Submit Competition_Coding-3 PR

Problem_1.py

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+# https://leetcode.com/problems/pascals-triangle/description/
+
+# Time complexity: O(n ^ 2) 
+# Space complexity: O(1)
+# Explanation: We can create rows one by one, and compute the in between values using the previous row. The first value and last value is always 1.
+
+class Solution:
+    def generate(self, numRows: int) -> List[List[int]]:
+        triangle = []
+
+        for r in range(numRows):
+            row = [1] * (r + 1)
+            for j in range(1, len(row) - 1):
+                row[j] = triangle[r - 1][j - 1] + triangle[r - 1][j]
+            triangle.append(row)
+
+        return triangle

Problem_2.py

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+# https://leetcode.com/problems/k-diff-pairs-in-an-array/description/
+
+# Time complexity: O(n) 
+# Space complexity: O(n)
+# Explanation: To avoid sorting we can use a hash map to keep a count of the frequencies of each number in nums, and then we have two cases
+# - if k > 0, then we need sum of current number and k to be present in hash map to get diff to k
+# - if k = 0, then we need more than 2 instances of a number to get diff to 0
+
+from collections import Counter
+
+class Solution:
+    def findPairs(self, nums: List[int], k: int) -> int:
+        counter = Counter(nums)
+        res = 0
+        for n in counter:
+            if k > 0 and n + k in counter:
+                res += 1
+            elif k == 0 and counter[n] > 1:
+                res += 1
+        return res