[:solved] two leetcode problem

Author: ppxyn1

contains-duplicate/ppxyn1.py

@@ -0,0 +1,18 @@
+# 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
+'''

two-sum/ppxyn1.py

@@ -0,0 +1,34 @@
+# 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
+