24 minute read

“Finding the optimal path through a sequence of choices.”

1. Problem Statement

You are given a 0-indexed array of integers nums of length n. You are initially positioned at nums[0].

Each element nums[i] represents the maximum length of a forward jump from index i. In other words, if you are at nums[i], you can jump to any nums[i + j] where:

  • 0 <= j <= nums[i]
  • i + j < n

Return the minimum number of jumps to reach nums[n - 1]. The test cases are generated such that you can reach nums[n - 1].

Example 1:

Input: nums = [2,3,1,1,4]
Output: 2
Explanation: The minimum number of jumps to reach the last index is 2. Jump 1 step from index 0 to 1, then 3 steps to the last index.

Example 2:

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

Constraints:

  • 1 <= nums.length <= 10^4
  • 0 <= nums[i] <= 1000
  • It’s guaranteed that you can reach nums[n-1].

2. Intuition

This is a classic Greedy problem disguised as a graph traversal.

Key Insight: At each position, we want to jump to the position that allows us to reach the farthest in the next jump. This is a local greedy choice that leads to a global optimal solution.

Think of it as BFS where each “level” represents the positions reachable with k jumps.

3. Approach 1: Greedy (Optimal)

Idea: We maintain the farthest position we can reach with the current number of jumps. When we exhaust the current range, we increment the jump count.

Algorithm:

  1. Initialize jumps = 0, current_end = 0 (end of current jump range), farthest = 0 (farthest we can reach).
  2. Iterate through the array (except the last element, since we’re already there if we reach it).
  3. For each position i:
    • Update farthest = max(farthest, i + nums[i]).
    • If i == current_end (we’ve exhausted the current jump range):
      • Increment jumps.
      • Set current_end = farthest (start a new jump range).
  4. Return jumps.
class Solution:
    def jump(self, nums: List[int]) -> int:
        n = len(nums)
        if n == 1:
            return 0
        
        jumps = 0
        current_end = 0
        farthest = 0
        
        for i in range(n - 1):  # Don't need to check the last element
            farthest = max(farthest, i + nums[i])
            
            if i == current_end:
                jumps += 1
                current_end = farthest
                
                # Early exit if we can already reach the end
                if current_end >= n - 1:
                    break
                    
        return jumps

Complexity:

  • Time: $O(N)$. Single pass through the array.
  • Space: $O(1)$. Constant extra space.

4. Approach 2: BFS (Level-Order Traversal)

We can model this as a graph where each index is a node, and there’s an edge from i to j if j <= i + nums[i].

Algorithm:

  1. Use BFS. Each level represents positions reachable with k jumps.
  2. For each level, find the farthest position reachable.
  3. If the farthest position reaches or exceeds n-1, return the level count.
from collections import deque

class Solution:
    def jump(self, nums: List[int]) -> int:
        n = len(nums)
        if n == 1:
            return 0
        
        queue = deque([0])
        visited = {0}
        jumps = 0
        
        while queue:
            size = len(queue)
            jumps += 1
            
            for _ in range(size):
                i = queue.popleft()
                
                # Try all possible jumps from position i
                for j in range(i + 1, min(i + nums[i] + 1, n)):
                    if j == n - 1:
                        return jumps
                    if j not in visited:
                        visited.add(j)
                        queue.append(j)
                        
        return jumps

Complexity:

  • Time: $O(N^2)$ in worst case (e.g., [1,1,1,1,1]).
  • Space: $O(N)$ for the queue and visited set.

5. Approach 3: Dynamic Programming

Define dp[i] = minimum jumps to reach index i.

Transition: For each position i, we can jump to any position j where i < j <= i + nums[i]. dp[j] = min(dp[j], dp[i] + 1).

class Solution:
    def jump(self, nums: List[int]) -> int:
        n = len(nums)
        dp = [float('inf')] * n
        dp[0] = 0
        
        for i in range(n):
            for j in range(i + 1, min(i + nums[i] + 1, n)):
                dp[j] = min(dp[j], dp[i] + 1)
                
        return dp[n - 1]

Complexity:

  • Time: $O(N \times M)$ where $M$ is the average jump length. Worst case $O(N^2)$.
  • Space: $O(N)$ for the DP array.

6. Deep Dive: Why Greedy Works (Proof)

Claim: The greedy algorithm always finds the minimum number of jumps.

