[mand2, yapp] week1 답안 제출

best-time-to-buy-and-sell-stock/mand2.py

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+# 문제: https://leetcode.com/problems/best-time-to-buy-and-sell-stock/description/
+
+def solution(prices):
+    profit, buy = 0, prices[0]
+    for price in prices:
+        diff = price - buy
+        buy = min(price, buy)
+        profit = max(profit, diff)
+    return profit
+
+
+# 시간복잡도: O(n)
+# 공간복잡도: O(1)
+
+
+answer_1 = solution([7, 1, 5, 3, 6, 4])
+answer_2 = solution([7, 6, 4, 3, 1])
+
+print(answer_1 == 5)
+print(answer_2 == 0)

contains-duplicate/mand2.py

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+# 문제: https://leetcode.com/problems/contains-duplicate/
+def containsDuplicate(nums) -> bool:
+    done = set()
+    for num in nums:
+        if num in done:
+            return True
+        done.add(num)
+    return False
+
+
+# 시간복잡도: O(n)
+# 공간복잡도: O(n)
+
+print((containsDuplicate([1, 2, 3, 1]) is True))
+print((containsDuplicate([1, 2, 3, 4]) is False))
+print((containsDuplicate([1, 1, 1, 3, 3, 4, 3, 2, 4, 2]) is True))

two-sum/mand2.py

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+# 문제: https://leetcode.com/problems/two-sum/description/
+
+# targetNum - list[i] 값이 list에 있는지 확인만 하면 끝. -> 이 아니고 i. j 리턴
+def solutions(nums, target_num):
+    table = {num: idx for idx, num in enumerate(nums)}
+
+    for i, value in enumerate(nums):
+        look = target_num - value
+        # value -> idx로 바로 치환하기가 어렵..
+        if look in table and i != table[look]:
+            look_idx = table[look]
+            return [i, look_idx]
+
+
+# 시간복잡도: O(n)
+# 공간복잡도: O(n)
+
+answer_1 = solutions([2, 7, 11, 15], 9)
+answer_2 = solutions([3, 3], 6)  # 중복된수가나오면..?!?!?!?!
+answer_3 = solutions([3, 2, 4], 6)
+
+print(answer_1 == [0, 1])
+print(answer_2 == [0, 1])
+print(answer_3 == [1, 2])

valid-anagram/mand2.py

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+# 문제: https://leetcode.com/problems/valid-anagram/
+from collections import defaultdict
+
+
+# 풀이: s 와 t 에 입력된 알파벳의 갯수를 체크한다. ...
+# 애너그램이면 T
+def is_anagram(s, t) -> bool:
+    # 글자수가 같다는 조건이 없음.
+    if len(s) != len(t):
+        return False
+
+    word_counter = defaultdict(int)
+    # s 문자열 분해
+    for alpha in s:
+        word_counter[alpha] += 1
+
+    for beta in t:
+        if beta not in word_counter or word_counter[beta] == 0:
+            return False
+        word_counter[beta] -= 1
+    return True
+
+# 시간복잡도: O(n)
+# 공간복잡도: O(n)
+
+tc_1 = is_anagram("anagram", "nagaram") is True
+tc_2 = is_anagram("rat", "car") is False
+tc_3 = is_anagram("a", "ab") is False
+tc_4 = is_anagram("aacc", "ccac") is False
+
+print('tc_1', tc_1)
+print('tc_2', tc_2)
+print('tc_3', tc_3)
+print('tc_4', tc_4)

valid-palindrome/mand2.py

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+# 문제: https://leetcode.com/problems/valid-palindrome/description/
+
+def solution(sentence):
+    # removing all non-alphanumeric characters (숫자는 ㅇㅋ)
+    selected = ''
+    for s in sentence:
+        if s.isalnum():
+            selected += s.lower()
+    return selected == selected[::-1]
+
+
+# 시간복잡도: O(n)
+# 공간복잡도: O(n)
+
+
+answer_1 = solution("A man, a plan, a canal: Panama")
+answer_2 = solution("0P")
+answer_3 = solution("race a car")
+answer_4 = solution(" ")
+answer_5 = solution("a")
+
+print(answer_1 is True)
+print(answer_2 is False)
+print(answer_3 is False)
+print(answer_4 is True)
+print(answer_5 is True)