Avoiding multiple nested for-loops in python

Question

How to avoid multiple nested for-loops when one nested for-loop has range up to the current iteration of the outer for-loop? For example, consider the following code: This program returns a triplet from a list arr such that arr[i] - arr[j] = arr[j] - arr[k] = d and i<j<k.

d =3
arr = [1, 2, 4, 5, 7, 8, 10]
list1 = []

for biggest in range(0, len(arr)):
    for bigger in range(0, biggest):
        for big in range(0, bigger):
            if abs(arr[big] - arr[bigger]) == d and abs(arr[bigger] - arr[biggest]) == d:
                list1.append([arr[big], arr[bigger], arr[biggest]])
print(list1))

Are there any other alternatives to using multiple nested loops?

Answer 1

You can replace the three loops with:

from itertools import combinations

for big, bigger, biggest in combinations(range(0, len(arr)), 3):

You can replace all the code with:

print([t for t in combinations(arr, 3)
       if t[2] - t[1] == t[1] - t[0] == d])

Hope this helps!!

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Thanks!

Answer 2

Instead of multi-loop, If you can categorize all the threads into a loop, you can easily go with the less complexity with the in python, and for the nested loop, it is total standing with the loop in between the loop.

Comments

Can you explain somewhat deeply using syntax bro please

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Hii @Nikhil,

Example code:

d =3
arr = [1, 2, 4, 5, 7, 8, 10]
list1 = []

for biggest in range(0, len(arr)):
    for bigger in range(0, biggest):
        for big in range(0, bigger):
            if abs(arr[big] - arr[bigger]) == d and abs(arr[bigger] - arr[biggest]) == d:
                list1.append([arr[big], arr[bigger], arr[biggest]])
print(list1))

You can replace the three loops with:

from itertools import combinations

for big, bigger, biggest in combinations(range(0, len(arr)), 3):

You can replace all the code with:

print([t for t in combinations(arr, 3)
       if t[2] - t[1] == t[1] - t[0] == d])

---

Hey, @Nikhil,

Could you please make your query a bit clear that how you want to use syntax and where? Please post your workarounds.