Question

1160. Find Words That Can Be Formed by Characters

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Description

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.

Solutions

Solution 1: Counting

We can use an array $cnt$ of length $26$ to count the occurrence of each letter in the string $chars$.

Then we traverse the string array $words$. For each string $w$, we use an array $wc$ of length $26$ to count the occurrence of each letter in the string $w$. If for each letter $c$, $wc[c] \leq cnt[c]$, then we can spell the string $w$ with the letters in $chars$, otherwise we cannot spell the string $w$. If we can spell the string $w$, then we add the length of the string $w$ to the answer.

After the traversal, we can get the answer.

The time complexity is $O(L)$, and the space complexity is $O(C)$. Here, $L$ is the sum of the lengths of all strings in the problem, and $C$ is the size of the character set. In this problem, $C = 26$.

class Solution:
  def countCharacters(self, words: List[str], chars: str) -> int:
    cnt = Counter(chars)
    ans = 0
    for w in words:
      wc = Counter(w)
      if all(cnt[c] >= v for c, v in wc.items()):
        ans += len(w)
    return ans

class Solution {
  public int countCharacters(String[] words, String chars) {
    int[] cnt = new int[26];
    for (int i = 0; i < chars.length(); ++i) {
      ++cnt[chars.charAt(i) - 'a'];
    }
    int ans = 0;
    for (String w : words) {
      int[] wc = new int[26];
      boolean ok = true;
      for (int i = 0; i < w.length(); ++i) {
        int j = w.charAt(i) - 'a';
        if (++wc[j] > cnt[j]) {
          ok = false;
          break;
        }
      }
      if (ok) {
        ans += w.length();
      }
    }
    return ans;
  }
}

class Solution {
public:
  int countCharacters(vector<string>& words, string chars) {
    int cnt[26]{};
    for (char& c : chars) {
      ++cnt[c - 'a'];
    }
    int ans = 0;
    for (auto& w : words) {
      int wc[26]{};
      bool ok = true;
      for (auto& c : w) {
        int i = c - 'a';
        if (++wc[i] > cnt[i]) {
          ok = false;
          break;
        }
      }
      if (ok) {
        ans += w.size();
      }
    }
    return ans;
  }
};

func countCharacters(words []string, chars string) (ans int) {
  cnt := [26]int{}
  for _, c := range chars {
    cnt[c-'a']++
  }
  for _, w := range words {
    wc := [26]int{}
    ok := true
    for _, c := range w {
      c -= 'a'
      wc[c]++
      if wc[c] > cnt[c] {
        ok = false
        break
      }
    }
    if ok {
      ans += len(w)
    }
  }
  return
}

function countCharacters(words: string[], chars: string): number {
  const idx = (c: string) => c.charCodeAt(0) - 'a'.charCodeAt(0);
  const cnt = new Array(26).fill(0);
  for (const c of chars) {
    cnt[idx(c)]++;
  }
  let ans = 0;
  for (const w of words) {
    const wc = new Array(26).fill(0);
    let ok = true;
    for (const c of w) {
      if (++wc[idx(c)] > cnt[idx(c)]) {
        ok = false;
        break;
      }
    }
    if (ok) {
      ans += w.length;
    }
  }
  return ans;
}

class Solution {
  /**
   * @param String[] $words
   * @param String $chars
   * @return Integer
   */
  function countCharacters($words, $chars) {
    $sum = 0;
    for ($i = 0; $i < strlen($chars); $i++) {
      $hashtable[$chars[$i]] += 1;
    }
    for ($j = 0; $j < count($words); $j++) {
      $tmp = $hashtable;
      $sum += strlen($words[$j]);
      for ($k = 0; $k < strlen($words[$j]); $k++) {
        if (!isset($tmp[$words[$j][$k]]) || $tmp[$words[$j][$k]] === 0) {
          $sum -= strlen($words[$j]);
          break;
        }
        $tmp[$words[$j][$k]] -= 1;
      }
    }
    return $sum;
  }
}