What is the brute force approach and complexity for Jump Game II?

Recursively try every landing spot and keep the smallest jump count with time complexity O(2^N) and space complexity O(N).

What is the optimal approach and complexity for Jump Game II?

Scan once; add a jump each time the scan reaches the end of the current round with optimal time complexity O(N) and space complexity O(1).

How do you identify the pattern for Jump Game II?

Look for key signals in the prompt: "minimum number of jumps" indicates Each jump is one more round. The answer is how many rounds it takes to reach the end.; "maximum jump length" indicates Every index in a round can start the next round, so the rounds are blocks that touch each other.; "you can always reach the last index" indicates The rounds keep growing until they cover the last index, so the loop always finishes..

Jump Game II

Medium·LeetCode #45 Pro OnlyBasic Reinforcement

LeetCode

Problem Statement

You start on index 0 of an array nums. Each number is the longest jump you can make from that index: from index i you may land on any index from i + 1 up to i + nums[i], as long as you stay inside the array.

The last index can always be reached. Return the fewest jumps that get you from index 0 to the last index.

Input

An array of integers nums. nums[i] is the furthest forward jump from index i.

Output

The fewest number of jumps needed to reach the last index.

Example 1

Input: nums = [2, 3, 1, 1, 4]

Output: 2

Jump 1 step from index 0 to index 1, then 3 steps from index 1 to the last index.

Example 2

Input: nums = [2, 3, 0, 1, 4]

Output: 2

Jump from index 0 to index 1, then from index 1 to index 4. The zero at index 2 is never needed.

Constraints

• 1 ≤ nums.length ≤ 10^4
• 0 ≤ nums[i] ≤ 1000
• The last index can always be reached.

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“minimum number of jumps”

→ Each jump is one more round. The answer is how many rounds it takes to reach the end.

“maximum jump length”

→ Every index in a round can start the next round, so the rounds are blocks that touch each other.

“you can always reach the last index”

→ The rounds keep growing until they cover the last index, so the loop always finishes.

Approach Complexity Ladder

Brute Force — try every jump, keep the fewestO(2^N)O(N)

Recursively try every landing spot and keep the smallest jump count

Better — minJumps[i] table over earlier indexesO(N²)O(N)

minJumps[i] is the fewest jumps to reach i, taken from the best earlier index that jumps to it

Optimal — count the rounds with curEnd and farO(N)O(1)

Scan once; add a jump each time the scan reaches the end of the current round

Brute Force Intuition

Let fewestFrom(i) be the fewest jumps from index i to the last index. If i is the last index, the answer is 0. Otherwise try each landing spot i + len for len from 1 up to nums[i], take 1 + fewestFrom(i + len), and keep the smallest.

Recognition rule for this sub-pattern:

Track the farthest point you can reach so far; the answer depends only on that frontier.

Remember this · Greedy

Trigger

Maximise count / minimise cost with a natural "best next step", sorting makes the choice obvious, reachability.

Think

Is there a choice I can always make safely without looking ahead?

Key question

Why can't the greedy choice be worse than the optimal one? (Try to break it with a small example.)

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