I'm currently tackling a Python assignment where I need to create a program that returns a list of integers forming the largest sum of non-consecutive values from a given list. Additionally, any negative values should be skipped. My current implementation calculates the maximum sum correctly, but I'm not sure how to modify the code to track and store each individual value that contributes to that sum. I'd appreciate any guidance you can provide! Here's the current code I'm working with:
```python
def setmax(nums, n, memo, store):
if n-1 == 0:
return nums[n-1]
elif n-2 == 0 and nums[0] >= nums[1]:
return nums[n-2]
elif n-2 == 0:
return nums[n-1]
else:
memo[n] = max(setmax(nums, n-2, memo, store) + nums[n-1], setmax(nums, n-1, memo, store))
return memo[n]
def max_independent_set(nums):
n = len(nums)
if n == 0:
return 0
if n == 1:
return nums[0]
if n == 2:
return max(nums[0], nums[1])
store = []
memo = []
for y in range(n+1):
memo.append(-1)
return setmax(nums, n, memo, store)
```
Thanks for any advice you can share!
3 Answers
Also, don't forget to check boundary conditions when you start with negative numbers. In your `max_independent_set`, make sure that if all values are negative, it returns an empty list instead of a negative number. This way, you'll effectively skip those values in your results. Good luck!
It sounds like you're on the right track! To track the elements that contribute to the maximum sum, you could modify the `setmax` function to add each selected number to the `store` list. Instead of just returning the sum, you can also return the list of numbers that make up that sum. For each decision point (whether to include the current number or not), you can append it to `store` only if you choose to include it in the sum. This way, you’ll end up with the complete list of non-consecutive numbers that generate the largest sum.
By the way, to get the highlight syntax for your code, make sure to wrap it with triple backticks like this:
```
# Your code here
``` That way, it’s easier to read and others can help you more effectively!

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