I just gave a function that takes two arrays named magazine and note

Code is:

def checkMagazine(magazine, note):
        l1 = len(note)
        l2 = len(magazine)
        magazine.sort()
        note.sort()
        i = 0
        j = 0
        count = 0
        while i<l1 and j<l2:
                if note[i] == magazine[j]:
                        count += 1
                        i += 1
                j += 1
        if count == l1:
                print ('Yes')
        else:
                print('No')

We have to find all words of note in magazine (for answer to be 'yes'). The words in note may not follow the same order in magazine (check eg 2). Also try to check the difference between eg1 and eg2.

eg 1:

    magazine: I understood what you said. 

    note: I you said

eg 2:

    magazine: check the words carefully in this sentence hope you check properly

    note: words this properly check check

our code should work for both eg1 and eg2, So, I sorted both magazine and note, by magazine.sort() and note.sort() it becomes:

    magazine: carefully check check hope in properly sentence the this words you

    note: check check properly this words

keep i = 0, j= 0,count = 0

while loop(i < length of note , j < length of magazine):

if note[i] == magazine[j] increase 'i' and 'count'; else do nothing

but increase j every time you enter the loop

if every word of note is in magazine, our count will be length of 'note' array, so we print 'yes' if count == l1; else the answer is 'no'

using either 'i' or 'count' is enough, I used both to make code understandable.

May be I didn't explain well, but this code worked for all test cases.