Adding Simple Solution for Problem - 15 - 3Sum

Author: Vanditg

3_Sum/SimpleSolution.py

@@ -0,0 +1,40 @@
+##==================================
+## Leetcode
+## Student: Vandit Jyotindra Gajjar
+## Year: 2020
+## Problem: 15
+## Problem Name: 3Sum 
+##===================================
+#Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
+#
+#Note:
+#
+#The solution set must not contain duplicate triplets.
+#
+#Example:
+#
+#Given array nums = [-1, 0, 1, 2, -1, -4],
+#
+#A solution set is:
+#[
+#  [-1, 0, 1],
+#  [-1, -1, 2]
+#]
+class Solution:
+    def threeSum(self, nums):
+        nums.sort()	#Sort the input array
+        result = set()	#Create empty set result
+        for i in range(len(nums)):	#Loop through list 
+            tmp = -nums[i]	#Create tmp which is negative of the list i's value
+            left, right = i + 1, len(nums) - 1	#Create left and right which stores given values in loop
+            while left < right:	#While left is less then right 
+            sum = nums[left] + nums[right]	#Create sum which stores the given equations value 
+            if sum < tmp:	#Condition-check: if sum's value is less then tmp
+                left += 1	#Increase the left value by 1.
+            elif sum > tmp	#Condition-check: else if sum's value is greater than tmp
+                right -= 1	#Decrease the right value by 1. 
+            else:	#Condition-check: If we reach the target getting zero.
+                result.add((nums[i], nums[left], nums[j]))	#Add those value in set one list by list. 
+                left += 1	#Increase left by 1
+                right -= 1	#Decrease right by 1. 
+        return result	#We return the final solution set.