Done Competitive_Coding-3

k-diff-pairs-in-an-array.py

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+#  ------ Solution 1 --- 
+'''   1. Sort array
+      2. Maintain left and right pointer
+      3. Append the result in set for unique
+
+ Time complexity : O(nlogn)
+ Space Complexity : O(n)
+'''
+class Solution:
+    def findPairs(self, nums: List[int], k: int) -> int:
+        n =len(nums)
+        nums.sort()
+        l ,r = 0, 1
+        res = set()
+        while l < n and r < n:
+            diff = nums[r] - nums[l] 
+            if diff == k:
+                res.add((nums[l],nums[r]))
+                l += 1
+                r += 1
+            elif diff < k:
+                r += 1
+            else:
+                l += 1
+
+            if l == r:
+                r += 1
+        print(res)
+        return len(res)
+
+#-------Soultion 2 : Creating hasmap with frequency
+''' Time complexity : O(n)
+    Space Complexity : O(n)
+ '''
+class Solution:
+    def findPairs(self, nums: List[int], k: int) -> int:
+        n =len(nums)
+        count = 0
+        hashmap = {}
+        for n in nums:
+            hashmap[n] = 1 + hashmap.get(n,0)
+        for n in hashmap:
+            if k == 0:
+                if hashmap[n] >= 2:
+                    count += 1
+            else:
+                comp = n + k 
+                if comp in hashmap:
+                    count += 1
+        return count
+
+
+

pascals-triangle.py

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+'''  
+ Time complexity : O(n ^ 2)
+ Space Complexity : O(1)
+
+ Approach :  1. Append 1 to start and end at each row
+             2. Middle element = Sum of j and j-1 of previous rows
+'''
+
+class Solution:
+    def generate(self, numRows: int) -> List[List[int]]:
+        result = []
+        for i in range(numRows):
+            l = []
+            for j in range(i+1):
+                if j == 0 or j == i:
+                    l.append(1)
+                else:
+                    l.append(result[i-1][j] + result[i-1][j-1])
+            result.append(l)
+        return result
+