# 383.Ransom-Note

## 383. Ransom Note

## 题目地址

<https://leetcode.com/problems/ransom-note/>

## 题目描述

```
Given an arbitrary ransom note string and another string containing letters from all the magazines, write a function that will return true if the ransom note can be constructed from the magazines ; otherwise, it will return false.

Each letter in the magazine string can only be used once in your ransom note.

Note:
You may assume that both strings contain only lowercase letters.

canConstruct("a", "b") -> false
canConstruct("aa", "ab") -> false
canConstruct("aa", "aab") -> true
```

## 代码

### Approach #1 Two HashMaps

```java
class Solution {

    // Takes a String, and returns a HashMap with counts of
    // each character.
    private Map<Character, Integer> makeCountsMap(String s) {
        Map<Character, Integer> counts = new HashMap<>();
        for (char c: s.toCharArray()) {
            int currentCount = counts.getOrDefault(c, 0);
            counts.put(c, currentCount + 1);
        }
        return counts;
    }


    public boolean canConstruct(String ransomNote, String magazine) {

        // Check for obvious fail case.
        if (ransomNote.length() > magazine.length()) {
            return false;
        }

        // Make the count maps.
        Map<Character, Integer> ransomNoteCounts = makeCountsMap(ransomNote);
        Map<Character, Integer> magazineCounts = makeCountsMap(magazine);

        // For each unique character, c, in the ransom note:
        for (char c : ransomNoteCounts.keySet()) {
            // Check that the count of char in the magazine is equal
            // or higher than the count in the ransom note.
            int countInMagazine = magazineCounts.getOrDefault(c, 0);
            int countInRansomNote = ransomNoteCounts.get(c);
            if (countInMagazine < countInRansomNote) {
                return false;
            }
        }

        // If we got this far, we can successfully build the note.
        return true;
    }
}
```

### Approach 3: One HashMap

```java
class Solution {

    // Takes a String, and returns a HashMap with counts of
    // each character.
    private Map<Character, Integer> makeCountsMap(String s) {
        Map<Character, Integer> counts = new HashMap<>();
        for (char c : s.toCharArray()) {
            int currentCount = counts.getOrDefault(c, 0);
            counts.put(c, currentCount + 1);
        }
        return counts;
    }


    public boolean canConstruct(String ransomNote, String magazine) {

        // Check for obvious fail case.
        if (ransomNote.length() > magazine.length()) {
            return false;
        }

        // Make a counts map for the magazine.
        Map<Character, Integer> magazineCounts = makeCountsMap(magazine);

        // For each character in the ransom note:
        for (char c : ransomNote.toCharArray()) {
            // Get the current count for c in the magazine.
            int countInMagazine = magazineCounts.getOrDefault(c, 0);
            // If there are none of c left, return false.
            if (countInMagazine == 0) {
                return false;
            }
            // Put the updated count for c back into magazineCounts.
            magazineCounts.put(c, countInMagazine - 1);
        }

        // If we got this far, we can successfully build the note.
        return true;
    }
}
```


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