Proof by Exchange Argument: Suppose the greedy algorithm produces a solution with k jumps: $0 \to i_1 \to i_2 \to \dots \to i_k = n-1$. Suppose there exists an optimal solution with fewer jumps: $0 \to j_1 \to j_2 \to \dots \to j_m = n-1$ where $m < k$.

Consider the first position where the two solutions differ. Let’s say the greedy chooses $i_1$ and the optimal chooses $j_1$. By the greedy choice, $i_1$ is the farthest position reachable from 0. Therefore, $j_1 \leq i_1$.

Now, from $j_1$, the optimal solution reaches $j_2$. But since $j_1 \leq i_1$, and the greedy algorithm considers all positions reachable from $i_1$, it must be that the greedy can also reach $j_2$ (or farther) in the next jump.

By induction, we can show that the greedy solution reaches at least as far as the optimal solution at each step. Since both reach $n-1$, and the greedy makes the farthest jump at each step, it cannot make more jumps than the optimal.

Contradiction. Therefore, the greedy algorithm is optimal.

7. Detailed Walkthrough

Let’s trace nums = [2, 3, 1, 1, 4].

Initial State:

  • jumps = 0, current_end = 0, farthest = 0.

Iteration:

  1. i = 0: farthest = max(0, 0 + 2) = 2. i == current_end, so jumps = 1, current_end = 2.
  2. i = 1: farthest = max(2, 1 + 3) = 4. i < current_end, so no jump yet.
  3. i = 2: farthest = max(4, 2 + 1) = 4. i == current_end, so jumps = 2, current_end = 4.
  4. i = 3: We’ve reached n - 1 = 4 with current_end = 4, so we stop.

Result: jumps = 2.

Path: $0 \to 1 \to 4$ (Jump to index 1, then jump to index 4).

8. Variant: Jump Game I (Can Reach?)

Problem: Given nums, return true if you can reach the last index.

Greedy Solution:

def canJump(nums):
    farthest = 0
    for i in range(len(nums)):
        if i > farthest:
            return False  # Can't reach position i
        farthest = max(farthest, i + nums[i])
        if farthest >= len(nums) - 1:
            return True
    return True

9. Variant: Jump Game III (Reach Zero)

Problem: Given arr and start, you can jump to start + arr[start] or start - arr[start]. Return true if you can reach any index with value 0.

BFS Solution:

from collections import deque

def canReach(arr, start):
    n = len(arr)
    queue = deque([start])
    visited = {start}
    
    while queue:
        i = queue.popleft()
        if arr[i] == 0:
            return True
        
        for next_i in [i + arr[i], i - arr[i]]:
            if 0 <= next_i < n and next_i not in visited:
                visited.add(next_i)
                queue.append(next_i)
                
    return False

10. System Design: Pathfinding in Games

Jump Game II is essentially a simplified version of pathfinding algorithms used in game AI.

Real-World Application: Platformer Games

  • Problem: Find the shortest sequence of jumps for a character to reach a goal.
  • Constraints: Jump height, gravity, obstacles.
  • Algorithm: A* search with heuristic = Manhattan distance to goal.

Optimization:

  • Precompute Reachability Graph: For static levels, precompute which platforms are reachable from each platform.
  • Caching: Cache optimal paths for frequently visited platform pairs.

11. Deep Dive: Jump Game with Costs

Problem: Each jump from i to j has a cost cost[i][j]. Find the minimum cost to reach the end.

Algorithm: Dijkstra’s Algorithm.

import heapq

def minCostJump(nums, cost):
    n = len(nums)
    dist = [float('inf')] * n
    dist[0] = 0
    heap = [(0, 0)]  # (cost, index)
    
    while heap:
        d, i = heapq.heappop(heap)
        if i == n - 1:
            return d
        if d > dist[i]:
            continue
            
        for j in range(i + 1, min(i + nums[i] + 1, n)):
            new_cost = d + cost[i][j]
            if new_cost < dist[j]:
                dist[j] = new_cost
                heapq.heappush(heap, (new_cost, j))
                
    return dist[n - 1]

12. Interview Questions

  1. Jump Game II (Classic): Solve in O(N) time and O(1) space.
  2. Jump Game I: Can you reach the last index?
  3. Jump Game III: Can you reach any index with value 0?
  4. Jump Game IV: Given arr, you can jump to i+1, i-1, or any j where arr[j] == arr[i]. Find minimum jumps to reach the last index.
  5. Frog Jump: A frog can jump k-1, k, or k+1 units. Can it cross the river?
  6. Minimum Jumps with Cost: Each jump has a cost. Find the minimum cost path.

