1160

1160. Find Words That Can Be Formed by Characters (Easy)

You are given an array of strings words and a string chars.

A string is good if it can be formed by characters from chars (each character can only be used once).

Return the sum of lengths of all good strings in words.

 

Example 1:

Input: words = ["cat","bt","hat","tree"], chars = "atach"
Output: 6
Explanation: The strings that can be formed are "cat" and "hat" so the answer is 3 + 3 = 6.

Example 2:

Input: words = ["hello","world","leetcode"], chars = "welldonehoneyr"
Output: 10
Explanation: The strings that can be formed are "hello" and "world" so the answer is 5 + 5 = 10.

 

Constraints:

• 1 <= words.length <= 1000
• 1 <= words[i].length, chars.length <= 100
• words[i] and chars consist of lowercase English letters.

Companies: Karat, Indeed, Amazon, Google

Related Topics:
Array, Hash Table, String

Similar Questions:

• Rearrange Characters to Make Target String (Easy)

Hints:

• Solve the problem for each string in words independently.
• Now try to think in frequency of letters.
• Count how many times each character occurs in string chars.
• To form a string using characters from chars, the frequency of each character in chars must be greater than or equal the frequency of that character in the string to be formed.

Solution 1.

// OJ: https://leetcode.com/problems/find-words-that-can-be-formed-by-characters/
// Author: github.com/lzl124631x
// Time: O(N) where N is the length of all the contents in `words` and `chars`
// Space: O(1)
class Solution {
public:
    int countCharacters(vector<string>& words, string chars) {
        int cnt[26] = {0}, ans = 0;
        for (char c : chars) cnt[c - 'a']++;
        for (auto w : words) {
            int c[26] = {0};
            bool good = true;
            for (char ch : w) {
                if (++c[ch - 'a'] <= cnt[ch - 'a']) continue;
                good = false;
                break;
            }
            if (good) ans += w.size();
        }
        return ans;
    }
};