Merge pull request #5 from camilosalazar98/master

My solution for 1.1

Author: fvntr

Python/chapter01/1.1 - Is Unique/camilo_sol.py

@@ -0,0 +1,31 @@
+# Implement an algorithm to determine if a string has all unique charactersself.
+# What if you cannot use additional data structures.
+
+
+def isUnique(string_):
+    for i in range(len(string_)):
+        i = 0
+        for j in range(1,len(string_)):
+            if string_[i] == string_[j]:
+                return False
+    return True
+
+print(isUnique("GeeksforGeeks"))
+#This soluation is O(n^2)
+
+#let's see if we can go faster
+
+def isUnique_(string_):
+    d = {}
+    for i in string_:
+        if i in d:
+            d[i] += 1
+
+        else:
+            d[i] = 1
+    for keys,value in d.items():
+        if value > 1:
+            return False
+    return True
+
+print(isUnique_("camilo"))

Python/chapter01/1.2 - Check Perm/camilo_sol_1.2.py

@@ -0,0 +1,29 @@
+def perm_check(str_1,str_2):
+    if len(str_1) != len(str_2):#If both str are diiferent lenghts return false
+        return False
+    d = {}
+    for i in str_1: #takes all the letter in string one and add them to the hash map
+        if i in d:  #counts the number of letter
+            d[i] += 1
+        else:
+            d[i] = 1
+
+    for j in str_2: #
+        if j in d:#Looks at the letter in the string and subtract the number
+                  #of the same letter from the value
+            d[j] -= 1
+        else:
+            d[j] = 1
+
+    for keys,value in d.items():
+        if value > 0:
+            return False
+    return True
+
+
+#O(n) sol?
+print(perm_check("camilo","pop")) #False
+
+print(perm_check("camilo","camplo")) #false
+
+print(perm_check("camilo","olimac")) #True