README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/1800-1899/1897.Redistribute%20Characters%20to%20Make%20All%20Strings%20Equal/README_EN.md	1309	Weekly Contest 245 Q1	Hash Table String Counting

1897. Redistribute Characters to Make All Strings Equal

中文文档

Description

You are given an array of strings words (0-indexed).

In one operation, pick two distinct indices i and j, where words[i] is a non-empty string, and move any character from words[i] to any position in words[j].

Return true if you can make every string in words equal using any number of operations, and false otherwise.

 

Example 1:

Input: words = ["abc","aabc","bc"]
Output: true
Explanation: Move the first 'a' in words[1] to the front of words[2],
to make words[1] = "abc" and words[2] = "abc".
All the strings are now equal to "abc", so return true.

Example 2:

Input: words = ["ab","a"]
Output: false
Explanation: It is impossible to make all the strings equal using the operation.

 

Constraints:

• 1 <= words.length <= 100
• 1 <= words[i].length <= 100
• words[i] consists of lowercase English letters.

Solutions

Solution 1

Python3

class Solution:
    def makeEqual(self, words: List[str]) -> bool:
        counter = Counter()
        for word in words:
            for c in word:
                counter[c] += 1
        n = len(words)
        return all(count % n == 0 for count in counter.values())

Java

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

C++

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

Go

func makeEqual(words []string) bool {
	counter := [26]int{}
	for _, word := range words {
		for _, c := range word {
			counter[c-'a']++
		}
	}
	n := len(words)
	for _, count := range counter {
		if count%n != 0 {
			return false
		}
	}
	return true
}

TypeScript

function makeEqual(words: string[]): boolean {
    let n = words.length;
    let letters = new Array(26).fill(0);
    for (let word of words) {
        for (let i = 0; i < word.length; ++i) {
            ++letters[word.charCodeAt(i) - 97];
        }
    }

    for (let i = 0; i < letters.length; ++i) {
        if (letters[i] % n != 0) {
            return false;
        }
    }
    return true;
}