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' inwords[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 returntrue
.
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.
Solution: Hashtable
Count the frequency of each character, it must be a multiplier of n such that we can evenly distribute it to all the words.
e.g. n = 3, a = 9, b = 6, c = 3, each word will be “aaabbc”.
Time complexity: O(n)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: bool makeEqual(vector<string>& words) { vector<int> freq(26); for (const auto& word : words) for (char c : word) ++freq[c - 'a']; for (int f : freq) if (f % words.size()) return false; return true; } }; |
Python3 one-liner
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# Author: Huahua class Solution: def makeEqual(self, words: List[str]) -> bool: return all(c % len(words) == 0 for c in Counter(''.join(words)).values()) |
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