(n)
Space: O(1)
Keep maxReachable = 0. For each index i, if maxReachable < i, you are stranded and cannot proceed. Otherwise, update maxReachable = max(maxReachable, i + nums[i]). If at any point maxReachable >= n - 1, the destination is reachable.
Approach 2: Forward Greedy Maximum Reach (Optimal) Implementation
Approach 2: Forward Greedy Maximum Reach (Optimal)
class Solution {
public boolean canJump(int[] nums) {
int n = nums.length;
int maxReachable = 0;
for (int i = 0; i < n; i++) {
if (maxReachable < i) {
return false;
}
maxReachable = Math.max(maxReachable, i + nums[i]);
}
return true;
}
}class Solution {
public:
bool canJump(vector<int>& nums) {
int n = nums.size();
int maxReachable = 0;
for (int i = 0; i < n; i++) {
if (maxReachable < i) {
return false;
}
maxReachable = max(maxReachable, i + nums[i]);
}
return true;
}
};class Solution:
def canJump(self, nums: list[int]) -> bool:
max_reachable = 0
for i, val in enumerate(nums):
if i > max_reachable:
return False
max_reachable = max(max_reachable, i + val)
return Truefunction canJump(nums) {
let maxReachable = 0;
for (let i = 0; i < nums.length; i++) {
if (i > maxReachable) return false;
maxReachable = Math.max(maxReachable, i + nums[i]);
}
return true;
}Edge Cases to Watch
- Single element array (e.g. nums = [0]; already at destination, returns true)
- Zero preventing progress (e.g. [3, 2, 1, 0, 4]; cannot jump past index 3, returns false)
- Large jump from start (e.g. [5, 0, 0, 0, 0]; can jump directly to end, returns true)
- Zero at intermediate position but easily jumped over (e.g. [2, 0, 0]; returns true)
Algorithmic Pattern
View Pattern Guide →
Master the Greedy Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Greedy pattern.
How to Recognize This Pattern
- Can you reach the end?
- Greedy max reach
Complexity Analysis
Time Complexity
O(n)
Space Complexity
O(1)
Tags
Array Greedy DP
Frequently Asked Questions
What is the optimal time and space complexity for Jump Game?
The optimal solution for Jump Game achieves O(n) runtime complexity and requires O(1) auxiliary space. Time complexity is O(n) because we make a single pass through the array. Space complexity is O(1) as only a single integer tracking max reachable position is maintained.
How do I identify that Jump Game requires the Greedy pattern?
In technical interviews, look for key problem signals: Can you reach the end?; Greedy max reach. Recognizing these invariants allows you to bypass naive brute-force exhaustive search in favor of Greedy strategy.
What critical edge cases must be handled in Jump Game?
Essential edge cases to verify include: Single element array (e.g. nums = [0]; already at destination, returns true); Zero preventing progress (e.g. [3, 2, 1, 0, 4]; cannot jump past index 3, returns false); Large jump from start (e.g. [5, 0, 0, 0, 0]; can jump directly to end, returns true); Zero at intermediate position but easily jumped over (e.g. [2, 0, 0]; returns true). Always walk through these scenarios with your interviewer before coding.
Which tech companies frequently ask Jump Game?
Jump Game is a Medium-difficulty question tagged with high recurrence. It is actively asked in technical screening and onsite interview loops at Amazon, Microsoft, Apple.