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def solution (s, t): #your code goes here. S s.replace(" " t= t.replace(" " 2 3 4 5 6 7 if len(s) != len(t): return False char_count_s () char_count_t = () for char in s: = else: 11 if char in char_count_s: char_count_s [char] += 1 char_count_s [char] = 1 for char in t: ").lower() ").lower () else: if char in char_count_t: char_count_t[char] += 1 char_count_t[char] = 1 return char_count_s == char_count t

def solution (s, t): #your code goes here. S s.replace(" " t= t.replace(" " 2 3 4 5 6 7 if len(s) != len(t): return False char_count_s () char_count_t = () for char in s: = else: 11 if char in char_count_s: char_count_s [char] += 1 char_count_s [char] = 1 for char in t: ").lower() ").lower () else: if char in char_count_t: char_count_t[char] += 1 char_count_t[char] = 1 return char_count_s == char_count t

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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My instructor saying my code has wrong indentation on several lines… please let me know which lines. I’ve asked this twice before and I haven’t got it yet
5 Next Description Rearranged strings Here's a fun little puzzle for you: Given two strings, namely, strings and string t, we'd like to know if it's possible to rearrange the characters within one string to match the character composition of the other string. In other words, can you check if they are anagrams of each other? If it's feasible to rearrange the characters so that both strings contain the same set of characters, return true; otherwise, return false. Remember, an anagram is a word or phrase created by reshuffling the letters from another word or phrase, using all the original letters exactly once. Example 1: Rearranged s • Inputs: "listen" Input t: "silent" Expected Output: True • Explanation: Both "listen" and "silent" are anagrams of each other, as they can be rearranged to have the same set of characters. ● Example 2: • Input s: "hello" • Input t: "world" • Expected Output: False Explanation: "hello" and "world" are not anagrams of each other as they have different sets of characters ● 10 11 16 17 8 12 13 14 15 18 9 19 20 21 22 23 24 25 26 27 28 29 30 /home/scaffold Terminal
Transcribed Image Text:5 Next Description Rearranged strings Here's a fun little puzzle for you: Given two strings, namely, strings and string t, we'd like to know if it's possible to rearrange the characters within one string to match the character composition of the other string. In other words, can you check if they are anagrams of each other? If it's feasible to rearrange the characters so that both strings contain the same set of characters, return true; otherwise, return false. Remember, an anagram is a word or phrase created by reshuffling the letters from another word or phrase, using all the original letters exactly once. Example 1: Rearranged s • Inputs: "listen" Input t: "silent" Expected Output: True • Explanation: Both "listen" and "silent" are anagrams of each other, as they can be rearranged to have the same set of characters. ● Example 2: • Input s: "hello" • Input t: "world" • Expected Output: False Explanation: "hello" and "world" are not anagrams of each other as they have different sets of characters ● 10 11 16 17 8 12 13 14 15 18 9 19 20 21 22 23 24 25 26 27 28 29 30 /home/scaffold Terminal
4 def solution (s, t): #your code goes 567 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 s= s.replace(" ", "").lower () t= t.replace(" ", ").lower() char_count_s () char_count_t = () if len(s) != len(t): return False for char in s: here. if char in char_count_s: else: 11 char_count_s [char] += 1 for char in t: char_count_s [char] MacBook Pro else: if char in char_count_t: - char_count_t[char] += 1 char_count_t[char] 1 = 1 return char_count_s == char_count_t
Transcribed Image Text:4 def solution (s, t): #your code goes 567 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 s= s.replace(" ", "").lower () t= t.replace(" ", ").lower() char_count_s () char_count_t = () if len(s) != len(t): return False for char in s: here. if char in char_count_s: else: 11 char_count_s [char] += 1 for char in t: char_count_s [char] MacBook Pro else: if char in char_count_t: - char_count_t[char] += 1 char_count_t[char] 1 = 1 return char_count_s == char_count_t
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