13. Common Mistakes

  • Off-by-One: Iterating through the entire array including the last element (unnecessary).
  • Not Handling Single Element: nums = [0] should return 0 jumps.
  • Greedy Choice: Thinking we should always jump to the farthest position immediately (wrong! We should jump to the position that allows the farthest next jump).
  • BFS Optimization: Not using the “level-by-level” optimization, leading to $O(N^2)$ instead of $O(N)$.

14. Performance Comparison

import time
import random

def benchmark():
    sizes = [100, 1000, 10000]
    
    for size in sizes:
        nums = [random.randint(1, 10) for _ in range(size)]
        
        # Greedy
        start = time.time()
        # greedy_solution(nums)
        greedy_time = time.time() - start
        
        # DP
        start = time.time()
        # dp_solution(nums)
        dp_time = time.time() - start
        
        print(f"Size {size}: Greedy={greedy_time:.6f}s, DP={dp_time:.6f}s")

# Expected: Greedy is 10-100x faster than DP for large inputs.

15. Deep Dive: Jump Game IV (BFS with HashMap)

Problem: Given an array arr, you can jump from index i to:

  • i + 1
  • i - 1
  • Any index j where arr[j] == arr[i] and i != j

Find the minimum number of jumps to reach the last index.

Example:

Input: arr = [100,-23,-23,404,100,23,23,23,3,404]
Output: 3
Explanation: 0 -> 4 -> 3 -> 9

Algorithm (BFS with Optimization):

from collections import deque, defaultdict

def minJumps(arr):
    n = len(arr)
    if n == 1:
        return 0
    
    # Build value -> indices mapping
    graph = defaultdict(list)
    for i, val in enumerate(arr):
        graph[val].append(i)
    
    queue = deque([0])
    visited = {0}
    steps = 0
    
    while queue:
        size = len(queue)
        steps += 1
        
        for _ in range(size):
            i = queue.popleft()
            
            # Try all three types of jumps
            # 1. i + 1
            if i + 1 < n and i + 1 not in visited:
                if i + 1 == n - 1:
                    return steps
                visited.add(i + 1)
                queue.append(i + 1)
            
            # 2. i - 1
            if i - 1 >= 0 and i - 1 not in visited:
                visited.add(i - 1)
                queue.append(i - 1)
            
            # 3. Same value jumps
            for j in graph[arr[i]]:
                if j not in visited:
                    if j == n - 1:
                        return steps
                    visited.add(j)
                    queue.append(j)
            
            # CRITICAL: Clear the list to avoid revisiting
            graph[arr[i]].clear()
    
    return steps

Optimization: After visiting all indices with value arr[i], we clear the list. This prevents revisiting the same value group multiple times.

Complexity:

  • Time: $O(N)$. Each index is visited at most once.
  • Space: $O(N)$ for the graph and visited set.

16. Deep Dive: Frog Jump (DP with Set)

Problem: A frog is crossing a river by jumping on stones. The frog can jump k - 1, k, or k + 1 units where k is the last jump distance. Can the frog cross?

Example:

Input: stones = [0,1,3,5,6,8,12,17]
Output: true
Explanation: 0 -> 1 (1 unit) -> 3 (2 units) -> 5 (2 units) -> 6 (1 unit) -> 8 (2 units) -> 12 (4 units) -> 17 (5 units)

Algorithm (DP with HashMap):

def canCross(stones):
    if stones[1] != 1:
        return False  # First jump must be 1 unit
    
    stone_set = set(stones)
    dp = {}  # (position, last_jump) -> bool
    
    def dfs(pos, k):
        if pos == stones[-1]:
            return True
        if (pos, k) in dp:
            return dp[(pos, k)]
        
        for next_k in [k - 1, k, k + 1]:
            if next_k > 0:
                next_pos = pos + next_k
                if next_pos in stone_set:
                    if dfs(next_pos, next_k):
                        dp[(pos, k)] = True
                        return True
        
        dp[(pos, k)] = False
        return False
    
    return dfs(1, 1)

Complexity:

  • Time: $O(N^2)$. At most $N$ positions and $N$ possible jump distances.
  • Space: $O(N^2)$ for memoization.

17. Production Application: Route Optimization

Scenario: Delivery truck routing (Amazon, UPS).

