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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
**Rearranged Strings** Here's a fun little puzzle for you: Given two strings, namely, string s 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:** - Input s: "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.
Transcribed Image Text:**Rearranged Strings** Here's a fun little puzzle for you: Given two strings, namely, string s 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:** - Input s: "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.
```python def solution(s, t): # your code goes here. s = s.replace(" ", "").lower() t = t.replace(" ", "").lower() if len(s) != len(t): return False char_count_s = {} char_count_t = {} for char in s: if char in char_count_s: char_count_s[char] += 1 else: char_count_s[char] = 1 for char in t: if char in char_count_t: char_count_t[char] += 1 else: char_count_t[char] = 1 return char_count_s == char_count_t ``` ### Explanation This Python function, `solution(s, t)`, checks if two strings `s` and `t` are anagrams of each other, ignoring spaces and case sensitivity. 1. **String Normalization**: - Spaces are removed from both strings with `s.replace(" ", "").lower()` and `t.replace(" ", "").lower()`. - Both strings are converted to lowercase to ensure that character comparison is case insensitive. 2. **Length Check**: - The lengths of the normalized strings are compared. If they are not equal, the function immediately returns `False`, indicating that the strings cannot be anagrams. 3. **Character Counting**: - Two dictionaries, `char_count_s` and `char_count_t`, are used to count the occurrences of each character in `s` and `t`. 4. **Comparison**: - Finally, the function returns `True` if `char_count_s` equals `char_count_t`, meaning both strings have the same characters in the same frequency, confirming they are anagrams.
Transcribed Image Text:```python def solution(s, t): # your code goes here. s = s.replace(" ", "").lower() t = t.replace(" ", "").lower() if len(s) != len(t): return False char_count_s = {} char_count_t = {} for char in s: if char in char_count_s: char_count_s[char] += 1 else: char_count_s[char] = 1 for char in t: if char in char_count_t: char_count_t[char] += 1 else: char_count_t[char] = 1 return char_count_s == char_count_t ``` ### Explanation This Python function, `solution(s, t)`, checks if two strings `s` and `t` are anagrams of each other, ignoring spaces and case sensitivity. 1. **String Normalization**: - Spaces are removed from both strings with `s.replace(" ", "").lower()` and `t.replace(" ", "").lower()`. - Both strings are converted to lowercase to ensure that character comparison is case insensitive. 2. **Length Check**: - The lengths of the normalized strings are compared. If they are not equal, the function immediately returns `False`, indicating that the strings cannot be anagrams. 3. **Character Counting**: - Two dictionaries, `char_count_s` and `char_count_t`, are used to count the occurrences of each character in `s` and `t`. 4. **Comparison**: - Finally, the function returns `True` if `char_count_s` equals `char_count_t`, meaning both strings have the same characters in the same frequency, confirming they are anagrams.
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