diff --git a/MinRows.py b/MinRows.py
new file mode 100644
index 0000000..3ae6ab9
--- /dev/null
+++ b/MinRows.py
@@ -0,0 +1,13 @@
+def minRows (hs):
+    rowMins = [-1 for i in range (len (hs))] 
+    rowCount = 0
+    for num in hs:
+        for i in range (len (rowMins)):
+            if (rowMins[i] == -1):
+                rowMins[i] = num
+                rowCount +=1
+                break
+            elif (rowMins[i] > num):
+                rowMins[i] = num
+                break
+    return rowCount
diff --git a/palindrome_checker.py b/palindrome_checker.py
new file mode 100644
index 0000000..44c5c0a
--- /dev/null
+++ b/palindrome_checker.py
@@ -0,0 +1,15 @@
+'''
+Following function checks if a word is a palindrome
+Takes in user input
+'''
+def isPalindromic (word):
+    for i in range (0, len (word) // 2):
+        if (word[i].lower() != word[len(word) - 1 - i].lower()):
+            return False
+        return True
+word = input("Enter a word you want to check: ")
+
+if (isPalindromic (word)):
+    print ("{} is a planindrome!".format(word))
+else:
+    print ("{} is not a palindrome".format(word))
\ No newline at end of file
diff --git a/passcracker.py b/passcracker.py
new file mode 100644
index 0000000..8bdc778
--- /dev/null
+++ b/passcracker.py
@@ -0,0 +1,16 @@
+'''
+following code prints out all the possilble alphabetic combination for string length of n
+characters are limited to a-z for now
+'''
+
+def print_all_combs (n):
+    def print_combs_recur (str):
+        if (len (str) == n):
+            print (str)
+        else:
+            for i in range (0, 26):
+                print_combs_recur (str + chr (ord ('a') + i))
+    print_combs_recur ("")
+
+# print all the possible 4 letter words using [a-z]
+print_all_combs (4)
diff --git a/permutation_checker.py b/permutation_checker.py
new file mode 100644
index 0000000..f750d0a
--- /dev/null
+++ b/permutation_checker.py
@@ -0,0 +1,32 @@
+'''
+Following code checks whether or not
+two strings are permutaitons of one another
+Not case sensetive
+'''
+
+def arePerms (str1, str2):
+    if (len (str1) != len (str2)):
+        return False
+    # Create a histogram of str1 and cancel it out using str2
+    # If there are any left overs, not permuations of each other
+    histogram = [0 for i in range (0, 26)]
+    
+    for i in range (0, len (str1)):
+        # creating a histogram for how many times each letter in str1 appears
+        # subtracting 97 to normalize characters by taking the value of 'a' on each character
+        # decrease each count of str2
+        histogram[ord(str1[i].lower()) - 97] +=1
+        histogram[ord(str2[i].lower()) - 97] -= 1
+
+    #check if there are any leftovers
+    for count in histogram:
+        if count != 0:
+            return False
+    return True
+
+#Tests
+print(arePerms ("abc", "cba")) #True
+print(arePerms ("lol", "llz")) #False
+print(arePerms ("ABcd", "cBaD")) #True
+print(arePerms ("", "")) #True
+print(arePerms ("oofdf", "oofd")) #False
\ No newline at end of file