Problem: Given a list of delivery locations and the maximum distance the truck can travel from each location, find the minimum number of “hops” (refueling stops) to deliver all packages.

Mapping to Jump Game:

  • nums[i] = maximum distance from location i.
  • Goal: Reach the last location with minimum refueling stops.

Extensions:

  • Time Windows: Each location has a delivery time window.
  • Capacity Constraints: Truck has limited capacity.
  • Multiple Vehicles: Coordinate multiple trucks.

Algorithm: Jump Game II + Constraint Satisfaction.

18. Production Application: Network Packet Routing

Scenario: Data packet routing in a network.

Problem: A packet needs to travel from source to destination. Each router can forward the packet to routers within a certain “hop distance”. Find the minimum number of hops.

Mapping:

  • Nodes = routers.
  • nums[i] = maximum hop distance from router i.
  • Goal: Minimum hops from source to destination.

Real-World Constraints:

  • Congestion: Some routers are overloaded (higher cost).
  • Latency: Each hop has a latency.
  • Reliability: Some links may fail.

Algorithm: Dijkstra’s Algorithm with dynamic weights.

19. Advanced Variant: Jump Game with Obstacles

Problem: Some positions are obstacles (cannot land on them). Find the minimum jumps.

Algorithm (Modified BFS):

def jumpWithObstacles(nums, obstacles):
    n = len(nums)
    obstacle_set = set(obstacles)
    
    if 0 in obstacle_set or n - 1 in obstacle_set:
        return -1  # Can't start or can't finish
    
    queue = deque([0])
    visited = {0}
    jumps = 0
    
    while queue:
        size = len(queue)
        jumps += 1
        
        for _ in range(size):
            i = queue.popleft()
            
            for j in range(i + 1, min(i + nums[i] + 1, n)):
                if j in obstacle_set:
                    continue  # Skip obstacles
                if j == n - 1:
                    return jumps
                if j not in visited:
                    visited.add(j)
                    queue.append(j)
    
    return -1  # Can't reach

20. Mathematical Analysis: Expected Jumps

Question: If nums[i] is uniformly random in [1, k], what is the expected number of jumps for an array of length n?

Analysis:

  • Average jump length: $\frac{k+1}{2}$.
  • Expected number of jumps: $\approx \frac{n}{\frac{k+1}{2}} = \frac{2n}{k+1}$.

Example: $n = 100$, $k = 10$.

  • Expected jumps: $\frac{200}{11} \approx 18$.

21. Parallel Algorithm: Jump Game on GPU

Problem: Solve Jump Game II for millions of arrays in parallel (batch processing).

Algorithm (CUDA):

  1. Kernel: Each thread processes one array.
  2. Shared Memory: Store the array in shared memory for fast access.
  3. Reduction: Use parallel reduction to find the farthest reachable position.

Pseudocode:

__global__ void jumpGameKernel(int* arrays, int* results, int n, int batch_size) {
    int tid = blockIdx.x * blockDim.x + threadIdx.x;
    if (tid >= batch_size) return;
    
    int* nums = arrays + tid * n;
    int jumps = 0, current_end = 0, farthest = 0;
    
    for (int i = 0; i < n - 1; i++) {
        farthest = max(farthest, i + nums[i]);
        if (i == current_end) {
            jumps++;
            current_end = farthest;
        }
    }
    
    results[tid] = jumps;
}

22. Interview Deep Dive: Jump Game V

Problem: Given arr and d, you can jump at most d indices away. You can only jump to indices with smaller values. Find the maximum number of indices you can visit.

Example:

Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2
Output: 4
Explanation: 6 -> 4 -> 8 -> 6 (indices 0 -> 1 -> 4 -> 3)

Algorithm (DP with Sorting):

def maxJumps(arr, d):
    n = len(arr)
    dp = [1] * n  # dp[i] = max visits starting from i
    
    # Sort indices by value (process smaller values first)
    indices = sorted(range(n), key=lambda i: arr[i])
    
    for i in indices:
        # Try jumping left
        for j in range(i - 1, max(-1, i - d - 1), -1):
            if arr[j] >= arr[i]:
                break  # Can't jump to taller
            dp[i] = max(dp[i], dp[j] + 1)
        
        # Try jumping right
        for j in range(i + 1, min(n, i + d + 1)):
            if arr[j] >= arr[i]:
                break
            dp[i] = max(dp[i], dp[j] + 1)
    
    return max(dp)

Complexity: $O(N \log N + N \cdot d)$.

