[bhyun-kim, 김병현] Week 5 Solutions

Author: bhyun-kim

3sum/bhyun-kim.py

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+"""
+15. 3Sum
+https://leetcode.com/problems/3sum/description/
+
+Solution: 
+    - Sort the list
+    - Iterate through the list
+    - For each element, find the two elements that sum up to -element
+    - Use two pointers to find the two elements
+    - Skip the element if it is the same as the previous element
+    - Skip the two elements if they are the same as the previous two elements
+    - Add the set to the output list
+
+    Example:
+        -----------------------------------------
+        low :    |
+        high:              |   
+            i : |
+                [-4,-1,-1,0,1,2]
+
+        -----------------------------------------
+        low :      |
+        high:              |   
+            i : |
+                [-4,-1,-1,0,1,2]
+
+                ... no possible set with i=-4, and high=2
+
+        -----------------------------------------
+        low :       |
+        high:              |   
+            i :     |
+                [-4,-1,-1,0,1,2]
+
+
+Time complexity: O(n^2)
+    - O(nlogn) for sorting
+    - O(n^2) for iterating through the list and finding the two elements
+    - Total: O(n^2)
+Space complexity: O(n)
+    - O(n) for the output list
+    - O(n) for the prevs dictionary
+    - O(n) for the prevs_n set
+    - Total: O(n)
+"""
+
+
+from typing import List
+
+
+class Solution:
+    def threeSum(self, nums: List[int]) -> List[List[int]]:
+        nums = sorted(nums)
+        output = []
+        prevs = dict()
+        prevs_n = set()
+
+        for i in range(len(nums) - 2):
+            n_i = nums[i]
+
+            if n_i in prevs_n:
+                continue
+            else:
+                prevs_n.add(n_i)
+
+            low, high = i + 1, len(nums) - 1
+            while low < high:
+                n_low = nums[low]
+                n_high = nums[high]
+                if n_i + n_low + n_high == 0:
+                    if not f"[{n_i},{n_low},{n_high}]" in prevs:
+                        prevs[f"[{n_i},{n_low},{n_high}]"] = 1
+                        output.append([n_i, n_low, n_high])
+                    low += 1
+                    high -= 1
+
+                elif n_i + n_low + n_high < 0:
+                    low += 1
+
+                elif n_i + n_low + n_high > 0:
+                    high -= 1
+
+        return output

encode-and-decode-strings/bhyun-kim.py

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+"""
+271. Encode and Decode Strings
+https://leetcode.com/problems/encode-and-decode-strings/description/
+
+Solution:
+    - Concatenate the strings with a special character
+    - Split the string by the special character
+
+Time complexity: O(n)
+    - Concatenating the strings: O(n)
+    - Splitting the string: O(n)
+    - Total: O(n)
+
+Space complexity: O(n)
+    - Storing the output: O(n)
+    - Total: O(n)
+"""
+from typing import List
+
+
+class Codec:
+    def encode(self, strs: List[str]) -> str:
+        """Encodes a list of strings to a single string."""
+        output = strs[0]
+
+        for i in range(1, len(strs)):
+            output += "é" + strs[i]
+
+        return output
+
+    def decode(self, s: str) -> List[str]:
+        """Decodes a single string to a list of strings."""
+
+        return s.split("é")

longest-consecutive-sequence/bhyun-kim.py

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+"""
+128. Longest Consecutive Sequence
+https://leetcode.com/problems/longest-consecutive-sequence/description/
+
+Solution:
+    - Create a set of the input list
+    - Iterate through the set
+    - For each element, find the consecutive elements
+    - Use a while loop to find the consecutive elements
+    - Update the longest length
+
+Time complexity: O(n)
+    - O(n) for iterating through the set
+
+Space complexity: O(n)
+    - O(n) for the set
+    - Total: O(n)
+
+"""
+
+from typing import List
+
+
+class Solution:
+    def longestConsecutive(self, nums: List[int]) -> int:
+        longest_len = 0
+        nums_set = set(nums)
+
+        for n in nums_set:
+            if n - 1 not in nums_set:
+                current_n = n
+                current_len = 1
+
+                while current_n + 1 in nums_set:
+                    current_n += 1
+                    current_len += 1
+
+                longest_len = max(longest_len, current_len)
+
+        return longest_len

