My epic padlet by rianna l

Hello

riannalei — 2024-02-23 02:09:41 UTC

text

should we start with problem 1

2024-02-23 02:10:15 UTC

2024-02-23 02:10:38 UTC

for number 1 we can iterate through array and use a set to keep track of visited values

yeah, and return true and break when it finds a value twice

2024-02-23 02:12:03 UTC

2024-02-23 02:12:19 UTC

yea that sounds good

riannalei — 2024-02-23 02:12:53 UTC

def containsDuplicate(self, nums):
       hashset = set()  #Initialize hashset as a set
       if len(nums) == 0:
           return False
       else:
           for n in nums:
               if n in hashset:
                   return True
               else:
                   hashset.add(n)  # Add unique elements to hashset
           return False  # No duplicates found, return False

2024-02-23 02:14:45 UTC

how were you able to write code in the box

riannalei — 2024-02-23 02:17:10 UTC

i used leetcodes platform to write it and then just pasted it here

2024-02-23 02:18:14 UTC

for problem 2

I was thinking we take the first num, add it to each number in the list using a for loop, comparing the result with target, then if not found, goes back, takes the second element does the same thing, basically iterating. It is a tedious process and has high cost, what do u think, any other ideas?

a hash map would be better I think

2024-02-23 02:19:36 UTC

2024-02-23 02:20:20 UTC

both approaches would work, but hash map might be better for Time complexity

yeah ur right

2024-02-23 02:20:45 UTC

riannalei — 2024-02-23 02:30:10 UTC

do u use enumerate?

yeah

2024-02-23 02:36:06 UTC

2024-02-23 02:36:25 UTC

def solution(array, target):
    #Using hashmap to store the sum complements
    hashmap ={}

    for i in range(len(array)):
        #if the target 
        if target - array[i] in hashmap:
            return [i, hashmap[target - array[i]]]
        else:
            hashmap[target - array[i]] = i

corrected version

2024-02-23 02:40:55 UTC

def solution(array, target):
    #Using hashmap to store the sum complements
    hashmap ={}

    for i in range(len(array)):
        #if the target 
        if target - array[i] in hashmap:
            return [i, hashmap[target - array[i]]]
        else:
            hashmap[array[i]] = i

hashtable, we check the difference between each entry and see if its one?

2024-02-23 02:45:43 UTC

do we leave it unsorted or can we sort it first?

2024-02-23 02:46:31 UTC

no no 3rd

2024-02-23 02:47:04 UTC

my bad

2024-02-23 02:47:09 UTC

it has to be sorted for the second example to work right

2024-02-23 02:47:47 UTC