23. Conclusion

Jump Game II is a beautiful problem that teaches us the power of greedy algorithms. The key insight—that we can make locally optimal choices to achieve a globally optimal solution—is a recurring theme in algorithm design.

Key Takeaways:

  • Greedy > DP: For this problem, greedy is simpler and faster.
  • BFS Perspective: Thinking in terms of “levels” helps visualize the solution.
  • Proof Techniques: Exchange arguments are powerful for proving greedy correctness.
  • Real-World Applications: Routing, pathfinding, resource allocation.

The variants (Jump Game I, III, IV, V, Frog Jump) test your ability to adapt the core algorithm to different constraints. Mastering these variations prepares you for a wide range of interview questions.

24. Advanced Variant: Jump Game VI (DP with Deque)

Problem: Given nums and k, you can jump at most k steps. Each position has a score. Maximize the total score.

Example:

Input: nums = [1,-1,-2,4,-7,3], k = 2
Output: 7
Explanation: 0 -> 3 -> 5 (scores: 1 + 4 + 3 = 8, but we start at 1, so 1 + 4 + 3 = 8... actually the path is 0->3->5 with scores 1+4+3=8)

Algorithm (DP with Monotonic Deque):

from collections import deque

def maxResult(nums, k):
    n = len(nums)
    dp = [float('-inf')] * n
    dp[0] = nums[0]
    dq = deque([0])  # Stores indices
    
    for i in range(1, n):
        # Remove indices that are out of range
        while dq and dq[0] < i - k:
            dq.popleft()
        
        # dp[i] = max(dp[j]) + nums[i] for j in [i-k, i-1]
        dp[i] = dp[dq[0]] + nums[i]
        
        # Maintain decreasing deque
        while dq and dp[dq[-1]] <= dp[i]:
            dq.pop()
        dq.append(i)
    
    return dp[n - 1]

Complexity: $O(N)$ using monotonic deque.

25. Advanced Variant: Jump Game VII (String with Constraints)

Problem: Given a binary string s and integers minJump and maxJump. You start at index 0. You can jump to index j if:

  • s[j] == '0'
  • i + minJump <= j <= min(i + maxJump, n - 1)

Can you reach the last index?

Algorithm (BFS with Prefix Sum):

from collections import deque

def canReach(s, minJump, maxJump):
    n = len(s)
    if s[-1] == '1':
        return False
    
    queue = deque([0])
    farthest = 0
    
    while queue:
        i = queue.popleft()
        
        # Jump to range [i + minJump, i + maxJump]
        start = max(i + minJump, farthest + 1)
        end = min(i + maxJump, n - 1)
        
        for j in range(start, end + 1):
            if s[j] == '0':
                if j == n - 1:
                    return True
                queue.append(j)
        
        farthest = max(farthest, i + maxJump)
    
    return False

Optimization: Use a “visited” array to avoid revisiting indices.

26. Competitive Programming: Jump Game Speedrun

Problem: Given 1000 test cases, each with an array of length 10,000. Solve Jump Game II for all.

Optimization Techniques:

1. Fast I/O:

import sys
input = sys.stdin.readline

def solve():
    n = int(input())
    nums = list(map(int, input().split()))
    # ... greedy solution

2. Avoid Unnecessary Checks:

# Early exit if we can already reach the end
if current_end >= n - 1:
    break

3. Use Arrays Instead of Lists:

import array
nums = array.array('i', map(int, input().split()))

4. Inline Functions:

# Instead of max(a, b), use:
farthest = a if a > b else b

27. Interview Strategy: Recognizing Jump Game Patterns

Pattern Recognition:

  • Greedy: If the problem asks for “minimum jumps” and you can jump anywhere within a range.
  • BFS: If the problem has constraints on where you can jump (e.g., only to specific values).
  • DP: If the problem asks for “maximum score” or “number of ways”.

Common Variations:

  1. Can Reach? → Greedy (Jump Game I).
  2. Minimum Jumps? → Greedy (Jump Game II).
  3. Reach Specific Value? → BFS (Jump Game III).
  4. Jump with Constraints? → BFS with HashMap (Jump Game IV).
  5. Maximum Visits? → DP with Sorting (Jump Game V).
  6. Maximum Score? → DP with Deque (Jump Game VI).

28. Code Template: Universal Jump Game Solver

def jump_game_template(nums, target_condition, jump_rules):
    """
    Universal template for Jump Game problems.
    
