diff --git a/Contributor.md b/Contributor.md
index 82262f8..44acc14 100644
--- a/Contributor.md
+++ b/Contributor.md
@@ -7,3 +7,4 @@
 - [Shaurya026](https://github.com/Shaurya026)
 - [Jeevesh-Joshi](https://github.com/Jeevesh-Joshi)
 - [Shivgopal07](https://github.com/Shivgopal07)
+- [varishtsg](https://github.com/varishtsg)
diff --git a/unique_element.py b/unique_element.py
new file mode 100644
index 0000000..bafff6f
--- /dev/null
+++ b/unique_element.py
@@ -0,0 +1,45 @@
+# Problem Statement: Find the most unique non repeating element from a list
+# a = [1,2,2,4,4,3,3]
+# Output: 1
+# Using Numpy
+# Initialization Part
+import numpy as np
+ 
+a = [1,2,1,2,3,4,4,5,6,6,5]
+ 
+# Using numpy.unique() we'll find the elements and their frequency distribution aka counts
+nums, counts = np.unique(a,return_counts=True)
+ 
+# The minimum a number will repeat will be 1. It was given in the problem statement that the number will be only a single number that doesn't repeat. Else this method returns all numbers that don't repeat
+# This gives us the position of all elements that do not repeat
+pos = np.where(counts == 1)
+ 
+# With the position(s) we can now print out the non repeating numbers
+print(nums[pos])
+# [3]
+
+# Using Collections 
+# Initialization Part
+import collections
+ 
+a = [1,2,1,2,3,4,4,5,6,6,5]
+ 
+y = Counter(a)
+# This gives a counter object like this
+# Counter({1: 2, 2: 2, 3: 1, 4: 2, 5: 2, 6: 2})
+ 
+# Doesn't this look like a dictionary, we can now use a for loop and find the non repeating element. But wait, collections has other things too.
+ 
+y.most_common()
+# So this gives us a very convenient list like this
+# [(1, 2), (2, 2), (4, 2), (5, 2), (6, 2), (3, 1)]
+ 
+# The last element here is the non repeating one, which we will access like this
+ 
+y.most_common()[-1][0]
+# 3
+ 
+# The whole thing can be written in one line like this:
+ 
+collections.Counter(a).most_common()[-1][0]
+# 3