Java solution, HashMap & HashSet

  • 1

    Time complexity: O(nk), space complexity O(nk). n is number of words, k is the average length of words.

    class Solution {
        public boolean areSentencesSimilar(String[] words1, String[] words2, String[][] pairs) {
            if (words1.length != words2.length) return false;
            Map<String, Set<String>> map = new HashMap<>();
            for (String[] p : pairs) {
                Set<String> set0 = map.getOrDefault(p[0], new HashSet<>());
                map.put(p[0], set0);
                Set<String> set1 = map.getOrDefault(p[1], new HashSet<>());
                map.put(p[1], set1);
            for (int i = 0; i < words1.length; i++) {
                if (words1[i].equals(words2[i])) continue;
                if (map.containsKey(words1[i]) && map.get(words1[i]).contains(words2[i])) continue;
                return false;
            return true;

  • 0

    why time complexity is nk? because equals() method cost average O(k) time?

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