Merge pull request #16 from mihirs16/master

added leetcode questions

Author: shushrutsharma

14. Questions/leetcode 290 - word pattern.py

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+# word pattern | leetcode 290 | https://leetcode.com/problems/word-pattern/
+# create a vocabulary to match pattern and a seen hashset to record seen words
+
+class Solution:
+    def wordPattern(self, pattern: str, s: str) -> bool:
+        vocab = dict()
+        seens = dict()
+        sent = s.split(" ")
+        
+        if len(sent) != len(pattern):
+            return False
+
+        for i in range(len(pattern)):
+            i_patt = pattern[i]
+            i_sent = sent[i]
+
+            if vocab.get(i_patt):
+                if vocab[i_patt] != i_sent:
+                    return False
+            else:
+                if seens.get(i_sent):
+                    return False
+                vocab[i_patt] = i_sent
+                seens[i_sent] = True
+
+        return True

14. Questions/leetcode 347 - top k frequent elements.py

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+# top k frequency elements | leetcode 347 | https://leetcode.com/problems/top-k-frequent-elements/
+# use buckets with each bucket being the frequency of an element
+
+from collections import Counter
+
+class Solution:
+    def topKFrequent(self, nums: list[int], k: int) -> list[int]:
+        freq = Counter(nums)
+        N = len(nums)
+
+        # create buckets where index = frequency of element
+        buckets = [[] for x in range(N + 1)]
+        for f in freq:
+            buckets[freq[f]].append(f)
+
+        # get k elements starting from the end of the bucket
+        k_mf = []
+        for x in buckets[::-1]:
+            if k > 0:
+                if x != []:
+                    k_mf += x
+                    k -= len(x)
+            else:
+                return k_mf
+                

14. Questions/leetcode 49 - group anagrams.py

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+# group anagrams | leetcode 49 | https://leetcode.com/problems/group-anagrams/
+# method: dictionary with char counter as key
+
+from collections import defaultdict
+
+class Solution:
+    def groupAnagrams(self, strs):
+        grouped = defaultdict(list)
+        
+        for each_word in strs:
+            count_of_ch = [0] * 26
+            for each_ch in each_word:
+                count_of_ch[ord(each_ch) - ord("a")] += 1
+            grouped[tuple(count_of_ch)].append(each_word)
+            
+        return grouped.values()