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algorithm – A solution N ^ 2 to 3Sum

I've tried to solve a 3Sum problem in leetcode

Given a table nums of not are there any elements a, b, c in nums such as a + b + c = 0? Find all the unique triplets in the array that gives the sum of zero.

Note:

The solution set must not contain duplicate triplets.

Example:

Given table numbers = [-1, 0, 1, 2, -1, -4],

A set of solutions is:
[
  [-1, 0, 1],
  [-1, -1, 2]
]

I take all day to get such a solution

class Solution: Single-pass approach

def threeSum (self, nums, target = 0) -> List[List[int]]:
"" "
: type nums: list[int]
        : target type: int
: triplets of type: list[List[int]]"" "
nums.sort ()
triplets = []

        def twoSum (nums, target, lo, hi) -> List[List[int]]:
"" "
: type nums: list[int]
            : target type: int
"" "
search = {}

if hi - lo < 2: return []

            for lo in range(lo, hi):
                if lo > 1 and numbers[lo] == numbers[lo-2]: continue #skip the duplicates
if nums[lo] > target: pause

num2 = nums[lo]
                complement = target - num 2
logging.debug (f " n  nsub_nums: {nums[lo:hi]} ")
logging.debug (f "num2: {num2}, that is complement: {complement}")

if lookup.get (complement)! = None:
num3 = complement
triplets.append ([num1, num2, num3])

Search[nums[nums[nums[nums[lo]]= lo
logging.debug (f "lookup: {lookup}")
logging.debug (f "triplets: {triplets}")


if len (nums) < 3: return [] 

        lo = 0
        hi = len(nums)
        for lo in range(lo, hi):
            if lo > 2 and numbers[lo] == numbers[lo-3]: continue #skip the duplicates
if nums[lo] > target: pause

num1 = nums[lo]        

            twoSum (nums, target-num1, lo + 1, hi)

return the triplets

Unfortunately, the answer is wrong.

With regard to the deuxSum solution to find all possible unique couples

class Solution: Single-pass approach
def twoSum (self, nums, target) -> List[List[int]]:
"" "
: type nums: list[int]
        : target type: int
"" "
nums.sort ()
nums_d: dict = {}
couples = []

        if len (nums) < 2:
            return []

        for i in range(len(nums)):
            if i > 1 and numbers[i] == numbers[i-2]: continue #skip the duplicates

complement = target - nums[i]
            logging.debug (f "complement: {complement}")

if nums_d.get (complement)! = None:
couples.append ([nums[nums[nums[nums[i], supplement])
nums_d[nums[nums[nums[nums[i]]= i
logging.debug (f "nums_d: {nums_d}")
logging.debug (f "couples: {couples}")
return couples 

It works properly

Target: 9
nums: [4, 7, 6, 3, 5]
DEBUG complement: 6
DEBUG nums_d: {3: 0}
DEBUG couples: []
DEBUG complement: 5
DEBUG nums_d: {3: 0, 4: 1}
DEBUG couples: []
DEBUG complement: 4
DEBUG nums_d: {3: 0, 4: 1, 5: 2}
DEBUG couples: [[5, 4]]DEBUG complement: 3
DEBUG nums_d: {3: 0, 4: 1, 5: 2, 6: 3}
DEBUG couples: [[5, 4], [6, 3]]DEBUG complement: 2
DEBUG nums_d: {3: 0, 4: 1, 5: 2, 6: 3, 7: 4}
DEBUG couples: [[5, 4], [6, 3]]result: [[5, 4], [6, 3]].
target: 2
nums: [0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1]
DEBUG complement: 2
DEBUG nums_d: {0: 0}
DEBUG couples: []
DEBUG complement: 2
DEBUG nums_d: {0: 1}
DEBUG couples: []
Complement DEBUG: 1
DEBUG nums_d: {0: 1, 1: 4}
DEBUG couples: []
Complement DEBUG: 1
DEBUG nums_d: {0: 1, 1: 5}
DEBUG couples: [[1, 1]]result: [[1, 1]].
-------------------------------------------------- --------------------
Rated 2 tests in 0.001s

Could you please help one advice to my threeSum solution?

Is there a solution to the problem of the traveling traveler with the binary weights?

No, because if each edge has a weight equal to 1, it remains to determine if there is such a circuit, which is the problem of the Hamiltonian cycle, and it is always difficult. (The link leads to a Hamiltonian Path Wikipedia page.) The path and cycle versions of the problem are difficult.)

Hypergeometric Functions – A Special Solution to the Hermite Differential Equation

I know that the general form solution to the differential equation of Hermit
$$ y & # 39; – 2xy & # 39; + 2 lambda y = 0 $$
is
$$ y (x) = a_1 M (- frac { lambda} {2}, frac {1} {2}, x ^ 2) + a_2 H ( lambda, x), $$
or $ M ( cdot, cdot, cdot) $ is a confluent hypergeometric function of the first type, and $ H ( cdot, cdot) $ is a Hermite polynomial.

For a general value of $ lambda $ (real value negative and not integer), is there a special solution to Hermit's differential equation such that its first-order derivative goes to zero for $ x rightarrow- infty $? In other words, is there a parametric characterization of $ a_2 $ as a function of $ a_1 $ and $ lambda $ such as $ y & # 39; (x) rightarrow 0 $ as $ x rightarrow- infty $?

From the numerical calculations of Mathematica, it seems that such a case exists. However, I could not characterize it explicitly. I would appreciate any help on this. Thank you!

[GET][NULLED] – JNews – Unique Solution for Web Publishing v4.0.2

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[GET][NULLED] – JNews – Unique Solution for Web Publishing v4.0.2

[GET][NULLED] – JNews – Unique Solution for Web Publishing v4.0.2

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[GET][NULLED] – JNews – Unique Solution for Web Publishing v4.0.2

Theory of complexity – What's wrong with this solution for $ mathcal {O} ({ log ({n choose frac {n} {2}})}) $?

In this recitation on MIT OCW, the instructor uses the Stirling approximation to calculate that

$ mathcal {O} ({ log ({n choose frac {n} {2}})}) = mathcal {O} (n) $.

However, I followed the following steps to conclude that $ mathcal {O} ({ log ({n choose frac {n} {2}})}) = mathcal {O} ( log {n}) $. Where did I go wrong?

First of all, note that $ {n choose frac {n} {2}} = frac {n!} { frac {n} {2}! frac {n} {2}!}. In basic logarithmic laws, we get that this is equal to $ log {(n!)} – log {( frac {n} {2}! frac {n} {2}!)} $. It follows that:

$$
mathcal {O} ( log {(n!)} – log {( frac {n} {2}! frac {n} {2}!)}) \
= mathcal {O} ( log {(n!)}) \
= mathcal {O} Big ( log { big (n (1) (n-2) cdots (1) big)} Big) \
= mathcal {O} Big ( log {n} + log {(n-1)} + ldots + log {(1))} Big)
= mathcal {O} ( log {n})
$$

So, what's wrong here? I have gone through it for a while and I do not see any mistakes. In charting these functions, however, I can see very clearly that there must be an error.

Advertising blocker solution

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What do you think of Copper's name field solution?

I am looking for several CRMs and have not met anyone who uses the unique name field as does Copper. I think it's brilliant and that it can solve a situation in a design I'm currently working on. There are no reviews (that I've seen) for this implementation. What is the UX community thinking here? Advantages vs. disadvantages.

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