diff --git a/coding_freshmen/JAVA/Swayam/Anagram in Java b/coding_freshmen/JAVA/Swayam/Anagram in Java
new file mode 100644
index 0000000..db5ad11
--- /dev/null
+++ b/coding_freshmen/JAVA/Swayam/Anagram in Java	
@@ -0,0 +1,32 @@
+import java.io.*;
+import java.util.Arrays;
+import java.util.Collections;
+class Main {
+static boolean checkAnagram(char[] strana1, char[] strana2)
+{
+
+int len1 = strana1.length;
+int len2 = strana2.length;
+
+if (len1 != len2)
+return false;
+
+Arrays.sort(strana1);
+Arrays.sort(strana2);
+
+for (int i = 0; i < len1; i++)
+if (strana1[i] != strana2[i])
+return false;
+return true;
+}
+
+public static void main (String args[])
+{
+char strana1[] = { 't', 'e', 's', 't' };
+char strana2[] = { 't', 't', 'e', 'w' };
+if (checkAnagram(strana1, strana2))
+System.out.println("The strings to be checked are" + " anagram of each other");
+else
+System.out.println("The strings to be checked are not" + " anagram of each other");
+}
+}
diff --git a/coding_freshmen/JAVA/Swayam/Anagram.md b/coding_freshmen/JAVA/Swayam/Anagram.md
new file mode 100644
index 0000000..56c814a
--- /dev/null
+++ b/coding_freshmen/JAVA/Swayam/Anagram.md
@@ -0,0 +1,19 @@
+# <Title of the Problem>
+Anagram 
+
+# Problem Explanation 🚀
+An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.
+
+# Your logic 🤯
+* Approach: Created two arrays to hold the string and compared the array one by one
+* Own test cases if any
+* Code Structure and Libraries used
+
+# Time Complexity and Space Complexity
+```cpp
+Example 
+
+Time Complexity -> O(n^2)
+Space Complexity -> O(1)
+
+```
diff --git a/coding_freshmen/PYTHON/Sort_1.py b/coding_freshmen/PYTHON/Sort_1.py
new file mode 100644
index 0000000..52dd11f
--- /dev/null
+++ b/coding_freshmen/PYTHON/Sort_1.py
@@ -0,0 +1,11 @@
+import array as arr
+arrr = arr.array('i',[4,2,5,9,1,8,6])
+def bubblesort(arrr):
+for j in range(len(arrr)-1,0,-1):
+for i in range(j):
+if arrr[i]> arrr[i+1]:
+temp = arrr[i]
+arrr[i] = arrr[i+1]
+arrr[i+1] = temp
+bubblesort(arrr)
+print(arrr)
diff --git a/coding_freshmen/PYTHON/YTERWQ/Product_Finder.py b/coding_freshmen/PYTHON/YTERWQ/Product_Finder.py
new file mode 100644
index 0000000..8bf5466
--- /dev/null
+++ b/coding_freshmen/PYTHON/YTERWQ/Product_Finder.py
@@ -0,0 +1,12 @@
+def product(ar, n):
+
+	result = 1
+	for i in range(0, n):
+		result = result * ar[i]
+	return result
+ar = [ 1, 2, 3, 4, 5 ]
+n = len(ar)
+
+print(product(ar, n))
+
+
diff --git a/coding_freshmen/PYTHON/YTERWQ/Readme.md b/coding_freshmen/PYTHON/YTERWQ/Readme.md
new file mode 100644
index 0000000..c077100
--- /dev/null
+++ b/coding_freshmen/PYTHON/YTERWQ/Readme.md
@@ -0,0 +1,25 @@
+# <Title of the Problem>
+Pivot_Product_Array
+
+# Problem Explanation 🚀
+you are given an array of size n.
+
+you need to find the product of the elements of the array except the pivot element.
+
+Input array   =  [0, 1, 0, 1, 0, 0, 1, 1, 1, 0] 
+
+Output array =  [0, 0, 0, 0, 0, 1, 1, 1, 1, 1]
+
+# Your logic 🤯
+* Approach: Compared each element of the array with the rest of the elements and sorted accordingly
+* Own test cases if any
+* Code Structure and Libraries used
+
+# Time Complexity and Space Complexity
+```cpp
+Example 
+
+Time Complexity -> O(n^2)
+Space Complexity -> O(1)
+
+```