Added the solutions

k-diff.py

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+# The code defines a findPairs method to count the number of unique k-diff pairs in a list of integers.
+# A k-diff pair is defined as two numbers in the list where the absolute difference between them is equal to k.
+
+# Initial Setup:
+#   - Use the Counter class to count the frequency of each number in nums and store it in 'count'.
+#       - This provides an efficient way to check for the existence of elements and their frequencies.
+
+# Case 1: k > 0
+#   - For a positive difference k, iterate over each unique number in 'count'.
+#   - Check if the current number + k exists in the Counter.
+#       - This means there exists a pair (i, i + k) with the desired difference.
+#   - Sum up the results of these checks (True counts as 1) to get the total number of such pairs.
+
+# Case 2: k == 0
+#   - For a zero difference, a valid pair exists only if the frequency of a number in the list is greater than 1.
+#   - Iterate over each number in 'count' and check if its frequency is greater than 1.
+#   - Sum up the results of these checks to get the count of such pairs.
+
+# Final Result:
+#   - The method returns the total number of unique k-diff pairs found in the list.
+
+# TC: O(n) - The time complexity is linear as the Counter is constructed in O(n), and each unique number in the Counter is processed once.
+# SC: O(n) - The space complexity is linear due to the storage of the Counter, which stores the frequency of each unique number.
+
+
+from typing import Counter, List
+
+
+class Solution:
+    def findPairs(self, nums: List[int], k: int) -> int:
+        count = Counter(nums)
+        if k > 0:
+            return sum([i + k in count for i in count])
+        else:
+            return sum([count[i] > 1 for i in count])

pascal.py

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+# The code defines a generate method to create Pascal's Triangle with the specified number of rows.
+# Pascal's Triangle is a triangular array where each element is the sum of the two elements directly above it in the previous row.
+
+# Initialization:
+#   - 'triangle' is an empty list that will store the rows of Pascal's Triangle.
+
+# Main Loop:
+#   - Iterate over the range from 0 to numRows - 1 (inclusive), where each iteration corresponds to constructing one row of the triangle:
+#       - Create a new row filled with ones, with a length equal to the current row index + 1.
+#       - For rows with more than two elements (i.e., i > 0):
+#           - Calculate the inner elements of the row (from index 1 to i - 1) using values from the previous row.
+#           - Each inner element at index j is computed as the sum of triangle[i - 1][j - 1] (left parent) and triangle[i - 1][j] (right parent).
+#       - Append the completed row to 'triangle'.
+
+# Final Result:
+#   - After all rows are constructed, 'triangle' contains the complete Pascal's Triangle with numRows rows, which is returned.
+
+# TC: O(n^2) - Constructing each row involves iterating over the row's length, resulting in quadratic time complexity for n rows.
+# SC: O(n^2) - The space complexity is proportional to the total number of elements in the triangle, which is the sum of the first n integers.
+
+
+from typing import List
+
+class Solution:
+    def generate(self, numRows: int) -> List[List[int]]:
+        triangle = []
+
+        for i in range(numRows):
+            row = [1] * (i + 1)
+
+            for j in range(1, i):
+                row[j] = triangle[i - 1][j - 1] + triangle[i - 1][j]
+
+            triangle.append(row)
+
+        return triangle