383. Ransom Note

Description

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
ransomNote and magazine consist of lowercase English letters.

Solutions

Solution 1: Hash Table or Array

We can use a hash table or an array \(cnt\) of length \(26\) to record the number of times each character appears in the string magazine. Then traverse the string ransomNote, for each character \(c\) in it, we decrease the number of \(c\) by \(1\) in \(cnt\). If the number of \(c\) is less than \(0\) after the decrease, it means that the number of \(c\) in magazine is not enough, so it cannot be composed of ransomNote, just return \(false\).

Otherwise, after the traversal, it means that each character in ransomNote can be found in magazine. Therefore, return \(true\).

The time complexity is \(O(m + n)\), and the space complexity is \(O(C)\). Where \(m\) and \(n\) are the lengths of the strings ransomNote and magazine respectively; and \(C\) is the size of the character set, which is \(26\) in this question.

Python3JavaC++GoTypeScriptC#PHP

class Solution:
    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
        cnt = Counter(magazine)
        for c in ransomNote:
            cnt[c] -= 1
            if cnt[c] < 0:
                return False
        return True

class Solution {
    public boolean canConstruct(String ransomNote, String magazine) {
        int[] cnt = new int[26];
        for (int i = 0; i < magazine.length(); ++i) {
            ++cnt[magazine.charAt(i) - 'a'];
        }
        for (int i = 0; i < ransomNote.length(); ++i) {
            if (--cnt[ransomNote.charAt(i) - 'a'] < 0) {
                return false;
            }
        }
        return true;
    }
}

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        int cnt[26]{};
        for (char& c : magazine) {
            ++cnt[c - 'a'];
        }
        for (char& c : ransomNote) {
            if (--cnt[c - 'a'] < 0) {
                return false;
            }
        }
        return true;
    }
};

func canConstruct(ransomNote string, magazine string) bool {
    cnt := [26]int{}
    for _, c := range magazine {
        cnt[c-'a']++
    }
    for _, c := range ransomNote {
        cnt[c-'a']--
        if cnt[c-'a'] < 0 {
            return false
        }
    }
    return true
}

function canConstruct(ransomNote: string, magazine: string): boolean {
    const cnt: number[] = Array(26).fill(0);
    for (const c of magazine) {
        ++cnt[c.charCodeAt(0) - 97];
    }
    for (const c of ransomNote) {
        if (--cnt[c.charCodeAt(0) - 97] < 0) {
            return false;
        }
    }
    return true;
}

public class Solution {
    public bool CanConstruct(string ransomNote, string magazine) {
        int[] cnt = new int[26];
        foreach (var c in magazine) {
            ++cnt[c - 'a'];
        }
        foreach (var c in ransomNote) {
            if (--cnt[c - 'a'] < 0) {
                return false;
            }
        }
        return true;
    }
}

class Solution {
    /**
     * @param String $ransomNote
     * @param String $magazine
     * @return Boolean
     */
    function canConstruct($ransomNote, $magazine) {
        $arrM = str_split($magazine);
        for ($i = 0; $i < strlen($magazine); $i++) {
            $hashtable[$arrM[$i]] += 1;
        }
        for ($j = 0; $j < strlen($ransomNote); $j++) {
            if (!isset($hashtable[$ransomNote[$j]]) || $hashtable[$ransomNote[$j]] == 0) {
                return false;
            } else {
                $hashtable[$ransomNote[$j]] -= 1;
            }
        }
        return true;
    }
}

383. Ransom Note

Description

Solutions

Solution 1: Hash Table or Array

Comments