# 5 Best Ways to Find the Size of the Longest Sublist with Constant Speed in Python

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π‘ Problem Formulation: The challenge is to determine the length of the longest contiguous segment (sublist) in a list of car speeds where the speed is unchanging. For example, given a list of speeds `[10, 10, 20, 20, 20, 30, 30]`, the desired output would be `3`, correlating to the three consecutive speeds of `20`.

## Method 1: Iterative Comparison

This method involves iterating over the list of car speeds, comparing each element with its successor to determine the length of the longest constant-speed sublist. The function specification includes a simple loop, a conditional statement to check for constant speeds, and variables to track the maximum sublist size found.

Here’s an example:

```def longest_constant_speed(speeds):
max_length = cur_length = 1
for i in range(1, len(speeds)):
if speeds[i] == speeds[i-1]:
cur_length += 1
else:
max_length = max(max_length, cur_length)
cur_length = 1
return max(max_length, cur_length)

# Example usage:
print(longest_constant_speed([10, 10, 20, 20, 20, 30, 30]))```

Output: `3`

This code snippet sets up two tracking variables `max_length` and `cur_length` and increments `cur_length` every time a constant speed is found. When a change in speed is detected, it updates `max_length` if necessary and resets `cur_length`.

## Method 2: Using itertools.groupby

The `groupby` function from Python’s `itertools` module can be used to group consecutive elements in a list. This method processes those groups to find the longest one with constant elements. The advantage here is readability and leveraging Python’s built-in functionality for cleaner code.

Here’s an example:

```from itertools import groupby

def longest_constant_speed(speeds):
return max(len(list(group)) for key, group in groupby(speeds))

# Example usage:
print(longest_constant_speed([10, 10, 20, 20, 20, 30, 30]))```

Output: `3`

By using `groupby`, the code becomes concise. It takes the list of speeds, groups it by consecutive equal elements, converts these groups to lists (to be able to use len()), and then finds the maximum length.

## Method 3: Dynamic Programming Approach

Here, we apply a dynamic programming approach to find the longest sublist with constant speed. We use an auxiliary array to store lengths of constant-speed sublists ending at each index, updating the array based on the previous values. This method is efficient and can handle large lists well.

Here’s an example:

```def longest_constant_speed(speeds):
if not speeds:
return 0
lengths = [1] * len(speeds)
for i in range(1, len(speeds)):
if speeds[i] == speeds[i - 1]:
lengths[i] = lengths[i - 1] + 1
return max(lengths)

# Example usage:
print(longest_constant_speed([10, 10, 20, 20, 20, 30, 30]))```

Output: `3`

In this piece of code, an array called `lengths` is initialized to track constant-speed stretches. As we scan through the speeds list, the lengths array is updated accordingly. Ultimately, the maximum value from the lengths array is returned as the result.

## Method 4: Enumerate and Two-Pointer Technique

Combining `enumerate` with the two-pointer technique can be employed to iterate through the list while keeping track of the start of a constant-speed segment. This method is straightforward and eliminates the need for an auxiliary array unlike dynamic programming.

Here’s an example:

```def longest_constant_speed(speeds):
max_length = start = 0
for i, speed in enumerate(speeds[1:], start=1):
if speed != speeds[start]:
start = i
max_length = max(max_length, i - start + 1)
return max_length

# Example usage:
print(longest_constant_speed([10, 10, 20, 20, 20, 30, 30]))```

Output: `3`

This code uses enumeration to get both the index and value while iterating. When a new speed is detected, the start index is moved. The length of each speed streak is compared to the previously found maximum to continually find the longest one.

## Bonus One-Liner Method 5: Functional Approach with max and map

For a concise solution, we can use `max` and `map` alongside `groupby` for a one-liner that accomplishes the same task. This method is not only succinct but also takes advantage of Pythonβs functional programming capabilities.

Here’s an example:

```from itertools import groupby

longest_constant_speed = lambda speeds: max(map(len, (list(group) for _, group in groupby(speeds))))

# Example usage:
print(longest_constant_speed([10, 10, 20, 20, 20, 30, 30]))```

Output: `3`

This one-liner defines a lambda function that embeds the logic from Method 2. It achieves the same result while retaining readability and conciseness. It cleverly uses a generator expression within `map` to obtain lengths of constant-speed sublists.

## Summary/Discussion

• Method 1: Iterative Comparison. Simple and intuitive. Efficiency may drop for very large lists.
• Method 2: Using `itertools.groupby`. Elegant and concise. Relatively slower due to the generation of temporary lists.
• Method 3: Dynamic Programming Approach. Efficient for large datasets. Uses extra space for the lengths array.
• Method 4: Enumerate and Two-Pointer. Memory efficient and straightforward. Slightly less readable due to manual index tracking.
• Method 5: Functional One-Liner. Extremely concise. Least explicit and may be less approachable for beginners.