Approach Summary
BFS-like greedy: track current window end and farthest reach. When reaching window end, increment jumps and extend window.
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: Dynamic Programming Tabulation | O(n^2) | O(n) | dp[i] represents minimum jumps to reach index i by checking all preceding indices j < i. |
| 2 Approach 2: Greedy Implicit BFS (Optimal) | O(n) | O(1) | Track current jump boundary and max reach, incrementing jumps whenever index reaches current. |
Step 1: Baseline
Approach 1: Dynamic Programming Tabulation
Time: O(n^2)
Space: O(n)
Initialize array dp of size n with infinity and dp[0] = 0. For each index i from 1 to n - 1, iterate through all j from 0 to i - 1: if j + nums[j] >= i, relax dp[i] = min(dp[i], dp[j] + 1). Return dp[n - 1]. This quadratic method is intuitive but slower on large inputs.
Approach 1: Dynamic Programming Tabulation Implementation
Approach 1: Dynamic Programming Tabulation
class Solution {
public int jump(int[] nums) {
int n = nums.length;
int[] dp = new int[n];
Arrays.fill(dp, Integer.MAX_VALUE);
dp[0] = 0;
for (int i = 1; i < n; i++) {
for (int j = 0; j < i; j++) {
if (j + nums[j] >= i && dp[j] != Integer.MAX_VALUE) {
dp[i] = Math.min(dp[i], dp[j] + 1);
}
}
}
return dp[n - 1];
}
}class Solution {
public:
int jump(vector<int>& nums) {
int n = nums.size();
vector<int> dp(n, INT_MAX);
dp[0] = 0;
for (int i = 1; i < n; i++) {
for (int j = 0; j < i; j++) {
if (j + nums[j] >= i && dp[j] != INT_MAX) {
dp[i] = min(dp[i], dp[j] + 1);
}
}
}
return dp[n - 1];
}
};class Solution:
def jump(self, nums: list[int]) -> int:
n = len(nums)
dp = [float('inf')] * n
dp[0] = 0
for i in range(1, n):
for j in range(i):
if j + nums[j] >= i:
dp[i] = min(dp[i], dp[j] + 1)
return dp[n - 1]function jump(nums) {
const n = nums.length;
const dp = new Array(n).fill(Infinity);
dp[0] = 0;
for (let i = 1; i < n; i++) {
for (let j = 0; j < i; j++) {
if (j + nums[j] >= i) {
dp[i] = Math.min(dp[i], dp[j] + 1);
}
}
}
return dp[n - 1];
} Step 2: Most Optimized
Approach 2: Greedy Implicit BFS (Optimal)
Time: O(n)
Space: O(1)
Maintain current (the furthest index reachable with current jump count) and maxReachable (the furthest index reachable with one additional jump). As we scan through indices 0 to n - 2, extend maxReachable. When i reaches current, we must take another jump: current = maxReachable and jumps++.
Approach 2: Greedy Implicit BFS (Optimal) Implementation
Approach 2: Greedy Implicit BFS (Optimal)
class Solution {
public int jump(int[] nums) {
int maxReachable = 0;
int jumps = 0;
int current = 0;
int n = nums.length;
for (int i = 0; i < n - 1; i++) {
maxReachable = Math.max(maxReachable, i + nums[i]);
if (current == i) {
current = maxReachable;
jumps++;
}
}
return jumps;
}
}class Solution {
public:
int jump(vector<int>& nums) {
int maxReachable = 0;
int jumps = 0;
int current = 0;
int n = nums.size();
for (int i = 0; i < n - 1; i++) {
maxReachable = max(maxReachable, i + nums[i]);
if (current == i) {
current = maxReachable;
jumps++;
}
}
return jumps;
}
};class Solution:
def jump(self, nums: list[int]) -> int:
n = len(nums)
max_reachable = 0
current = 0
jumps = 0
for i in range(n - 1):
max_reachable = max(max_reachable, i + nums[i])
if current == i:
current = max_reachable
jumps += 1
return jumpsfunction jump(nums) {
let maxReachable = 0;
let jumps = 0;
let current = 0;
for (let i = 0; i < nums.length - 1; i++) {
maxReachable = Math.max(maxReachable, i + nums[i]);
if (current === i) {
current = maxReachable;
jumps++;
}
}
return jumps;
}Edge Cases to Watch
- Length 1 array (nums = [0]; 0 jumps required, loop from 0 to n - 2 does not run, returns 0)
- Array where first element can reach the end in 1 jump (e.g. [5, 1, 1, 1, 1]; returns 1)
- Array requiring multiple short hops (e.g. [2, 3, 1, 1, 4]; returns 2)
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
- Minimum jumps to reach end
- BFS-like greedy
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 II?
The optimal solution for Jump Game II achieves O(n) runtime complexity and requires O(1) auxiliary space. Time complexity is O(n) because each index is processed exactly once in a single forward pass. Space complexity is O(1) since only three tracking variables (maxReachable, current, jumps) are maintained.
How do I identify that Jump Game II requires the Greedy pattern?
In technical interviews, look for key problem signals: Minimum jumps to reach end; BFS-like greedy. 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 II?
Essential edge cases to verify include: Length 1 array (nums = [0]; 0 jumps required, loop from 0 to n - 2 does not run, returns 0); Array where first element can reach the end in 1 jump (e.g. [5, 1, 1, 1, 1]; returns 1); Array requiring multiple short hops (e.g. [2, 3, 1, 1, 4]; returns 2). Always walk through these scenarios with your interviewer before coding.
Which tech companies frequently ask Jump Game II?
Jump Game II is a Medium-difficulty question tagged with high recurrence. It is actively asked in technical screening and onsite interview loops at Amazon, Google, Facebook.