Merge pull request #7 from camilosalazar98/master

Updates to some solutions

Author: fvntr

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

@@ -1,5 +1,7 @@
-# Implement an algorithm to determine if a string has all unique charactersself.
-# What if you cannot use additional data structures.
+"""
+Is Unique: Implement an algorithm to determine if a string has all unique
+characters. What if you cannot use additional data structures?
+"""
 
 
 def isUnique(string_):

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

@@ -1,3 +1,11 @@
+"""
+Check Permutation: Given two strings,write a method to decide if
+one is a permutation of the other.
+"""
+
+
+
+
 def perm_check(str_1,str_2):
     if len(str_1) != len(str_2):#If both str are diiferent lenghts return false
         return False

Python/chapter01/1.3 - URLify/camilo_solution_1.3.py

@@ -0,0 +1,11 @@
+"""
+URLify: Write a method to replace all spaces in a string with '%20'. You may
+ assume that the string has sufficient space at the end to hold the additional
+ characters,and that you are given the "true" length of the string.
+ (Note: If implementing in Java,please use a character array so that you can
+ perform this operation in place.)
+EXAMPLE
+Input: "Mr John Smith ", 13 Output: "Mr%20John%20Smith"
+"""
+
+def urlify(str):

Python/chapter01/1.3 - URLify/solution.py

@@ -0,0 +1,28 @@
+"""
+Palindrome Permutation: Given a string, write a function to check if it is a
+permutation of a palin­ drome. A palindrome is a word or phrase that is the same
+forwards and backwards. A permutation is a rearrangement of letters.
+The palindrome does not need to be limited to just dictionary words.
+EXAMPLE
+Input: Tact Coa
+Output: True (permutations: "taco cat", "atco eta", etc.)
+"""
+
+def PalinPerm(str):
+    d = {}
+    str = str.replace(' ','').lower()#Take out the spaces
+    for i in str:
+        if i in d:
+            d[i] += 1
+        else:
+            d[i] = 1
+    odd_count = 0 # pop ppo opp
+    for keys,value in d.items():
+        if value % 2 != 0 and odd_count == 0:# Remeber there should only be 1 where key equals when val%2
+            odd_count += 1
+        elif value % 2 != 0 and odd_count != 0:
+            return False
+    return True
+
+print(PalinPerm('taco cat'))#True
+print(PalinPerm('taco catt'))#False

Python/chapter01/1.4 - PalinPerm/camilo_solution_1.4.py

@@ -0,0 +1,75 @@
+"""
+One Away: There are three types of edits that can be performed on strings:
+insert a character, remove a character, or replace a character. Given two
+strings, write a function to check if they are one edit (or zero edits) away.
+EXAMPLE
+pale, ple -> true
+pales, pale -> true
+pale, bale -> true
+pale, bake -> false
+Hints:#23, #97, #130
+"""
+
+def oneAway(str1,str2):
+    str1 = str1.lower()
+    str2 = str2.lower()
+    d = {}
+    for i in str1:
+        if i in d:
+            d[i] =+ 1
+        else:
+            d[i] = 1
+
+    for j in str2:
+        if j in d:
+            d[j] -= 1
+
+        else:
+            d[j] = 1
+    count = 0# count the number of times
+
+
+    """
+    My thought on how to do this:
+    So knowing the pale and ple
+    when we subtract it from the hash table we know there should be a value of 1
+    for removal of one letter
+
+    The insertation of one letter should only leave the key of one
+    pales and pale
+
+    The interseting stuff happens when you have a change of letter.
+    The letter gets counting when you place it in the hash table
+    pale bale.
+    We use count to count the number of time count can be one. You'll find
+    that replaceing a letter is going to make count two v's of one
+
+    replacing two letter will make v's count more onces
+
+    You'll see this when you print k and v
+    
+    """
+    for k,v in d.items():
+        #print(k,v)
+        if v == 1:
+            count += 1
+
+    if count <= 2 : #
+        return True
+
+    else:
+        return False
+
+
+
+
+
+
+
+
+
+print(oneAway('pale', 'ple'))#True
+print(oneAway('pales', 'pale'))#True
+print(oneAway('pale', 'bale'))#True
+
+print(oneAway('pale', 'bake'))#False

Python/chapter01/1.5 - OneAway/camilo_solution_1.5.py

@@ -0,0 +1,36 @@
+"""
+1.6:String Compression: Implement a method to perform basic string compression using
+ the counts of repeated characters. For example, the string aabcccccaaa would
+ become a2blc5a3. If the "compressed" string would not become smaller than the
+ original string, your method should return
+the original string. You can assume the string has only uppercase and lowercase
+letters (a - z).
+"""
+
+
+
+def string_compression(str1):
+    count = 1
+    new_str = ""
+    for i in range(len(str1)-1):
+        if(str1[i] == str1[i+1]):
+            count+=1
+        else:
+            new_str += str1[i] + str(count)
+            count = 1
+    new_str += str1[i] + str(count)
+
+    if len(new_str) < len(str1):
+        return new_str
+
+    else:
+        return str1
+
+
+
+
+
+
+
+
+print(string_compression('aabcccccaaa'))

Python/chapter01/1.6 - String Compression/camilo_solution_1.6.py

@@ -0,0 +1,27 @@
+"""
+Rotate Matrix: Given an image represented by an NxN matrix, where each pixel
+in the image is 4 bytes, write a method to rotate the image by 90 degrees.
+Can you do this in place?
+Hints:#51, #100
+
+"""
+
+
+def rotate_matrix(matrix):
+    matrix.reverse()# we frist reverse the matrix then we take its transpose
+
+    m = len(matrix)
+    n = len(matrix[0])
+    for i in range(m):
+        for j in range(i+1,n):
+            temp = matrix[i][j]
+            matrix[i][j] = matrix[j][i]
+            matrix[j][i] = temp
+
+
+matrix = [[1,2,3],[4,5,6],[7,8,9]]
+rotate_matrix(matrix)
+#[[7, 4, 1], [8, 5, 2], [9, 6, 3]]
+
+
+print(matrix)

Python/chapter01/1.7 - Rotate Matrix/camilo_soluation_1.7.py

@@ -0,0 +1,22 @@
+"""
+Zero Matrix: Write an algorithm such that if an element in an MxN matrix is 0,
+its entire row and column are set to 0.
+Hints:#17, #74, #702
+"""
+
+
+
+def zeroMatrix(matrix):
+    m = len(matrix) - 1
+    n = len(matrix[0]) - 1
+    for i in range(m):
+        for j in range(n):
+                if matrix[i][j]  == 0:
+                    matrix[i+1][j] = 0
+
+
+matrix = [[0,1,2],[3,4,5],[6,7,8]]
+
+
+zeroMatrix(matrix)
+print(matrix)