Complete CC-3

118. Pascal's Triangle.py

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+# Time: O(n^2) where n is numRows => The outer loop runs until n and the inner loop runs i+1 so n*(n+1)
+# space: O(n^2) because we have to store all rows tho we return res we still had to store arrays until we return which takes space while executing
+
+# Generat rows and then append rows into the resultant array
+# the no. rows would be from 0 to the numRows
+# and the elements in row would be i + 1 with the ends being 1
+# and rest of the elements being the addition of [i-1][j-1] and [i-1][j]
+
+class Solution:
+    def generate(self, numRows: int) -> List[List[int]]:
+        res = []
+
+        for i in range(numRows):
+            sub_arr = []
+            for j in range(i+1):
+                if j == 0 or j == i:
+                    sub_arr.append(1)
+                else:
+                    left = res[i-1][j-1]
+                    right = res[i-1][j]
+                    sub_arr.append(left + right)
+            res.append(sub_arr)
+
+        return res

532. K-diff Pairs in an Array.py

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+# Time: O(n^2)
+# Space: O(1)
+
+# Iterate through i and j looking for the absolute different between them which is equal to k
+# check if the pair is already in the set (sort the pair as [3,1] and [1,3] can be stored seperately but are the same)
+# If its not there increment the counter and add it
+
+class Solution:
+    def findPairs(self, nums: List[int], k: int) -> int:
+        res = 0
+        unique = set()
+        for i in range(len(nums)):
+            for j in range(i+1,len(nums)):
+                if abs(nums[i] - nums[j]) == k:
+                    pair = tuple(sorted([nums[i],nums[j]]))
+                    if pair not in unique:
+                        unique.add(pair)
+                        res += 1
+
+        return res
+
+# Time: O(nlogn) + O(n) => O(nlogn)
+# Space: O(1)
+
+# Using two pointers 
+# we first sort the input array
+# After sorting we have a left and right pointer which start from 0 and 1 index respectively
+# Checking if the val is greater or smaller than k we shift the pointers
+# If it's equal we shift both the pointers
+# Also check if left is equal to right as we don't want both of the pointers at the same element
+class Solution:
+    def findPairs(self, nums: List[int], k: int) -> int:
+        sort = sorted(nums)
+        left = 0
+        right = 1
+        res = 0
+
+        while right < len(sort):
+            if left == right:
+                right += 1
+            elif sort[right] - sort[left] < k:
+                right += 1
+            elif sort[right] - sort[left] > k:
+                left += 1
+            else:
+                res += 1
+                left += 1
+                right += 1
+                # skip duplicates for left
+                while left < len(sort) and nums[left] == nums[left - 1]:
+                    left += 1
+
+        return res
+
+# Time: O(n)
+# Space: O(n)
+# When k is 0 check if there is a number that is occuring more than 1 time using a frequency map
+# add all the elemets in a set so that we have track of all the elements in nums array
+# after adding we loop through the elements in the set (we don't loop on nums array as we don't want repeating elements)
+
+class Solution:
+    def findPairs(self, nums: List[int], k: int) -> int:
+        if k == 0:
+            freq = Counter(nums)
+            count = 0
+            for v in freq.values():
+                if v > 1:
+                    count += 1
+            return count
+
+        seen = set(nums)
+        count = 0
+        for i in seen:
+            if i + k in seen:
+                count += 1
+
+        return count