Python interview with a LinkedIn engineer: Matching pairs

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Rather confusing interview. We write python to literally provide micro services in numerous docker containers reading/writing to Qs/Topics/S3 buckets/Lambdas/FTP servers/RSDBs/plotting graphs and you spend this entire interview on find the sum of two numbers in a list?? This is a bash interview! If this was my interview I would fail it. If you asked me to write 1000 lines to contact a server reliably and download a terabyte of data, I am good.

c = [14,13,6,7,8,10,1,2]

from itertools import combinations

def comb(a,target):

solution = 'No valid pairs'

comb = combinations(a,2)

for i in list(comb):

if sum(i) == target:

solution = '{0} and {1}'.format(i[0],i[1])

print(solution)

comb(c,27)

#No need of dictionary. We can sort the list and iterate
#may be add one more step is to remove numbers that are already >=target from the list
def findPair(l,target):
l.sort()
for i,item in enumerate(l):
for j in range(i,len(l)):
if item + l[j] == target:
print (item,l[j])

l=[6,8,12,9,17,10,45]
target=21
findPair(l,target)

Code for this question:

x=[12,14,8,6,1,2]
y=x
for i in x:
for j in y:
if i+j==3:
result=f"{i},{j}"
break
print(result)

lol @ actually getting two sum in an interview

If only real job interviews were this easy. It's always "code a solution to the cure for cancer with O(log n)

my solution
num_list = [14, 13, 6, 7, 8, 10, 1, 2]
target = 3
num_set = []

filtered_num_list = [i for i in num_list if i <= target]

for num in filtered_num_list:
    for n in filtered_num_list[1:]:
        if num + n == target:
            if (num, n) not in num_set and (n, num) not in num_set:
                num_set.append((num, n))

Just one for loop needed and I guess it'd be more efficient to exclude values > target.

def find_pair(values, target):
val_dict = {}
for val in values:
if val < target:
if val_dict.get(val) in values:
print(f"pairs found {val} {val_dict[val]}")
return 0
val_dict[val] = target – val
print("None found")
return 1

The criticism in the comments are ridiculous. The whole purpose of this is to practice, learn, get better, and gather feedback. This is why young software developers have a hard time because of the rude personalities that's in the area. Golden rule is to treat others the way you want to be treated. Let's not act like day one geniuses and have never improved in this line of work. smh

Please change title to – Python interview with a LinkedIn engineer: Two Sum

This is literally Two Sums 😱 you need to study more my friend

ml1 = [14, 3, 6, 7, 8, 10, 1, 2]

def find_target(x, t):

v = x.copy()

for i in x:

for j in v:

if i + j == t:

return i, j, t

print(find_target(ml1, 17))

waiting for the day when interviewers asks me such questions lol

Typical hash map question. It can be solved with O(N) complexity.

Watching this in preparation for a test through codepad.io and I think it boosted my confidence. Even though he's better with terminology than I am, my face is red from facepalming from the True-value thing.

Here is my solution for this problem; No need for second loop:

def find_pair(values, target):
value_dict = {}
for value in values:
if (target – value) in value_dict:
return "{} and {}".format(target – value, value)
value_dict[value] = value
return "No Valid Pair"

Found a simpler way to do it.
def find_pairs(values,target):
for x in values:
for e in values[1:-1]:
if x + e == target:
print(x,"and",e)
print(True)
else:
print("not valid")

x = find_pairs([14,13,6,7,8,10,1,2],3)

arr = [14, 13, 6, 7, 8, 10, 1, 2 ]
target = 3
for i in arr:
for j in arr:
if j + i == target:
print(f'{i} and {j}')
break

This works right?

It's so much more easy to do with a set and you only need to loop once.

Define the complement of a number to be the number that you need to add to it to get the target,

Loop through the list and see if the set contains the complement of the current number, if not add the current number to the set. Return the numbers if find the complement and false if you finish the loop without.

Leetcode easy problems for noobs 101.

