Two Sum Problem (Python) - HOW TO NAIL LeetCode Interview Questions

[@ashleyhaa]

These videos are great! I suggest talking a bit slower though so less advanced coders like myself have an easier time following :)

[@MsNennifer]

You could always play the video at a slower speed.

[@NN-uq6de]

@@MsNennifer Or he can go a little slower and explain things more thoroughly on each topic

[@Code0866]

@@NN-uq6de This is for intermediate-expert level programmers. You need to study more if you cant follow.

[@rebeccatom2188]

@@Code0866 ok nerd

[@satyamagrawal6854]

your runtime will decrease significantly if you use for i in range(0,len(nums)): rather than enumerate

[@cvxcfv]

Probably the best explanation out of the 4 videos I've invested watching

[@1pomii]

so what was the point of this? to show everyone you can talk fast?

[@njokuiyke01]

Hi, at what point did the elements get added to the dictionary

[@justinnguyen8049]

it was at line 18, values[elem] = i, that the element got put into the dictionary.

[@RegiiPad]

You can have more than one pair for the same target. The enhanced version of your code follows: 1 #!/usr/bin/python 2 3 from collections import namedtuple 4 import os 5 import sys 6 7 Pair = namedtuple('Pair', 'x1, x2, v1, v2') 8 9 def two_sum(list, target): 10 values = dict() 11 pairs = [] 12 for i, elem in enumerate(list): 13 comp = target - elem 14 if comp in values: 15 pairs.append(Pair(values[comp], i, list[values[comp]], list[i])) 16 pass 17 values[elem] = i 18 pass 19 return pairs 20 pass 21 22 def main(): 23 arr = [1, 2, 3, 4, 5, 6, 7] 24 tgt = 8 25 results = two_sum(arr, tgt) 26 27 for p in results: 28 print("Numbers are [{}]={} and [{}]={} to add {}".format( 29 p.x1, p.v1, p.x2, p.v2, tgt)) 30 pass 31 pass 32 33 if __name__ == '__main__': 34 main() 35 print("** END **") 36 pass 37 pass

[@KiteHQ]

! you should join our fb group and help out others looking to learn Python facebook.com/groups/505658083720291

[@RegiiPad]

Kite applied today

[@codewithchandan6146]

Hy can you solve my code please

[@KiteHQ]

If you liked this video, join the Kite Developer Community on Facebook for access to more resources + support from fellow Python developers. Time to level up! facebook.com/groups/505658083720291

[@damanvirdi168]

nice and crisp explanation

[@Samantha-no4es]

Can someone explain line 14? When did anything get added to the dictionary? To me, it seems to be empty so I don't yet understand why we're checking if the compliment is there :o

[@SouthernHadoken]

It got added on line 18. That the else in if/else.

[@universal4334]

@@SouthernHadoken then all the elements added to dic . Finally returns empty list.

[@faisalabdulkadir9439]

Why do we map the elements and indexes to the dict after our if statement??

[@SayfSentinel]

Hi Your solution does not apply for an array like [3,2,4] and a target of 6 When you enter your if statement target minus 3 return 3 which is already in your array and stop the loop

[@cristianmurrachavez5380]

No, because its searching in the dict not in the array and the number is only added after asking

[@BrendanMetcalfe]

Where is the code from the example?

[@harrisonooi296]

@0:56

[@frenchmike]

reading the answers at that pace with not detail or visuals it no better than handing me the code, no need for a video for that.

[@aditiparashar9171]

you are amazing

[@BrendanMetcalfe]

Thanks!

[@TheJmization]

wtf this is on easy mode wtf

[@seemeyeah1362]

I was under the impression line 18 would include the elif statement, and an else statement for line 20. How comes this isn't the case?

[@judahmcnicholl8334]

I know its a bit late for an answer, but when returning out of a function nothing below it gets executed, so if that statement is hit you have found the solution.

[@marufmazumder8294]

values[elem] = i ; ====> why is that?

[@killagarcia392]

by doing so we map the key (element from our list) to the key value (index) {element:index,...}

[@aakashjana6225]

I think you have over complicated the problem just run a nested loop where the second loop starts from one position after the current iterator value of the outer loop add the values stored in both the indexes and check if the equal your target that's it for i in range(0,len(arr)): for j in range(i, len(arr)): If are[i]+arr[j]==target: return i, j return False

[@manastrivedi3841]

It might seem like the solution you gave us optional but actually, it has a complexity of O(n^2)

[@7minutesdead]

Since this has a complexity of O(n^2), Leetcode won't accept it because your time limit will exceed on the later test cases that have very large input lists (I tried to do the same thing more or less). The solution in the video is much faster and won't time out.

[@ras_mag]

This solution is not optimal you are scanning the whole list for the solution.

[@scorpions7153]

Why are you in a hurry?