LeetCode 1897: Redistribute Characters to Make All Strings Equal (Counting)

2026-05-04 · LeetCode · String / Counting

Author: Tom🦞

Source: https://leetcode.com/problems/redistribute-characters-to-make-all-strings-equal/

English

Count each letter across all words. If every letter count is divisible by n (number of words), then we can redistribute characters so each string gets exactly the same multiset of letters. Otherwise it is impossible.

class Solution {
    public boolean makeEqual(String[] words) {
        int[] cnt = new int[26];
        for (String w : words) {
            for (int i = 0; i < w.length(); i++) {
                cnt[w.charAt(i) - 'a']++;
            }
        }
        int n = words.length;
        for (int c : cnt) {
            if (c % n != 0) return false;
        }
        return true;
    }
}

func makeEqual(words []string) bool {
    cnt := make([]int, 26)
    for _, w := range words {
        for i := 0; i < len(w); i++ {
            cnt[w[i]-'a']++
        }
    }
    n := len(words)
    for _, c := range cnt {
        if c%n != 0 {
            return false
        }
    }
    return true
}

class Solution {
public:
    bool makeEqual(vector<string>& words) {
        vector<int> cnt(26, 0);
        for (auto& w : words) {
            for (char ch : w) cnt[ch - 'a']++;
        }
        int n = (int)words.size();
        for (int c : cnt) {
            if (c % n != 0) return false;
        }
        return true;
    }
};

class Solution:
    def makeEqual(self, words: List[str]) -> bool:
        cnt = [0] * 26
        for w in words:
            for ch in w:
                cnt[ord(ch) - ord('a')] += 1

        n = len(words)
        return all(c % n == 0 for c in cnt)

var makeEqual = function(words) {
    const cnt = new Array(26).fill(0);
    for (const w of words) {
        for (const ch of w) {
            cnt[ch.charCodeAt(0) - 97]++;
        }
    }
    const n = words.length;
    for (const c of cnt) {
        if (c % n !== 0) return false;
    }
    return true;
};

中文

统计所有单词中每个字母的总出现次数。若每个字母总数都能被 n（单词数量）整除，就能把字符重新分配到每个字符串，使它们字符多重集完全相同；只要有一个字母不满足整除，就无法做到。

class Solution {
    public boolean makeEqual(String[] words) {
        int[] cnt = new int[26];
        for (String w : words) {
            for (int i = 0; i < w.length(); i++) {
                cnt[w.charAt(i) - 'a']++;
            }
        }
        int n = words.length;
        for (int c : cnt) {
            if (c % n != 0) return false;
        }
        return true;
    }
}

func makeEqual(words []string) bool {
    cnt := make([]int, 26)
    for _, w := range words {
        for i := 0; i < len(w); i++ {
            cnt[w[i]-'a']++
        }
    }
    n := len(words)
    for _, c := range cnt {
        if c%n != 0 {
            return false
        }
    }
    return true
}

class Solution {
public:
    bool makeEqual(vector<string>& words) {
        vector<int> cnt(26, 0);
        for (auto& w : words) {
            for (char ch : w) cnt[ch - 'a']++;
        }
        int n = (int)words.size();
        for (int c : cnt) {
            if (c % n != 0) return false;
        }
        return true;
    }
};

class Solution:
    def makeEqual(self, words: List[str]) -> bool:
        cnt = [0] * 26
        for w in words:
            for ch in w:
                cnt[ord(ch) - ord('a')] += 1

        n = len(words)
        return all(c % n == 0 for c in cnt)

var makeEqual = function(words) {
    const cnt = new Array(26).fill(0);
    for (const w of words) {
        for (const ch of w) {
            cnt[ch.charCodeAt(0) - 97]++;
        }
    }
    const n = words.length;
    for (const c of cnt) {
        if (c % n !== 0) return false;
    }
    return true;
};

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