1897. Redistribute Characters to Make All Strings Equal - Explanation

1. Frequency Count (Hash Map)

class Solution:
    def makeEqual(self, words: List[str]) -> bool:
        char_cnt = defaultdict(int)

        for w in words:
            for c in w:
                char_cnt[c] += 1

        for c in char_cnt:
            if char_cnt[c] % len(words):
                return False
        return True

Time & Space Complexity

Time complexity: $O(n * m)$
Space complexity: $O(1)$ since we have at most $26$ different characters.

Where $n$ is the number of words and $m$ is the average length of each word.

2. Frequency Count (Array)

class Solution:
    def makeEqual(self, words: List[str]) -> bool:
        freq = [0] * 26
        flag = 0
        n = len(words)

        for w in words:
            for c in w:
                i = ord(c) - ord('a')
                if freq[i] != 0:
                    freq[i] += 1
                    if freq[i] % n == 0:
                        flag += 1
                else:
                    freq[i] += 1
                    if freq[i] % n != 0:
                        flag -= 1
                freq[i] %= n

        return flag == 0

Time & Space Complexity

Time complexity: $O(n * m)$
Space complexity: $O(1)$ since we have at most $26$ different characters.

Where $n$ is the number of words and $m$ is the average length of each word.