Solving Leetcode Interviews in Seconds with AI: Minimum Length of Anagram Concatenation
Introduction
In this blog post, we will explore how to solve the LeetCode problem "3138" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.
Problem Statement
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. Example 2: Input: s = "abcbcacabbaccba" Output: 3 Constraints: 1 <= s.length <= 105 s consist only of lowercase English letters.
Explanation
Here's a solution approach:
- Character Frequency Analysis: Calculate the frequency of each character in the input string
s. The length of the repeating anagramtcan be determined by figuring out the smallest number of distinct characters such that their counts divide all the character counts ins. - GCD Calculation: The length of
tis equal to the number of distinct characters in the original stringt. If we consider the counts of each character ins, then the length oftis the greatest common divisor (GCD) of these counts. Iterate and Validate: Start with a candidate length
l(which must divide the GCD calculated earlier). For each potentiall, form a stringtusing the firstldistinct characters in sorted order and ensure thatscan be constructed using anagrams oft.Runtime Complexity: O(N + K log K), where N is the length of the string s, and K is the number of distinct characters in s (at most 26). Storage Complexity: O(K), mainly for storing character counts.
Code
from math import gcd
def min_anagram_length(s):
"""
Finds the minimum possible length of the string t, such that s is a concatenation of anagrams of t.
Args:
s: The input string.
Returns:
The minimum possible length of t.
"""
char_counts = {}
for char in s:
char_counts[char] = char_counts.get(char, 0) + 1
if not char_counts:
return 0
distinct_chars = list(char_counts.keys())
num_distinct = len(distinct_chars)
if num_distinct == 0:
return 0
g = 0
for count in char_counts.values():
g = gcd(g, count)
min_len = num_distinct
for length in range(1, num_distinct + 1):
if g % length == 0:
min_len = length
break
return min_len