filmov
tv
🔀✅ Check if Two Strings are Close | Leetcode 75 | String Transformation Problem | #Python #Leetcode

Показать описание
🔍 Understanding the Problem:
Two strings are considered close if one can be transformed into the other using these operations:
1️⃣ Swap any two characters:
Example: "abc" → "acb" → "bca" (Characters can be rearranged)
2️⃣ Change all occurrences of one character to another existing character:
Example: "aacabb" → "bbcbaa" (All 'a's → 'b', all 'b's → 'a')
Using these two operations, we need to check if word1 can be transformed into word2.
📌 Approach to Solve the Problem Efficiently:
1️⃣ Check Length: If word1 and word2 have different lengths, they can never be transformed into each other. Return False.
2️⃣ Check Character Sets: Both words must contain the same unique characters. If one contains a character that the other doesn’t, return False.
3️⃣ Check Character Frequencies:
The frequency of characters in both words must be the same when sorted.
Since we can transform all occurrences of one character into another, the order of the character itself doesn’t matter, but the count does.
📌 Code Explanation:
from collections import Counter
class Solution:
def closeStrings(self, word1, word2):
# Step 1: If the lengths are different, return False immediately
if len(word1) != len(word2):
return False
# Step 2: Count character frequencies in both words
freq1 = Counter(word1)
freq2 = Counter(word2)
# Step 3: Check if both words have the same unique characters
return False
# Step 4: Check if the sorted frequency counts match
# Example Test Cases
sol = Solution()
💡 Why This Works:
✔️ Same unique characters: Ensures we can swap characters.
✔️ Same character frequency pattern: Ensures we can transform one into another.
⏳ Time Complexity: O(n)
📦 Space Complexity: O(1) (since there are only 26 lowercase English letters)
🔥 This method efficiently determines whether two strings can be transformed into each other using O(n) time complexity.
#Leetcode #Python #String #HashMap #Frequency #Algorithm #DataStructures #InterviewPreparation
Two strings are considered close if one can be transformed into the other using these operations:
1️⃣ Swap any two characters:
Example: "abc" → "acb" → "bca" (Characters can be rearranged)
2️⃣ Change all occurrences of one character to another existing character:
Example: "aacabb" → "bbcbaa" (All 'a's → 'b', all 'b's → 'a')
Using these two operations, we need to check if word1 can be transformed into word2.
📌 Approach to Solve the Problem Efficiently:
1️⃣ Check Length: If word1 and word2 have different lengths, they can never be transformed into each other. Return False.
2️⃣ Check Character Sets: Both words must contain the same unique characters. If one contains a character that the other doesn’t, return False.
3️⃣ Check Character Frequencies:
The frequency of characters in both words must be the same when sorted.
Since we can transform all occurrences of one character into another, the order of the character itself doesn’t matter, but the count does.
📌 Code Explanation:
from collections import Counter
class Solution:
def closeStrings(self, word1, word2):
# Step 1: If the lengths are different, return False immediately
if len(word1) != len(word2):
return False
# Step 2: Count character frequencies in both words
freq1 = Counter(word1)
freq2 = Counter(word2)
# Step 3: Check if both words have the same unique characters
return False
# Step 4: Check if the sorted frequency counts match
# Example Test Cases
sol = Solution()
💡 Why This Works:
✔️ Same unique characters: Ensures we can swap characters.
✔️ Same character frequency pattern: Ensures we can transform one into another.
⏳ Time Complexity: O(n)
📦 Space Complexity: O(1) (since there are only 26 lowercase English letters)
🔥 This method efficiently determines whether two strings can be transformed into each other using O(n) time complexity.
#Leetcode #Python #String #HashMap #Frequency #Algorithm #DataStructures #InterviewPreparation