Completed Competitive Coding-3

k-diff_pairs_in_an_array.py

@@ -0,0 +1,24 @@
+#Time Complexity- O(n) Space Complexity- O(n)
+class Solution:
+    def findPairs(self, nums: list[int], k: int) -> int:
+        s = set()
+        map = { }
+        for i in range(len(nums)):
+            map[nums[i]] = i
+        for i in range(len(nums)):
+            y = nums[i] - k
+
+            if y in map and map[y] != i:
+                if y > nums[i]:
+                    s.add((y, nums[i]))
+                else:
+                    s.add((nums[i],y))
+            y = nums[i] + k
+
+            if y in map and map[y] != i:
+                if y > nums[i]:
+                    s.add((y, nums[i]))
+                else:
+                    s.add((nums[i], y))
+
+        return len(s)

pascals_triangle.py

@@ -0,0 +1,22 @@
+# https://leetcode.com/problems/pascals-triangle/description/
+
+class Solution:
+    def generate(self, numRows: int) -> list[list[int]]:
+        # creating the first of the triangle which is always 1. It's list inside a list
+        first_row = [[1]]
+        # as first row is already made above so we need to made remaining rows
+        for i in range(numRows -1):
+            # creating new temp list
+            # we are taking 0 for our understanding to build the next row
+            temp = [0] +    first_row[-1] + [0]
+            # creating an empty row for the next for loop
+            row = []
+            # building the next row and it's length would be the previous row+1
+            for j in range(len(first_row[-1]) + 1):
+                # apending the row
+                row.append(temp[j] + temp[j + 1])
+            first_row.append(row)
+        return(first_row)
+
+solution = Solution()
+print(solution.generate(5))