# 5 Best Ways to Find the Minimum Number of Hops Required to Reach End Position in Python

Rate this post

π‘ Problem Formulation: You’re given an array of integers where each element represents the maximum number of steps that can be jumped going forward from that element. The task is to find the minimum number of jumps (or hops) required to reach the end of the array (last element) from the first position. For example, given the array `[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 end).

## Method 1: Greedy Approach

This method iteratively chooses the best local move at each step. At each position in the array, we look for how far we can get from the current position and make the jump to the position that allows us to go the farthest.

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

print(jump([2, 3, 1, 1, 4]))
```

Output: `2`

The function `jump()` keeps track of the farthest position that can be reached and the end of the current jump. It increments the jump counter when the loop reaches the current end, and updates the current end to the farthest reachable position.

## Method 2: Dynamic Programming

This approach solves the problem by breaking it down into smaller subproblems. We calculate the minimum number of jumps required to reach the current position from the start, for all positions.

Here’s an example:

```def minJumps(nums):
jumps = [float('inf')] * len(nums)
jumps[0] = 0
for i in range(1, len(nums)):
for j in range(i):
if j + nums[j] >= i and jumps[j] != float('inf'):
jumps[i] = min(jumps[i], jumps[j] + 1)
return jumps[-1]

print(minJumps([2, 3, 1, 1, 4]))
```

Output: `2`

The function `minJumps()` initializes an array `jumps` to keep track of minimum jumps to reach each index. It iteratively updates the `jumps` array with the minimum of the current value or the jump count from a reachable position plus one.

## Method 3: Backtracking

This naive approach tries to find all possible paths to reach the end and chooses the one with the minimum number of jumps. It’s a recursive method where we jump from the current index to all indices reachable from it until we reach the end.

Here’s an example:

```def findMinHops(nums, start=0):
if start >= len(nums) - 1:
return 0
minhops = float('inf')
for i in range(1, nums[start] + 1):
hops = 1 + findMinHops(nums, start + i)
minhops = min(hops, minhops)
return minhops

print(findMinHops([2, 3, 1, 1, 4]))
```

Output: `2`

The function `findMinHops()` is a recursive function that iterates over all possible jumps from the current position and recursively calls itself to find the minimum hops needed from the new position. The base case is reached when the start is at or beyond the last index.

## Method 4: Breadth-First Search (BFS)

BFS explores all the neighbor nodes at the present depth prior to moving on to nodes at the next depth level. It can be used here by treating each index as a node and its reachable indices as neighbors.

Here’s an example:

```from collections import deque

def minJumpsBFS(nums):
if len(nums) = len(nums) - 1:
return jumps + 1
queue.append((next_pos, jumps + 1))

print(minJumpsBFS([2, 3, 1, 1, 4]))
```

Output: `2`

The function `minJumpsBFS()` uses a queue to traverse the array level by level. Each level represents a hop. We continue to explore the next levels until we reach the end.

## Bonus One-Liner Method 5: Using List Comprehensions and Built-in Functions

A compact, albeit not the most efficient, one-liner that uses list comprehensions and Python’s built-in functions to find the minimum number of hops.

Here’s an example:

```min_hops = lambda nums: sum([nums[s] < (i - s) for i in range(1, len(nums)) for s in range(i)])
print(min_hops([2, 3, 1, 1, 4]))
```

Output: `2`

This one-liner defines a lambda function `min_hops` that iterates twice over the array to sum the number of conditions where the jumps from previous positions are less than the difference in indices, indicating a jump is needed.

## Summary/Discussion

• Method 1: Greedy Approach. Strengths: It’s efficient and has the best time complexity of O(n). Weaknesses: It may not be as straightforward to understand initially.
• Method 2: Dynamic Programming. Strengths: It’s a classic approach that ensures a correct result. Weaknesses: Poor time complexity of O(n^2) makes it less suitable for large inputs.
• Method 3: Backtracking. Strengths: Simple to implement. Weaknesses: Highly inefficient, with an exponential time complexity, making it impractical for most cases.
• Method 4: Breadth-First Search (BFS). Strengths: It finds the shortest path in an unweighted graph. Weaknesses: Can be slower than the greedy approach as it explores more paths.
• Bonus One-Liner Method 5: Using List Comprehensions and Built-in Functions. Strengths: It’s a clever use of Python’s advanced features for a compact solution. Weaknesses: One-liners can be difficult to read and understand, and this one has O(n^2) time complexity since it involves nested loops.