1897. Redistribute Characters to Make All Strings Equal

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

• 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;
}
}

• 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;
}
};

• 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())


• 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
}

• 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;
}