💡 Problem Formulation: Given an array of non-negative integers where each number represents the maximum you can jump forward in the array, the challenge is to find the minimum number of jumps you need to reach the last index starting from the first. For instance, given the input [2,3,1,1,4], the minimum number of jumps to reach the last index is 2 (jump 2 steps from index 0 to 1, then 3 steps to the end).
Method 1: Greedy Approach
This method involves identifying the furthest reachable index at each point and making a jump to the position that gives the optimal distance coverage. It is an iterative method that updates the steps and maximum reach with each iteration until the end is reached or surpassed. This is generally the preferred method for its efficiency.
Here’s an example:
def minJumps(arr):
jumps = max_reach = curr_end = 0
for i in range(len(arr) - 1):
max_reach = max(max_reach, i + arr[i])
if i == curr_end:
jumps += 1
curr_end = max_reach
return jumps
# Example usage
arr = [2, 3, 1, 1, 4]
print(minJumps(arr))Output: 2
This snippet shows a `minJumps` function using a greedy approach. It iterates over the array while keeping track of the current furthest reach and the end of the current jump. When the end of the current jump is reached, the function increments the jump counter and sets a new target for the current jump’s end.
Method 2: Dynamic Programming
Dynamic Programming (DP) method involves solving the problem by breaking it down into simpler subproblems. It builds a table where each entry represents the minimum number of jumps to reach that index. Then, it fills in the table by comparing and selecting the minimum jumps from all the previous positions available.
Here’s an example:
def minJumps(arr):
if len(arr) == 1:
return 0
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 usage
arr = [2, 3, 1, 1, 4]
print(minJumps(arr))Output: 2
This snippet illustrates the DP methodology in action. The `minJumps` function initializes a list of “infinite” values that will eventually contain the minimum jumps needed to reach each index. The function systematically fills the table, assessing the jumps needed to get from each previous index to the current one.
Method 3: BFS (Breadth First Search) Approach
BFS can also be applied where the array is thought of as a graph with indices as nodes and edges between nodes of reachable indices. This method takes each node level by level and finds the quickest route to the end node akin to the shortest path in a graph.
Here’s an example:
from collections import deque
def minJumps(arr):
if len(arr) == 1:
return 0
queue = deque([(0, 0)])
visited = set()
while queue:
index, jumps = queue.popleft()
for i in range(index + 1, min(len(arr), index + arr[index] + 1)):
if i == len(arr) - 1:
return jumps + 1
if i not in visited:
visited.add(i)
queue.append((i, jumps + 1))
return -1
# Example usage
arr = [2, 3, 1, 1, 4]
print(minJumps(arr))Output: 2
The `minJumps` function uses BFS to explore the array. It utilizes a queue to keep track of the positions and the number of jumps taken to get there. If the end of the array is reachable from the current position, it returns the total number of jumps taken to reach that last index.
Method 4: Recursive Backtracking
This recursive method explores all paths, trying all possible jumps at each step. It is not efficient for large input sizes due to its exponential time complexity. However, for smaller inputs, this method is easy to implement and understand, as it simply explores the problem’s natural recursive structure.
Here’s an example:
def minJumps(arr, start=0):
if start >= len(arr) - 1:
return 0
if arr[start] == 0:
return float('inf')
min_jmp = float('inf')
for i in range(start + 1, start + arr[start] + 1):
jumps = minJumps(arr, i)
if jumps != float('inf'):
min_jmp = min(min_jmp, jumps + 1)
return min_jmp
# Example usage
arr = [2, 3, 1, 1, 4]
print(minJumps(arr))Output: 2
In this code snippet, `minJumps` function recurses with updated start positions, trying all jump distances possible from the current position. Each call returns the minimum jumps required from that point, and the minimum of these is taken to find the solution.
Bonus One-Liner Method 5: Pythonic Approach with List Comprehensions
For those who enjoy concise Python code, this one-liner uses list comprehensions to solve the problem. Note that it may not be the most efficient, but it is the most compact representation of a solution using Python’s capabilities.
Here’s an example:
def minJumps(arr):
return 0 if len(arr) <= 1 else 1 + min(minJumps(arr[i:]) for i in range(1, arr[0] + 1))
# Example usage
arr = [2, 3, 1, 1, 4]
print(minJumps(arr))Output: 2
This minimalistic solution involves a `minJumps` function that uses recursion and list comprehension. The function makes a jump at the current step and then makes a recursive call to find out the minimum number of jumps from there on.
Summary/Discussion
- Method 1: Greedy Approach. Efficient for large datasets. May not be straightforward to understand initially.
- Method 2: Dynamic Programming. More intuitive than Greedy for some, with a simple filling table mechanism. However, not as efficient for large datasets due to its polynomial time complexity.
- Method 3: BFS Approach. Similar time complexity to Greedy but uses extra space. It provides a different perspective in treating the array as a graph.
- Method 4: Recursive Backtracking. Easiest to reason about for small inputs. Highly inefficient for large arrays due to its exponential time complexity.
- Bonus Method 5: Pythonic Approach. Compact code for small inputs. Inefficient for larger inputs and difficult to debug or understand at a glance.