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Merge pull request #16 from mihirs16/master

added leetcode questions

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	`1`	`+# word pattern \| leetcode 290 \| https://leetcode.com/problems/word-pattern/`
	`2`	`+# create a vocabulary to match pattern and a seen hashset to record seen words`
	`3`	`+`
	`4`	`+class Solution:`
	`5`	`+ def wordPattern(self, pattern: str, s: str) -> bool:`
	`6`	`+ vocab = dict()`
	`7`	`+ seens = dict()`
	`8`	`+ sent = s.split(" ")`
	`9`	`+`
	`10`	`+ if len(sent) != len(pattern):`
	`11`	`+ return False`
	`12`	`+`
	`13`	`+ for i in range(len(pattern)):`
	`14`	`+ i_patt = pattern[i]`
	`15`	`+ i_sent = sent[i]`
	`16`	`+`
	`17`	`+ if vocab.get(i_patt):`
	`18`	`+ if vocab[i_patt] != i_sent:`
	`19`	`+ return False`
	`20`	`+ else:`
	`21`	`+ if seens.get(i_sent):`
	`22`	`+ return False`
	`23`	`+ vocab[i_patt] = i_sent`
	`24`	`+ seens[i_sent] = True`
	`25`	`+`
	`26`	`+ return True`

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	`1`	`+# top k frequency elements \| leetcode 347 \| https://leetcode.com/problems/top-k-frequent-elements/`
	`2`	`+# use buckets with each bucket being the frequency of an element`
	`3`	`+`
	`4`	`+from collections import Counter`
	`5`	`+`
	`6`	`+class Solution:`
	`7`	`+ def topKFrequent(self, nums: list[int], k: int) -> list[int]:`
	`8`	`+ freq = Counter(nums)`
	`9`	`+ N = len(nums)`
	`10`	`+`
	`11`	`+ # create buckets where index = frequency of element`
	`12`	`+ buckets = [[] for x in range(N + 1)]`
	`13`	`+ for f in freq:`
	`14`	`+ buckets[freq[f]].append(f)`
	`15`	`+`
	`16`	`+ # get k elements starting from the end of the bucket`
	`17`	`+ k_mf = []`
	`18`	`+ for x in buckets[::-1]:`
	`19`	`+ if k > 0:`
	`20`	`+ if x != []:`
	`21`	`+ k_mf += x`
	`22`	`+ k -= len(x)`
	`23`	`+ else:`
	`24`	`+ return k_mf`
	`25`	`+`

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	`1`	`+# group anagrams \| leetcode 49 \| https://leetcode.com/problems/group-anagrams/`
	`2`	`+# method: dictionary with char counter as key`
	`3`	`+`
	`4`	`+from collections import defaultdict`
	`5`	`+`
	`6`	`+class Solution:`
	`7`	`+ def groupAnagrams(self, strs):`
	`8`	`+ grouped = defaultdict(list)`
	`9`	`+`
	`10`	`+ for each_word in strs:`
	`11`	`+ count_of_ch = [0] * 26`
	`12`	`+ for each_ch in each_word:`
	`13`	`+ count_of_ch[ord(each_ch) - ord("a")] += 1`
	`14`	`+ grouped[tuple(count_of_ch)].append(each_word)`
	`15`	`+`
	`16`	`+ return grouped.values()`

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