One interesting real-world challenge is the painting fence problem in Python. This problem delves into combinatorial mathematics, presenting a scenario where the task is to determine the number of ways to paint a series of fences with a limited set of colors, ensuring that, at most, two adjacent fences share the same color. While simple, this problem finds applications in many domains. From graphic design and urban planning to scheduling algorithms.

The painting fence problem involves determining the number of ways to paint `n`

fences with `k`

different colors, ensuring that no two adjacent fences are the same color. Given the constraints of `n`

fences and `k`

colors, we aim to compute the total number of possible color combinations with our conditions.

Some important points that need to be remembered while solving this problem are given below:

The function we have to define is

`numWays`

, which takes in two parameters,`n`

and`k`

.The

`n`

parameter denotes the number of fences, and`k`

denotes the number of available colors.We have

`n`

fences, each of which can be painted in`k`

ways.We must return the total number of ways we can paint the fence, ensuring that, at most, two adjacent fences are the same color.

This Answer will provide a solution that effectively determines the overall count of acceptable color arrangements for a specified number of fences and available colors.

Now that we’ve understood the problem let’s understand it programmatically. __Educative 99__ may aid you in solving complex coding problems like this problem.

Here’s how the algorithm works:

Begin by defining a function

`numWays`

, which takes in two parameters:`n`

, denoting the number of fences, and`k`

, indicating the number of available colors.Handle the base case and check if either

`n`

or`k`

is`0`

. If so, it’s impossible to paint any fences without colors or fences, so return`0`

.Handle the case with only one fence (

`n == 1`

). If this is the case, the number of ways to paint is`k`

, as any available colors can be chosen.Initialize two variables (

`identical`

and`different`

) representing the number of ways to paint the first two fences with the same and different colors. Set`identical`

to`k`

(as there’s no constraint in the first fence) and`different`

to`k * (k - 1)`

(the second fence should have a different color than the first).Iterate through the fences to calculate the number of ways to paint each fence without violating the adjacent color constraint. At each iteration, temporarily store the count of ways to paint the fences with different colors in

`temp1`

and compute the count of ways to paint the fences with the next color combination in`temp2`

.Update

`identical`

and`different`

with the temporary values for the next iteration.Return the value in the sum of

`identical`

and`different`

when all the iterations are complete.

The iteration begins with 3 and ends at n+1. In this example, there will be two iterations. The values for the variables are given in the slide.

Now that we have understood the problem, let’s test our knowledge below:

1

What does `numWays`

return if the number of colors or fences is `0`

?

A)

`None`

B)

`0`

C)

`k`

D)

An error

Question 1 of 30 attempted

The code for this problem is provided below:

def numWays(n, k):if k == 0 or n == 0:return 0if n == 1:return kidentical, different = k, k * (k - 1)for _ in range(3, n + 1):temp1 = differenttemp2 = (identical + different) * (k - 1)identical, different = temp1, temp2return identical + differentn = 3k = 2result = numWays(n, k)print(f"Ways to paint {n} fences with {k} colors is: {result}")

This code can be explained as follows:

**Line 1:**Define a function`numWays`

that takes two parameters,`n`

and`k`

.**Lines 2–3:**Check if the number of colors or fences is`0`

. If either is, return`0`

as it’s impossible to paint anything without colors or fences.**Lines 4–5:**If there’s only one fence, return`k`

as with just one fence; we can use any available colors.**Line 7:**Initialize`identical`

and`different`

which represents the number of ways to paint the first two fences with the same color (`identical`

) and different colors (`different`

).**Line 8:**Iterate from the third fence to the`n`

-th fence.**Line 9:**Temporarily store the number of ways to paint the fences with different colors.**Line 10:**Calculate the number of ways to paint the fences with the next color combination. Multiply this sum by`(k - 1)`

to ensure the next fence has a different color from the previous one.**Line 11:**Update`identical`

and`different`

for the next iteration.**Line 13:**Return the total number of ways.**Lines 15–18:**Code to test our function.

Note:You may alter the code at the end to test with your own values.

The overall time complexity of the algorithm is

The space complexity of the provided solution is constant, denoted as

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