    Args:
        nums: Input array
        target_condition: Function that checks if we've reached the goal
        jump_rules: Function that returns valid next positions
    """
    from collections import deque
    
    n = len(nums)
    queue = deque([0])
    visited = {0}
    steps = 0
    
    while queue:
        size = len(queue)
        steps += 1
        
        for _ in range(size):
            i = queue.popleft()
            
            if target_condition(i, n):
                return steps - 1
            
            for j in jump_rules(i, nums, n):
                if j not in visited:
                    visited.add(j)
                    queue.append(j)
    
    return -1  # Can't reach

# Example usage for Jump Game II:
def solve_jump_game_ii(nums):
    return jump_game_template(
        nums,
        target_condition=lambda i, n: i == n - 1,
        jump_rules=lambda i, nums, n: range(i + 1, min(i + nums[i] + 1, n))
    )

29. Testing Strategy

Test Cases:

  1. Single Element: [0] → 0 jumps.
  2. All Ones: [1,1,1,1,1] → 4 jumps.
  3. Large Jump: [10,1,1,1,1] → 1 jump.
  4. Optimal Path Not Obvious: [2,3,1,1,4] → 2 jumps.
  5. Maximum Constraints: Array of length 10,000 with random values.

Edge Cases:

  • Empty array (if allowed).
  • Array with zeros in the middle (should still be reachable per problem statement).
  • Very large jump values (e.g., nums[0] = 10000).

30. Common Interview Follow-ups

Q1: What if we need to return the actual path, not just the number of jumps? A: Modify the greedy algorithm to store the path.

def jump_with_path(nums):
    n = len(nums)
    if n == 1:
        return 0, [0]
    
    jumps = 0
    current_end = 0
    farthest = 0
    path = [0]
    
    for i in range(n - 1):
        farthest = max(farthest, i + nums[i])
        
        if i == current_end:
            jumps += 1
            current_end = farthest
            path.append(current_end)
            
            if current_end >= n - 1:
                path[-1] = n - 1
                break
    
    return jumps, path

Q2: What if some positions are blocked (obstacles)? A: Use BFS and skip blocked positions.

Q3: What if each jump has a cost, and we want to minimize total cost? A: Use Dijkstra’s algorithm.

Q4: What if we can jump backwards? A: Use BFS (greedy won’t work).

31. Optimization: Space-Efficient DP

For DP solutions, we often only need the last k values. Use a sliding window.

def jump_game_dp_optimized(nums):
    n = len(nums)
    # Instead of dp = [inf] * n, use a deque of size k
    from collections import deque
    window = deque([0])  # dp[0] = 0
    
    for i in range(1, n):
        # Remove old values outside the window
        while window and window[0][0] < i - max_jump_distance:
            window.popleft()
        
        # dp[i] = min(window) + 1
        dp_i = window[0][1] + 1
        
        # Add to window
        while window and window[-1][1] >= dp_i:
            window.pop()
        window.append((i, dp_i))
    
    return window[-1][1]

32. Conclusion & Summary

Jump Game II is more than just a coding problem—it’s a gateway to understanding greedy algorithms, graph traversal, and dynamic programming. The key insights:

  1. Greedy Works: When we can make locally optimal choices that lead to global optimality.
  2. BFS is Versatile: Model as a graph and use level-order traversal.
  3. DP for Variants: When we need to track scores or counts.
  4. Proof Matters: Always verify that greedy is correct (exchange argument).

Mastery Checklist:

  • Solve Jump Game I (Can Reach?)
  • Solve Jump Game II (Minimum Jumps) in O(N) time, O(1) space
  • Solve Jump Game III (Reach Zero)
  • Solve Jump Game IV (BFS with HashMap)
  • Solve Jump Game V (DP with Sorting)
  • Solve Jump Game VI (DP with Deque)
  • Solve Jump Game VII (String Constraints)
  • Solve Frog Jump
  • Explain why greedy works (proof)
  • Implement the path reconstruction variant

Next Steps:

  • Practice on LeetCode: Problems 45, 55, 1306, 1345, 1340, 1696, 1871, 403.
  • Study related problems: Minimum Cost to Reach Destination, Cheapest Flights Within K Stops.
  • Explore advanced topics: A* search, Bidirectional BFS.

The journey from “Can I reach the end?” to “What’s the optimal path with constraints?” teaches us to think algorithmically and adapt solutions to new problems. This is the essence of problem-solving in computer science.