383. Ransom Note

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

Constraint:

• 1 <= ransomNote.length, magazine.length <= 10^5
• ransomNote and magazine consist of lowercase English letters.

Credits:

Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

Solution (Java)

class Solution {
    public boolean canConstruct(String ransomNote, String magazine) {
        int[] a = new int[26];
        int n = ransomNote.length();
        for (int i = 0; i < n; i++) {
            a[ransomNote.charAt(i) - 97]++;
        }
        for (int i = 0; i < magazine.length() && n != 0; i++) {
            if (a[magazine.charAt(i) - 97] > 0) {
                n--;
                a[magazine.charAt(i) - 97]--;
            }
        }
        return n == 0;
    }
}

Solution 1

/**
 * @param {string} ransomNote
 * @param {string} magazine
 * @return {boolean}
 */
var canConstruct = function(ransomNote, magazine) {
  const map = new Map();

  for (let i = 0; i < magazine.length; i++) {
    if (map.has(magazine[i])) {
      map.set(magazine[i], map.get(magazine[i]) + 1)
    } else {
      map.set(magazine[i], 1)
    }
  }

  for (let i = 0; i < ransomNote.length; i++) {
    if (!map.has(ransomNote[i]) || map.get(ransomNote[i]) === 0) {
      return false;
    }
    map.set(ransomNote[i], map.get(ransomNote[i]) - 1)
  }

  return true;
};