hash map / set

Question 1

find the difference of two arrays: intuition

Answer 1

You want to find the elements that are on only one list, not on both.

The unordered sets act like hash tables, allowing for quick checks to see if a specific marble exists in the other set.

We iterate through each element in one set, checking if it’s present in the other set. If not, it’s unique and gets added to the result.

Return the list ans

Question 2

find the difference of two arrays: code

Answer 2

def findDifference(nums1: List[int], nums2: List[int]) -> List[List[int]]:
set1, set2, result1, result2 = set(nums1), set(nums2), [], []

for i in nums1: 
    if i not in set2:
        result1.append(i)
    
for i in nums2:
    if i not in set1:
        result2.append(i)

return [list(set(result1)), list(set(result2))]

Question 3

unique number of occurrences: intuition

Answer 3

If we have the frequencies of all elements, we can put them in a hash set. If the size of the hash set is equal to the number of elements, it implies that the frequencies are unique. Hence, we will find the frequencies of all elements in a hash map and then put them in a hash set.

Algorithm:
Store the frequencies of elements in the array arr in the hash map freq.

Iterate over the hash map freq and insert the frequencies of all unique elements of array arr in the hash set freqSet.

Return true if the size of hash set freqSet is equal to the size of hash map freq, otherwise return false.

Question 4

unique number of occurrences: code

Answer 4

def uniqueOccurrences(arr: List[int]) -> bool:
freq = {}
for x in arr:
freq[x] = freq.get(x, 0) + 1

return len(freq) == len(set(freq.values()))

Question 5

determine if two strings are close: intuition

Question 6

determine if two strings are close: code

Question 7

equal row and column pairs: intuition

Question 8

equal row and column pairs: code