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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
``````

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
``````

## 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

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

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

[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.