💡 Problem Formulation: Jump Game II is a classic programming challenge where given an array of non-negative integers, each element represents your maximum jump length from that position. The goal is to reach the last index in the minimum number of jumps. For instance, given the input [2,3,1,1,4], the minimum number of jumps to reach the end is 2 (jump from index 0 to 1, then to the last index).
Method 1: Greedy Algorithm
This method uses a greedy approach to find the minimum number of jumps by iteratively calculating the furthest we can reach at the current step and updating our position. It’s efficient as it requires only O(n) time complexity, scanning through the array only once.
♥️ Info: Are you AI curious but you still have to create real impactful projects? Join our official AI builder club on Skool (only $5): SHIP! - One Project Per Month
Here’s an example:
def jump(nums):
jumps, current_end, farthest = 0, 0, 0
for i in range(len(nums) - 1):
farthest = max(farthest, i + nums[i])
if i == current_end:
jumps += 1
current_end = farthest
return jumps
# Example usage:
print(jump([2,3,1,1,4]))Output: 2
This code snippet first initializes jumps, current_end, and farthest to zero. It then iterates through the array (excluding the last element) and updates farthest with the maximum reach from the current position. When the index reaches the current_end, this indicates that a jump is necessary; we increase jumps by one and set current_end to farthest.
Method 2: Dynamic Programming
This approach uses dynamic programming to calculate the minimum number of jumps to reach each index. The algorithm builds up a solution by solving for all previous indices, which allows us to deduce the solution for the current index. This method, however, has a higher time complexity of O(n^2), which can be suboptimal for large inputs.
Here’s an example:
def jump(nums):
dp = [float('inf')] * len(nums)
dp[0] = 0
for i in range(1, len(nums)):
for j in range(i):
if j + nums[j] >= i:
dp[i] = min(dp[i], dp[j] + 1)
return dp[-1]
# Example usage:
print(jump([2,3,1,1,4]))Output: 2
In this code, an array dp is initialized to hold the minimum number of jumps needed to reach each index, with a default value of infinity. We then loop over the array and for each i, we look backwards to each j to check if that index can reach i. If it can, we update dp[i] to the minimum value of its current value and dp[j] + 1.
Method 3: BFS (Breadth-First Search)
The BFS method treats each index as a node in a graph and the jumps as edges. Starting from the first node, we traverse level by level, where each level represents a jump. This can be less efficient with a worst-case time complexity of O(n^2), but it is conceptually simpler as it relies on the well-known BFS graph traversal.
Here’s an example:
from collections import deque
def jump(nums):
if len(nums) <= 1: return 0
queue = deque([(0, 0)])
visited = set()
while queue:
index, steps = queue.popleft()
for i in range(index + 1, min(index + nums[index] + 1, len(nums))):
if i == len(nums) - 1: return steps + 1
if i not in visited:
visited.add(i)
queue.append((i, steps + 1))
# Example usage:
print(jump([2,3,1,1,4]))Output: 2
The code deploys a BFS traversal by initially enqueueing the first index with zero steps taken. For each node, we add its reachable nodes to the queue with an incremented step count. If we reach the last index, we return the total steps taken. We also maintain a visited set to avoid revisiting nodes.
Method 4: Bidirectional BFS
This method is an optimization of the traditional BFS by searching from both the start and the end, and meeting in the middle. The search space is reduced and often leads to a faster solution, especially for large arrays. However, it is more complex to implement and maintain than one-directional BFS.
Here’s an example:
def jump(nums):
# Implementation would be more complex
pass
# Example usage:
print(jump([2,3,1,1,4]))Output: To be implemented
This simplified code snippet suggests the concept of a bidirectional BFS without providing a full implementation. The idea is to alternately expand one layer from the start and one from the end, thus converging to a solution more efficiently than a unidirectional BFS.
Bonus One-Liner Method 5: List Comprehension with Recursion
A condensed and clever method using list comprehension and recursion to solve the problem with potentially exponential time complexity. Due to recursion, large inputs can cause a stack overflow. It’s more of an academic or interview answer rather than practical for production use due to its limitations.
Here’s an example:
def jump(nums):
return 0 if len(nums) <= 1 else 1 + min(jump(nums[i:]) for i in range(1, nums[0] + 1))
# Example usage:
print(jump([2,3,1,1,4]))Output: 2
This one-liner uses a recursive call for jump which returns 0 if there’s only one or no element. Otherwise, it makes a recursive call for each reachable index from the first element, effectively branching out to all possible paths and choosing the minimum from these.
Summary/Discussion
- Method 1: Greedy Algorithm. Offers O(n) time complexity and is efficient. It does not work well if you need to track the actual path taken.
- Method 2: Dynamic Programming. Guarantees the correct minimum number of jumps, but has higher time complexity (O(n^2)). It uses more memory compared to the greedy approach.
- Method 3: BFS (Breadth-First Search). Conceptually simpler using well-known BFS logic. Its performance can degrade with large input sizes, resulting in O(n^2) time complexity.
- Method 4: Bidirectional BFS. Potentially reduces search space significantly, providing faster results for large arrays. The complexity of the implementation is a notable weakness.
- Method 5: List Comprehension with Recursion. Highly compact and an interesting approach for small inputs or challenges. Practically, it’s not suitable for large inputs due to recursion limits and exponential time complexity.