How to Solve: Jump Game (Greedy vs DP) – Complete Guide for FAANG Interviews

How to Solve: Jump Game (Greedy vs DP) – Complete Guide for FAANG Interviews

? Introduction

"This single problem separates junior candidates from senior engineers—it tests greedy intuition, dynamic programming trade-offs, and edge-case awareness. Master it, and you’ll crush 1 in 4 onsite interviews where reachability or optimal path questions appear."

? WHAT YOU NEED TO KNOW FIRST

Before tackling Jump Game, ensure you understand: 1. Greedy Algorithms – Making locally optimal choices at each step to reach a global optimum. 2. Dynamic Programming (DP) – Breaking problems into overlapping subproblems and storing results to avoid recomputation. 3. Array Traversal – Efficiently iterating through arrays while tracking state (e.g., farthest reachable index).

? KEY TECHNIQUES & PATTERNS

? STEP-BY-STEP FRAMEWORK

Use this mental checklist for every Jump Game variant:

1. Clarify the Problem
2. Is the goal reachability (True/False) or minimum jumps?
3. Are jumps 0-based or 1-based? (Assume 0-based unless specified.)

4. Can you jump exactly nums[i] steps, or up to nums[i] steps?

5. Choose the Right Technique

6. Reachability? → Greedy (O(n) time, O(1) space).
7. Minimum jumps? → BFS (O(n) time, O(n) space) or DP (O(n²) time, O(n) space).

8. Follow-up with constraints? → DP or backtracking.

9. Greedy Approach (Reachability)

10. Initialize farthest = 0.

11. Iterate through the array:

    • If i > farthest, return False (can’t proceed further).
    • Update farthest = max(farthest, i + nums[i]).
    • If farthest >= last_index, return True.

12. DP Approach (Reachability)

13. Initialize dp[0] = True (starting point is reachable).
14. For each i from 1 to n-1:
    • For each j from 0 to i-1:
    • If dp[j] == True and j + nums[j] >= i, set dp[i] = True and break.

15. Return dp[n-1].

16. BFS Approach (Minimum Jumps)

17. Use a queue to track indices and their jump counts.
18. Mark visited indices to avoid cycles.

19. For each level (jump), explore all reachable indices and increment jump count.

20. Edge Cases

21. Empty array → True (trivial reachability).
22. Single element → True.
23. nums[0] = 0 → False (can’t move).

24. Large jumps (e.g., nums[i] = n-1) → Early termination.

25. Optimize

26. For DP, can you reduce space? (Yes, use a 1D array or even O(1) for reachability.)

27. For BFS, can you avoid revisiting indices? (Yes, track farthest like greedy.)

28. Test

29. Manually verify with small inputs (e.g., [2,3,1,1,4] → True).
30. Check edge cases (e.g., [3,2,1,0,4] → False).

? WORKED EXAMPLES

Example 1 – Basic (Reachability)

Problem: Given an array nums, determine if you can reach the last index starting from index 0. You can jump up to nums[i] steps from index i.

Input: nums = [2,3,1,1,4] Output: True (0 → 1 → 4)

Greedy Solution (Optimal)

def canJump(nums):

    farthest = 0

    for i in range(len(nums)):

        if i > farthest:

            return False

        farthest = max(farthest, i + nums[i])

        if farthest >= len(nums) - 1:

            return True

    return True  # Edge case: single element

Time: O(n) | Space: O(1) Why this works: At each step, we track the farthest reachable index. If we can’t proceed further (i > farthest), we return False. If we reach or exceed the last index, we return True.

Example 2 – Medium (Minimum Jumps)

Problem: Return the minimum number of jumps to reach the last index.

Input: nums = [2,3,1,1,4] Output: 2 (0 → 1 → 4)

BFS Solution (Optimal)

from collections import deque

def jump(nums):

    if len(nums) <= 1:

        return 0

    queue = deque([(0, 0)])  # (index, jumps)

    visited = set([0])

    while queue:

        idx, jumps = queue.popleft()

        for step in range(1, nums[idx] + 1):

            next_idx = idx + step

            if next_idx >= len(nums) - 1:

                return jumps + 1

            if next_idx not in visited:

                visited.add(next_idx)

                queue.append((next_idx, jumps + 1))

    return -1  # Unreachable

Time: O(n) | Space: O(n) Why this works: BFS explores all possible jumps level by level, ensuring the first time we reach the end is with the minimum jumps.

Greedy Optimization (O(1) Space)

def jump(nums):

    jumps = 0

    current_end = 0

    farthest = 0

    for i in range(len(nums) - 1):  # Don't need to jump from last index

        farthest = max(farthest, i + nums[i])

        if i == current_end:

            jumps += 1

            current_end = farthest

    return jumps

Time: O(n) | Space: O(1) Why this works: We track the farthest reachable index (farthest) and the end of the current jump (current_end). When we reach current_end, we increment jumps and extend current_end to farthest.

Example 3 – Hard (Follow-Up: Jump Game II with Obstacles)

Problem: Given nums where nums[i] is the maximum jump length, and an array obstacles where obstacles[i] = True means you cannot land on index i, return the minimum jumps to reach the end.

Input: nums = [2,3,1,1,4], obstacles = [False, False, True, False, False] Output: 2 (0 → 1 → 4, skipping index 2)

BFS with Obstacles

from collections import deque

def jumpWithObstacles(nums, obstacles):

    if len(nums) <= 1:

        return 0

    queue = deque([(0, 0)])  # (index, jumps)

    visited = set([0])

    while queue:

        idx, jumps = queue.popleft()

        for step in range(1, nums[idx] + 1):

            next_idx = idx + step

            if next_idx >= len(nums) - 1:

                return jumps + 1

            if next_idx not in visited and not obstacles[next_idx]:

                visited.add(next_idx)

                queue.append((next_idx, jumps + 1))

    return -1  # Unreachable

Time: O(n) | Space: O(n) Why this works: BFS ensures the shortest path, and we skip obstacles by checking obstacles[next_idx].

❌ Common Mistakes

? INTERVIEW TrapS

? 1-Minute Recap

"Alright, let’s lock this in. For Jump Game problems, start by clarifying: is it reachability or minimum jumps? For reachability, use greedy—track the farthest index you can reach at each step. If you can’t proceed, return False. If you reach the end, return True. For minimum jumps, BFS is your friend—treat each index as a node and explore jumps level by level. Optimize space by tracking the farthest reachable index instead of a queue. Watch out for edge cases: empty arrays, single elements, and nums[0] = 0. And if the interviewer throws in obstacles or negative jumps, fall back to BFS or DP. You’ve got this—now go crush that interview!

? Final Notes

• Greedy is king for reachability (O(n) time, O(1) space).
• BFS is optimal for minimum jumps (O(n) time, O(n) space).
• DP is a fallback for complex constraints (O(n²) time, O(n) space).
• Always test edge cases—they’re where most candidates fail.

Now go practice on LeetCode 55 (Jump Game I) and 45 (Jump Game II)! ?