You are given a string s, which is known to be a concatenation of anagrams of some string t.
Return the minimum possible length of the string t.
An anagram is formed by rearranging the letters of a string. For example, "aab", "aba", and, "baa" are anagrams of "aab".
Example 1:
Input: s = "abba"
Output: 2
Explanation:
One possible string t could be "ba".
Example 2:
Input: s = "cdef"
Output: 4
Explanation:
One possible string t could be "cdef", notice that t can be equal to s.
Constraints:
1 <= s.length <= 105s consist only of lowercase English letters.This approach involves calculating the frequency of each character in the string s. By computing the greatest common divisor (GCD) of these frequencies, we can determine the minimum length of the original string t, because t should be such that each anagram formed from it can be repeated equal times to form s. The GCD gives us the maximum number of times the anagram can be repeated evenly across all character types.
We utilize the Python Counter to calculate the frequency of each character in s. Using the math.gcd function, we iteratively calculate the GCD of all character frequencies. This gives us the smallest possible length of t.
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Time Complexity: O(n), where n is the length of the string s due to character counting.
Space Complexity: O(1), assuming only a constant number of English lowercase characters.
In this approach, we analyze how different counts of characters can be evenly divided to possibly rearrange into equal-sized parts forming the string s. This involves checking divisibility conditions and if all the character counts align to a common modularity that suggests a potential original t length. Essentially, we check if a string part can be divided in a way where each anagram fit the same pattern.
In this Python code, we calculate the maximum divisor k of a string that satisfies a condition where character frequencies modulo k equal zero, indicating each character fits perfectly into parts of size k. This is verified iteratively up to the length of s, providing our minimum t length.
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Time Complexity: O(n * sqrt(n)) due to divisibility checks for each divisor.
Space Complexity: O(1), as additional storage is minimal.
| Approach | Complexity |
|---|---|
| Frequency Analysis and GCD Method | Time Complexity: O(n), where n is the length of the string Space Complexity: O(1), assuming only a constant number of English lowercase characters. |
| Divisibility and Modular Arithmetic Analysis | Time Complexity: O(n * sqrt(n)) due to divisibility checks for each divisor. Space Complexity: O(1), as additional storage is minimal. |
Valid Anagram - Leetcode 242 - Python • NeetCode • 660,738 views views
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