product-of-array-except-self/bhyun-kim.py

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+"""
+238. Product of Array Except Self
+https://leetcode.com/problems/product-of-array-except-self/description/
+
+Solution:
+    - Create two lists to store the product of elements from the left and right
+    - Multiply the elements from the left and right lists to get the output
+
+    Example:
+        nums = [4, 2, 3, 1, 7, 0, 9, 10]
+
+        left is numbers to the left of the current index
+        right is numbers to the right of the current index
+        from_left is the product of the numbers to the left of the current index
+        from_right is the product of the numbers to the right of the current index
+
+        i = 0: (4) 2 3 1 7 0 9 10
+        i = 1: 4 (2) 3 1 7 0 9 10
+        i = 2: 4 2 (3) 1 7 0 9 10
+        i = 3: 4 2 3 (1) 7 0 9 10
+        i = 4: 4 2 3 1 (7) 0 9 10
+        i = 5: 4 2 3 1 7 (0) 9 10
+        i = 6: 4 2 3 1 7 0 (9) 10
+        i = 7: 4 2 3 1 7 0 9 (10)
+
+        from_left = [0, 4, 8, 24, 24, 168, 0, 0]
+        from_right = [0, 0, 0, 0, 0, 90, 10, 0]
+        output = [0, 0, 0, 0, 0, 15120, 0, 0]
+
+Time complexity: O(n)
+    - Calculating the product of the elements from the left: O(n)
+    - Calculating the product of the elements from the right: O(n)
+    - Calculating the output: O(n)
+    - Total: O(n)
+
+Space complexity: O(n)
+    - Storing the product of the elements from the left: O(n)
+    - Storing the product of the elements from the right: O(n)
+    - Storing the output: O(n)
+"""
+from typing import List
+
+
+class Solution:
+    def productExceptSelf(self, nums: List[int]) -> List[int]:
+        output = [0] * len(nums)
+
+        from_left = [0] * (len(nums))
+        from_right = [0] * (len(nums))
+
+        from_left[0] = nums[0]
+        from_right[-1] = nums[-1]
+
+        for i in range(1, len(nums) - 1):
+            from_left[i] = from_left[i - 1] * nums[i]
+
+        for i in reversed(range(1, len(nums) - 1)):
+            from_right[i] = from_right[i + 1] * nums[i]
+
+        output[0] = from_right[1]
+        output[-1] = from_left[-2]
+        for i in range(1, len(nums) - 1):
+            output[i] = from_left[i - 1] * from_right[i + 1]
+
+        return output

top-k-frequent-elements/bhyun-kim.py

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+"""
+347. Top K Frequent Elements
+https://leetcode.com/problems/top-k-frequent-elements/description/
+
+Solution:
+    - Count the frequency of each element in the list
+    - Sort the dictionary by the frequency in descending order
+    - Return the first k elements in the sorted dictionary
+
+Time complexity: O(nlogn)
+    - Counting the frequency of each element: O(n)
+    - Sorting the dictionary: O(nlogn)
+    - Returning the first k elements: O(k)
+        k <= n
+    - Total: O(nlogn)
+
+Space complexity: O(n)
+    - Storing the frequency of each element: O(n)
+    - Storing the sorted dictionary: O(n)
+    - Storing the output: O(k)
+        k <= n
+"""
+from collections import OrderedDict
+from typing import List
+
+
+class Solution:
+    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
+        n_dict = OrderedDict()
+
+        for n in nums:
+            if n in n_dict:
+                n_dict[n] += 1
+            else:
+                n_dict[n] = 1
+
+        n_dict = sorted(n_dict.items(), key=lambda x: x[1], reverse=True)
+
+        output = [0] * k
+        for i in range(k):
+            output[i] = n_dict[i][0]
+
+        return output