[ppxyn1] WEEK 01 solutions

contains-duplicate/ppxyn1.py

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+# idea: Hash
+
+class Solution:
+    def containsDuplicate(self, nums: List[int]) -> bool:
+        count_dict={}
+        for i in range(len(nums)):
+            # print(count_dict) 
+            if nums[i] in count_dict.keys():
+                return True
+            else:
+                 count_dict[nums[i]] = 1
+        return False
+
+
+'''
+Trial and error
+Printing I/O inside the loop may cause Output Limit Exceeded
+'''
+

house-robber/ppxyn1.py

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+# idea: DP Top-down / Bottom-up
+# DP was comp up on my head, but it is not easy for me to utilse recursive function. 
+
+
+class Solution:
+    def rob(self, nums: List[int]) -> int:
+        length = len(nums)
+        if length == 0:
+            return 0
+        if length == 1:
+            return nums[0]
+
+        lst = [0]*length 
+        lst[0] = nums[0]
+        lst[1] = max(nums[0], nums[1])
+
+        for i in range(2, length):
+            lst[i] = max(lst[i-1], lst[i-2] + nums[i])
+
+        return lst[-1]
+

longest-consecutive-sequence/ppxyn1.py

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+# idea: Hash
+class Solution:
+    def longestConsecutive(self, nums: List[int]) -> int:
+        if not nums:
+            return 0
+
+        num_set = set(nums)
+        max_len = 1  
+
+        for num in num_set:
+            if num - 1 not in num_set:
+                current = num
+                tmp = 1
+                while current + 1 in num_set:
+                    current += 1
+                    tmp += 1
+                if tmp > max_len:
+                    max_len = tmp
+        return max_len
+
+
+
+
+
+'''
+Trial and error
+
+The code below was passed on Leecode since their Constraints was 0 <= nums.length <= 10^5.
+But I realised that I missed another constraint they mentioned, which is "You must write an algorithm that runs in O(n) time."
+'''
+# class Solution:
+#     def longestConsecutive(self, nums: List[int]) -> int:
+#         if len(nums)==0:
+#             return 0
+#         sorted_nums = sorted(list(set(nums))) 
+#         # sorted_nums = sorted(nums) # duplicate
+#         max_len = 1
+#         tmp = 1
+
+#         for i in range(1, len(sorted_nums)):
+#             if sorted_nums[i] == sorted_nums[i - 1] + 1:
+#                 tmp += 1
+#             else:
+#                 max_len = max(max_len, tmp)
+#                 tmp = 1 
+#         ans = max(max_len, tmp)
+#         return  ans
+
+
+
+

top-k-frequent-elements/ppxyn1.py

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+# idea: dictonary
+
+class Solution:
+    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
+        count_dict = {}
+        ans = []
+        for idx, val in enumerate(nums):
+            if val not in count_dict:
+                count_dict[val] = 1
+            else:
+                count_dict[val] +=1
+        sorted_items = sorted(count_dict.items(), key=lambda x: x[1], reverse=True) #sorted return list / dict.items() is tuple
+        # print(sorted_items) 
+        for i in range(k):
+            ans.append(sorted_items[i][0])
+        return ans
+
+'''
+Similar way : Using Counter() function
+'''
+

two-sum/ppxyn1.py

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+# idea: For each number n in nums, check if (target - n) exists in the remaining elements.
+
+class Solution:
+    def twoSum(self, nums: List[int], target: int) -> List[int]:
+        for idx, num in enumerate(nums):
+            required_num = target - num
+            if required_num in nums[idx+1:]:
+                return [idx, nums.index(required_num, idx+1)]
+
+
+'''
+Trial and error
+idea : two pointer
+I struggled to handle the indices of the original array after sorting it.
+The code below fails when negative numbers are involved.
+I realized it would be tricky to solve this problem with the two-pointer.
+'''
+
+# class Solution:
+#     def twoSum(self, nums: List[int], target: int) -> List[int]:
+#         sorted_nums = sorted(nums)
+#         left, right = 0, len(nums) - 1
+
+#         while left < right:
+#             s = sorted_nums[left] + sorted_nums[right]
+#             if s == target:
+#                 left_idx = nums.index(sorted_nums[left])
+#                 right_idx = nums.index(sorted_nums[right], left_idx + 1)
+#                 return [left_idx, right_idx]
+#             elif s < target:
+#                 left += 1
+#             else:
+#                 right -= 1
+