1160. Find Words That Can Be Formed by Characters

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.

 

Note:

  1. 1 <= words.length <= 1000
  2. 1 <= words[i].length, chars.length <= 100
  3. All strings contain lowercase English letters only.

1160. Find Words That Can Be Formed by Characters
struct Solution;

use std::collections::HashMap;

impl Solution {
    fn str_2_hs(s: &str) -> HashMap<char, i32> {
        let mut hs: HashMap<char, i32> = HashMap::new();
        for c in s.chars() {
            *hs.entry(c).or_default() += 1;
        }
        hs
    }

    fn count_characters(words: Vec<String>, chars: String) -> i32 {
        let chars = Self::str_2_hs(&chars);
        let mut sum = 0;
        for w in words {
            let mut chars = chars.clone();
            let mut valid = true;
            for c in w.chars() {
                let count = chars.entry(c).or_default();
                *count -= 1;
                if *count < 0 {
                    valid = false;
                    break;
                }
            }
            if valid {
                sum += w.len();
            }
        }
        sum as i32
    }
}

#[test]
fn test() {
    let words = vec_string!["cat", "bt", "hat", "tree"];
    let chars = "atach".to_string();
    assert_eq!(Solution::count_characters(words, chars), 6);
    let words = vec_string!["hello", "world", "leetcode"];
    let chars = "welldonehoneyr".to_string();
    assert_eq!(Solution::count_characters(words, chars), 10);
}