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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

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• 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;
  }
  
  

• impl Solution {
      pub fn make_equal(words: Vec<String>) -> bool {
          let mut cnt = std::collections::HashMap::new();
  
          for word in words.iter() {
              for c in word.chars() {
                  *cnt.entry(c).or_insert(0) += 1;
              }
          }
  
          let n = words.len();
          cnt.values().all(|&v| v % n == 0)
      }
  }
  
  

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