💡 Problem Formulation: We are given an array of non-negative integers where each element represents the maximum number of steps we can jump forward from that element. The task is to find the minimum number of jumps required to reach the last index of the array from the first index. For example, given the input array [2,3,1,1,4], 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).
Method 1: Greedy Algorithm
Greedy algorithms make the best possible choice at each step. For this problem, we choose the furthest reachable index each time. We iterate over the array and update the current reach and the last jump position, keeping track of the jumps we’ve made.
Here’s an example:
def min_jumps(arr):
if len(arr) reach:
break
reach = max(reach, i + arr[i])
if i == last_jump:
last_jump = reach
jumps += 1
return jumps if reach >= len(arr) - 1 else float('inf')
# Example use case
arr = [2,3,1,1,4]
print(min_jumps(arr))
Output: 2
This code defines a function min_jumps that takes an array arr and returns the minimum number of jumps required to reach the end. It uses a greedy approach to update the reach and jumps. If the end is not reachable, it returns infinity (float('inf')).
Method 2: Dynamic Programming
Dynamic programming solves each subproblem only once and stores the results. In this context, it calculates the minimum number of jumps needed to reach each index from the start. It iterates over the array, for each index calculating the minimum jump required by considering all the indices that can reach it.
Here’s an example:
def min_jumps(arr):
jumps = [float('inf')] * len(arr)
jumps[0] = 0
for i in range(1, len(arr)):
for j in range(i):
if i <= j + arr[j] and jumps[j] != float('inf'):
jumps[i] = min(jumps[i], jumps[j] + 1)
break
return jumps[-1]
# Example use case
arr = [2,3,1,1,4]
print(min_jumps(arr))
Output: 2
The code snippet defines the min_jumps function which initializes a list of jumps with infinity. It then updates each element with the minimum number of jumps required to get to that index. It returns the last element which represents the minimum steps to reach the last index of the array.
Method 3: Breadth-First Search (BFS)
Breadth-First Search (BFS) operates by exploring all the neighbors of a node before moving to the next level. In the array jump problem, BFS can be used by considering each index as a node and exploring all reachable indices from that node in a breadth-first manner.
Here’s an example:
from collections import deque
def min_jumps(arr):
if len(arr) = len(arr) - 1:
return jumps + 1
while current_reach < max_reach:
current_reach += 1
jump_queue.append((current_reach, jumps + 1))
return float('inf')
# Example use case
arr = [2,3,1,1,4]
print(min_jumps(arr))
Output: 2
In this example, the min_jumps function uses BFS to find the minimum number of jumps. It uses a queue to track the indices and the number of jumps taken to reach them. It updates the maximum reach as it progresses through each level until the last index is reached.
Method 4: Bidirectional BFS
Bidirectional BFS executes BFS from both the start and end nodes and stops when the searches meet in the middle. This reduces the overall search space, potentially leading to faster solutions in some cases. For the array jump challenge, it applies this strategy to find the intersection point which can lead us to the end fastest.
Here’s an example:
from collections import deque
def min_jumps(arr):
if len(arr) len(backward_queue):
forward_queue, backward_queue = backward_queue, forward_queue
index, jumps = forward_queue.popleft()
for next_index in range(index + 1, min(index + arr[index] + 1, len(arr))):
if next_index in backward_queue:
return jumps + backward_queue[backward_queue.index(next_index)][1] + 1
if next_index not in visited:
visited.add(next_index)
forward_queue.append((next_index, jumps + 1))
return float('inf')
# Example use case
arr = [2,3,1,1,4]
print(min_jumps(arr))
Output: 2
This code implements bidirectional BFS for the minimum jumps required problem. It uses two queues to search from both ends and stops when they meet somewhere in the middle. To keep track of visited indices, it utilizes a visited set.
Bonus One-Liner Method 5: Recursive Backtracking (Not Recommended)
Recursive backtracking explores all paths from the current index to the end by jumping to every possible next index. It is not recommended for large arrays due to exponential time complexity but can be illustrative for small cases.
Here’s an example:
def min_jumps(arr, pos=0):
if pos >= len(arr) - 1:
return 0
return 1 + min((min_jumps(arr, pos + i) for i in range(1, arr[pos] + 1)), default=float('inf'))
# Example use case
arr = [2,3,1,1,4]
print(min_jumps(arr))
Output: 2
The function min_jumps uses recursion to explore all paths from the current position by using a generator expression inside min. It adds one to the result to account for the current jump and recurses forward.
Summary/Discussion
- Method 1: Greedy Algorithm. Efficient for large arrays. Doesn’t always work for arrays with zeroes or negative steps.
- Method 2: Dynamic Programming. Guaranteed to find the solution. Time complexity can be high for large arrays.
- Method 3: Breadth-First Search (BFS). A methodical approach that works well for most cases. Can be less efficient than the greedy algorithm.
- Method 4: Bidirectional BFS. Can provide performance improvements over traditional BFS in certain scenarios. Complexity can be difficult to handle.
- Bonus Method 5: Recursive Backtracking. Simple to understand, but impractical for non-trivial cases due to its exponential time complexity.