LeetCode 1153: String Transforms Into Another String Solution
Master LeetCode problem 1153 (String Transforms Into Another String), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
1153. String Transforms Into Another String
Problem Explanation
Explanation:
Given two strings str1 and str2, we need to determine if it is possible to transform str1 into str2 by performing the following operations:
- Replace any character in
str1with any other lowercase English alphabet character. - Transform
str1intostr2.
To solve this problem, we can use a mapping to keep track of the characters that have been replaced. We need to ensure that each character in str1 maps to only one character in str2.
Algorithmic Idea:
- If the lengths of
str1andstr2are not equal, returnfalse, as it is impossible to transformstr1intostr2. - Create a mapping to store the character replacements from
str1tostr2. - Iterate through each character in
str1andstr2. - If the character in
str1is already mapped to a different character instr2, returnfalse. - If the character in
str1is not mapped yet, add it to the mapping. - After processing all characters, if the size of the mapping is less than 26 (the number of lowercase English alphabet characters), return
true.
Time Complexity:
The time complexity of this algorithm is O(n), where n is the length of the input strings str1 and str2.
Space Complexity:
The space complexity of this algorithm is O(26) = O(1) since the mapping can have at most 26 entries.
:
Solution Code
class Solution {
public boolean canConvert(String str1, String str2) {
if (str1.equals(str2)) {
return true;
}
Map<Character, Character> mapping = new HashMap<>();
for (int i = 0; i < str1.length(); i++) {
char c1 = str1.charAt(i);
char c2 = str2.charAt(i);
if (mapping.containsKey(c1) && mapping.get(c1) != c2) {
return false;
}
mapping.put(c1, c2);
}
return new HashSet<Character>(mapping.values()).size() < 26;
}
}Try It Yourself
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