Combination Sum iii | LeetCode 216 | C++, Java, Python

[@MiteshTank1]

I was struggling with backtracking concept of this problem but your explanation is so easy to understand! Thank you

[@KnowledgeCenter]

Glad to hear.

[@siddharthsingh4330]

It was hard to understand but code made it all easy. Thank you Sir

[@rohithbharathi3664]

just adding this if len(combination) > k: return which is missing in the python solution saves a hell lot of time

[@NatureLover-oq6uc]

VERY VERY GOOD EXPLANATION SIR....

[@KnowledgeCenter]

Thanks for liking

[@stith_pragya]

Thank You So Much Brother...🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻

[@KnowledgeCenter]

Glad you liked the video.

[@rajeshg1737]

Hacky (Python): def combinationSum3(self, k: int, n: int) -> List[List[int]]: ans = [] numbers = [i for i in range(1,10) ] for combination in itertools.combinations(numbers, k): if sum(combination) == n: ans.append(list(combination)) return ans

[@pif5023]

Thank you for the video! I got it correct evaluating all permutation but LeetCode was not happy with the time complexity. Restarting the loop from the last number evaluated does force the loop to only evaluate combinations and only once. That was something I wasn't sure of.

[@ankoor]

Python using recursion: def combinationSum3(self, k, n): def combination(A, k, n, output, result): if len(A) == 0: if len(result) == k and sum(result) == n: output.append(result) return L = result + [A[0]] R = result A = A[1:] combination(A, k, n, output, L) combination(A, k, n, output, R) A = [1, 2, 3, 4, 5, 6, 7, 8, 9] result = [] output = [] combination(A, k, n, output, result) return output

[@Vijaysharma-nz8fi]

Thanks a lot for nice explanation. Sir could you please explain lete code problem#282 Expression add operators. Thanks.

[@devbhattacharya153]

Thanks a lot sir

[@KnowledgeCenter]

Welcome.

[@janmichaelaustria620]

thank you once again sir! your videos have a good balance of theory and implementation!

[@chloe3337]

hello i don't understand how the numbers will be popped off the list twice once the first combination has been found. From temp = [1,2,6] -> temp.pop() -> [1,2] and then 2 pops out again and it becomes [1]

[@paragroy5359]

Nice explanation sir.......

[@sainikhilpalukuri1373]

Sir, I think the time complexity is o(n3) sir because we are looking all the combinations. Sir please correct me if I am wrong

[@KnowledgeCenter]

You are assuming k = 3, it was just in example. K can be any value from 1 to 9. e.g., if k = 2, we just need 2 values. If k = 5 we need to find 5 values.

[@midhileshmomidi2434]

Sir I have two doubts 1)Why did you take combination.copy() instead of combination 2)In recursion even after the sum exceeds like for sum = 9 combination is 2,3,4 then function is calling for 2,3,5.. 2,3,6 How to stop these calls Please explain

[@midhileshmomidi2434]

I did with itertools as that was the first idea pop up in mind But while trying in the backtracking approach I struggled to write recursive call function How to improve writing recursive functions please help sir

[@annesyabanerjee1892]

which software you are using ??