chore(Python): add two sum for Dictionaries (#656)

algorithms/Python/README.md

@@ -11,6 +11,9 @@
 - [Singly](linked_lists/singly.py)
 - [Reverse List](linked_lists/reverse-linkedlist.py)
 
+## Dictionaries
+- [Two Sum](dictionaries/two-sum.py)
+
 ## Multiplication
 - [Karatsuba](multiplication/karatsuba.py)
 

algorithms/Python/dictionaries/two-sum.py

@@ -0,0 +1,44 @@
+"""
+Problem:
+Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to
+target. You can not use the same element twice.
+You can return any one of the answers in any order, if not found return [-1, -1].
+
+
+Solution:
+In this solution we check if each element's complement i.e. difference between the target and the element
+(target−nums[i]) exists in the dictionary. If it does exist, we return current element's index and its complement's
+index else we store the complement in the dictionary.
+Beware that the complement must not be nums[i] itself, because we can not use the same element twice.
+
+Complexity Analysis:
+Time complexity: O(n). We traverse the list containing nn elements only once. Each lookup in the dictionary costs only
+O(1) time.
+Space complexity: O(n). The extra space required depends on the number of items stored in the dictionary,
+which stores at most nn elements.
+"""
+
+
+def twoSum(nums, target):
+    dictionary = {}
+    # using enumerate to iterate through the list keeping track of both index and element
+    for i, val in enumerate(nums):
+        diff = target - val
+        if diff in dictionary:
+            return [dictionary[diff], i]
+        dictionary[val] = i
+    return [-1, -1]
+
+
+def main():
+    try:
+        arr = list(map(int, input("Enter space-seperated integers: ").split()))
+        tar = int(input("Enter target: "))
+    except ValueError:
+        arr = [2, 7, 11, 15]
+        tar = 9
+    print(twoSum(arr, tar))
+
+
+if __name__ == "__main__":
+    main()