LeetCode 3138: Minimum Length of Anagram Concatenation Solution

Master LeetCode problem 3138 (Minimum Length of Anagram Concatenation), a medium 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.

3138. Minimum Length of Anagram Concatenation

Medium

Hash Table String Counting

Problem Explanation

Explanation

To solve this problem, we can iterate through the given string s and count the frequency of each character. We can then find the greatest common divisor (GCD) of all the character frequencies. The minimum possible length of the string t will be the GCD of all character frequencies.

Time Complexity

The time complexity of this approach is O(n), where n is the length of the input string s.

Space Complexity

The space complexity is O(1) since we are only using a constant amount of space.

Solution Code

public int minLenAnagramConcatenation(String s) {
    int[] freq = new int[26];
    for (char c : s.toCharArray()) {
        freq[c - 'a']++;
    }
    
    int gcd = freq[0];
    for (int i = 1; i < 26; i++) {
        gcd = calculateGCD(gcd, freq[i]);
    }
    
    return gcd;
}

private int calculateGCD(int a, int b) {
    while (b > 0) {
        int temp = b;
        b = a % b;
        a = temp;
    }
    return a;
}

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What is the time complexity of LeetCode 3138 (Minimum Length of Anagram Concatenation)?

The time complexity for LeetCode 3138 (Minimum Length of Anagram Concatenation) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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