383, Ransom Note

I Problem

Given two strings ransomNote and magazine, return true if ransomNote can be constructed by using the letters from magazine and false otherwise.

Each letter in magazine can only be used once in ransomNote.

Example 1
Input: ransomNote = "a", magazine = "b"
Output: false

Example 2
Input: ransomNote = "aa", magazine = "ab"
Output: false

Example 3
Input: ransomNote = "aa", magazine = "aab"
Output: true

Constraints

• 1 <= ransomNote.length, magazine.length <= 10⁵
• ransomNoteandmagazineconsist of lowercase English letters.

Related Topics

• Hash Table
• String
• Counting

II Solution

Approach 1: Hash Table

pub fn can_construct(ransom_note: String, magazine: String) -> bool {
    let a_u8 = b'a';
    let mut arr = magazine.chars().fold([0; 26], |mut arr, c| {
        let i = (c as u8 - a_u8) as usize;
        arr[i] += 1;
        arr
    });

    for c in ransom_note.chars() {
        let i = (c as u8 - a_u8) as usize;
        arr[i] -= 1;
        if arr[i] < 0 {
            return false;
        }
    }
    
    true
}

public boolean canConstruct(String ransomNote, String magazine) {
    int[] arr = new int[26];
    for (char c : magazine.toCharArray()) {
        arr[c - 'a']++;
    }

    for (char c : ransomNote.toCharArray()) {
        int i = c - 'a';
        arr[i]--;
        if (arr[i] < 0) {
            return false;
        }
    }

    return true;
}