My func does it in 3.75 µs per loop in colab.research.google.com. On my i3 3220T it completes the task in 7.13 µs ± 701 ns per loop.
Is it good? 🙂

def find_pairs(values, target):
num_list = []
for value in values:
if value < target:
num_list.append(value)
for num in num_list:
for x in num_list:
if target == num + x:
return str(num) + " and " + str(x)

def findpairs(n,m):
m=int(m)
start=0
end=0
n.sort()
matchfound=False
if len(n) <=1 :
return "No valid pairs"
for i in range(len(n)):
for j in range(len(n)):
if j!=i :
if n[i]+n[j]==m :#match found
start=n[i]
end=n[j]
matchfound=True
break
if matchfound:
break
if matchfound:
return str(start) + " and " + str(end)
else:
return "No valid pairs"

def solution(l,n):
for i in l :
if n-i in l:
return str(i) + " and " + str(n-i)
else :
return "none!"

How about this:

def find(nums, target):
comp = { }
for num in nums:
try:
if comp[num] == num:
return [num, target – num]
except:
comp[target – num] = target – num
return []

time complexity O(log N)
worst case O(N) when no matching pairs

def find_pairs(arr, target):

d = {arr[i] : False for i in range(len(arr))}

ans = []

for num in arr:

reminder = target – num

if d[num] == False and reminder in d and num and num != reminder:

ans.append(f"{num} and {reminder}")

d[num] = True

d[target – num] = True

return ans if ans else "no valid pairs."

I am confused

How this is faster than a "nested for loop"?

""for tagert_compliment in val_dict"" — The in operator also uses a for loop btw.

def find_pairs(a,target):
dic = {i: target-i for i in a if target-i>=0}
for i in dic.keys():
if dic[i] in dic:
return i,dic[i]
return 'no pairs'

@13.55 that excitement when 'interviewer' finally 'understands' that -2 and says "its because 0 – 2 = -2 , True is 0" 🙁 , man please in what language True is 0 man its 1 – 3

I don’t understand why he is using dict[value] – target, (@ ; @) dict[value] = True which also = 1 and 1-3= -2 that’s why it’s constant -2 in print. It’s so easy and how could this question get fucked up….

can not believe he accepted using hash table, he just cut the corner, the actual answer is using sort then step through elements from start and end end to find the pair.

dan the solution could have done much more easily, like looping through the array 2 levels and compare each some 2 values with the target, the store them in a array

def pairs(nums, target):
for num in range(len(nums)):
if (target-nums[num]) in nums[num+1:]:
return(str(nums[num])+ " and " + str((target-nums[num])))
return("can't be done")

my solution:

target_num = 9 # change target_num to whatever you want
list = [1, 10, 4, 5] # change numbers in list to whatever you want

def find_pair(temp, ind1):
ind2 = 0
for num in list:
if ind2 != ind1:
if num + temp == target_num:
print(f"#'s that add up to {target_num}: ( {temp}, {num} ) ")
return True
ind2 += 1

ind = 0
for num_ in list:
if find_pair(num_, ind) == True:
break
elif ind == len(list)-1:
print(f"No pairs add up to {target_num}");
ind += 1

['{0} and {1}'.format(n, target – n) for n in values if target – n in values][0]

'Howboudat!'

the solution he found is extremely complicated and unecessary, this can be easily done with a double for interation, and then testing the sum of the numbers: for i in values:

for j in values:

if i+j==target:

list1.append(i)

list1.append(j)

a=sorted(set(list1))

list2.append(a)

list1.clear()

then wth a litlle tupple and sets adjustment, you can easily extract the numbers, without duplicates, i don;t understand why the long complicated coding, programers need to be simple and efficient.

how about this (nice, simple and short):
# Program to give valid pairs which add up to target

def get_pairs(nums, target):
sublist = list(filter(lambda x: x <= target, sorted(nums)))
for i in sublist:
for j in sublist:
if i + j == target:
return((i, j))

return 'No valid pairs'

nums = [14, 13, 6, 7, 8, 10, 1, 2]
target = 24
pairs = get_pairs(nums, target)
print(pairs)

I haven't gone through the whole video but thought of this solution.

def find_pairs(values, target):

pairs = {}

for index, x in enumerate(values):

if x not in pairs:

pairs[target-x] = x

else:

return [pairs[x], x]

return 'No Valid Pairs'

– Pairs will store (target-number) as the key and number of number as the value
Rest is easy to figure out.

this audio was pretty bad

This interviewer needs to stop eating…

Hi,all.
Whar about
a = [14,13,6,7,8,10,1,2]

def foo(a,v):

dct = {}

[dct.update({x:v-x}) for x in a if x<v and x*2!=v]

for i in dct.keys():

if i in dct.values(): return (i,v-i)

print(foo(a,3))