Merge pull request #19 from mihirs16/master

added leetcode questions

Author: shushrutsharma

14. Questions/leetcode 15 - three sum.py

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+# three sum | leetcode 15 | https://leetcode.com/problems/3sum/
+# - sorted; nested loop; outer loop for first element 
+# - inner loop for two sum on rest of list
+# - avoid duplicates by shifting window till last occurrence
+
+class Solution:
+    def threeSum(self, nums: list[int]) -> list[list[int]]:
+        nums.sort()
+        N = len(nums)
+        triplets = []
+        for i in range(N):
+            if i > 0 and nums[i] == nums[i - 1]:
+                continue
+
+            ptrL = i + 1
+            ptrR = N - 1
+            while ptrL < ptrR:
+                s = nums[i] + nums[ptrL] + nums[ptrR]
+                if s > 0:
+                    ptrR -= 1
+                elif s < 0:
+                    ptrL += 1
+                else:
+                    triplets.append([nums[i], nums[ptrL], nums[ptrR]])
+                    ptrL += 1
+                    while nums[ptrL] == nums[ptrL - 1] and ptrL < ptrR:
+                        ptrL += 1
+                
+        return triplets

14. Questions/leetcode 167 - two sum II.py

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+# two sum II - input array is sorted | leetcode 167 | https://leetcode.com/problems/two-sum-ii-input-array-is-sorted
+# use two pointers on sorted array; if sum > target slide window left, else slide window right
+
+class Solution:
+    def twoSum(self, numbers: list[int], target: int) -> list[int]:
+        ptrL = 0
+        ptrR = 1
+        N = len(numbers)
+
+        while ptrR < N:
+            s = numbers[ptrR] + numbers[ptrL]
+            if s == target:
+                return [ptrL + 1, ptrR + 1]
+            elif s < target:
+                ptrL += 1
+                ptrR += 1
+            else:
+                ptrL -= 1
+            
+        # unreachable for testcases with exactly one solution
+        return [-1, -1]

14. Questions/leetcode 2306 - naming a company.py

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+# naming a company | leetcode 2306 | https://leetcode.com/problems/naming-a-company
+# bucket by starting character to make it n(26^2.n) and compare each set with each other
+
+class Solution:
+    def distinctNames(self, ideas: list[str]) -> int:
+        buckets = dict()
+        num_distinct = 0
+
+        for idea in ideas:
+            if buckets.get(idea[0]) is None:
+                buckets[idea[0]] = {idea[1:]}
+            else:
+                buckets[idea[0]].add(idea[1:])
+
+        for prefix_i, suffix_i in buckets.items():
+            for prefix_j, suffix_j in buckets.items():
+                if prefix_i == prefix_j:
+                    continue
+                common = len(suffix_i & suffix_j)
+                common_i = len(suffix_i) - common
+                common_j = len(suffix_j) - common
+                num_distinct += common_i * common_j
+        
+        return num_distinct