Java O(n) solution using HashMap

  • 0
    public class Solution {
        public boolean canConstruct(String ransomNote, String magazine) {
            Map<Character, Integer> map = new HashMap<>();
            for (int i = 0; i < magazine.length(); i++) {
                if (!map.containsKey(magazine.charAt(i))) map.put(magazine.charAt(i), 0);
                map.put(magazine.charAt(i), map.get(magazine.charAt(i)) + 1);
            for (int i = 0; i < ransomNote.length(); i++) {
                if (!map.containsKey(ransomNote.charAt(i))) return false;
                int temp = map.get(ransomNote.charAt(i)) - 1;
                if (temp < 0) return false;
                map.put(ransomNote.charAt(i), temp);
            return true;

    My idea is record the times of appearance for every character in magazine, and check if enough for ransom note. Time complexity O(m + n) where m is the length of magazine and n is the length of ransom note.

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