Merge pull request #13 from camilosalazar98/master

Adding 2.1,2.2,2.3

Author: fvntr

Python/chapter01/1.8 - Zero Matrix/camilo_solution_1.8.py

@@ -2,21 +2,67 @@
 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
+
+
+
+I've been stuck on this for a couple of days..
+Really good resource:
+https://www.youtube.com/watch?v=ZzpJgRvqSJQ&t=113s
+https://www.youtube.com/watch?v=qWeNXOCff3o&t=408s
 """
 
 
 
 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
+    rowZero = False
+    colZero = False
+
+    for i in range(0,len(matrix)):
+        if matrix[i][0] == 0:
+            colZero = True
+
+    for i in range(0,len(matrix[0])):
+        if matrix[0][i] == 0:
+            rowZero = True
+
+    for i in range(1,len(matrix)):
+        for j in range(1,len(matrix[0])):
+            if matrix[i][j] == 0:
+                matrix[0][j] == 0
+                matrix[i][0] == 0
+
+
+    for i in range(1,len(matrix)):
+        if(matrix[i][0] == 0):
+            for j in range(1,len(matrix[0])):
+                matrix[i][j] = 0
+
+    for i in range(1,len(matrix[0])):
+        if(matrix[0][i] == 0):
+            for j in range(1,len(matrix)):
+                matrix[i][j] = 0
+
+    if rowZero:
+        for i in range(0,len(matrix[0])):
+            matrix[0][i] = 0
 
+    if colZero:
+        for i in range(0,len(matrix)):
+            matrix[i][0] = 0
 
-matrix = [[0,1,2],[3,4,5],[6,7,8]]
 
 
+
+
+
+
+
+
+matrix = [[0,1,1],
+          [1,1,1],
+          [1,1,1]]
+
 zeroMatrix(matrix)
+
+
 print(matrix)

Python/chapter01/1.9 - String Rotation/camilo_solution_1.9.py

@@ -0,0 +1,27 @@
+"""
+String Rotation:Assume you have a method isSubstring which checks if one word is a
+substring of another. Given two strings, sl and s2, write code to check if s2 is
+a rotation of sl using only one call to isSubstring (e.g.,"waterbottle" is a
+rotation of"erbottlewat").
+"""
+
+def isSubstring(s1,s2):
+    if s2 in s1: #check to see if the seqence of s2 is found in s1
+        return True
+    return False
+
+def Rotation(s1,s2):
+    """
+    if they are substring of each other then we first check its lenght
+    """
+    if len(s1) != len(s2): #if there not the same lenght then there different woulds
+        return False
+
+    else:
+        s1+=s1 #waterbottlewaterbottle Anyroataion can be found here
+        return isSubstring(s1,s2)
+
+s1 = "waterbottle"
+s2 = "erbottlewat"
+
+print(Rotation(s1,s2))

Python/chapter02/2.1 - Remove Dups/camilo_solution_2.1.py

@@ -0,0 +1,103 @@
+"""Romove Dups! Write code to remove duplicates from an unsorted linked list.
+FOLLOW UP
+How would you solve this problem if a temporary
+buffer is not allowed? Hints: #9, #40
+"""
+
+class Node:
+    #Singly link list
+    def __init__(self,data):
+        self.data = data
+        self.next = None
+
+class linklist:
+    #linkList class
+    def __init__(self):
+        self.head = None
+
+    def push(self,data):
+        node = Node(data)# create a new node
+
+        if self.head == None: #check to see if the head node is empty
+            self.head = node # If its empty add it to the new node
+            return
+        #if the head of the linklist is filled
+
+        current = self.head
+        while current.next is not None:#Check the current postion is empty
+        #Move to the next line if nothing is there
+            current = current.next
+
+        current.next = node #point self.head to a new node
+
+
+
+
+    def lenght(self):
+        #note the count doesn't start at zero
+        cur = self.head
+        counter = 0
+        while cur is not None:
+            counter+=1
+            cur = cur.next
+        print('Linklist len: '+str(counter))
+
+    def printList(self):
+        curr = self.head
+        elem = []
+
+        while(curr != None):
+            elem.append(curr.data)
+            curr = curr.next
+        print(elem)
+    #1->2->3
+    def remove_node(self,data):
+        #1->2->3
+        curr = self.head
+        if curr is not None and curr.data == data:
+            self.head = curr.next
+            curr = None
+        #look for the node that has the data we are looking for
+        while curr is not None:
+            if curr.data == data:
+                break
+            prev = curr
+            curr = curr.next
+
+        #if data isn't found just reutrn
+        if(curr == None):
+            return
+
+        #allow us to unlink the nodes
+        prev.next = curr.next
+        curr = None
+
+    def remove_dups(self):
+        d_ = set()# create a set where only unquine element can be in
+        curr = self.head
+        while curr is not None:
+            d_.add(curr.data)#add all the elements to the set
+            curr = curr.next
+        d_ = list(d_)#make theem in to a list
+        self.head = None#clear the orignal link list
+        for i in range(len(d_)):
+            self.push(d_[i])#Use the element in the set a build a linklist with no dups
+
+
+
+
+
+
+#Testing to see if my class works
+llist = linklist()
+llist.push(2)
+llist.push(2)
+llist.push(2)
+llist.push(2)
+llist.push(2)
+llist.printList()
+llist.lenght()
+#Testing 1->2->1->3->1 result should only be 1->2->3
+llist.remove_dups()
+llist.printList()
+llist.lenght()

Python/chapter02/2.2 - Return Kth to Last/camilo_solution_2.2.py

@@ -0,0 +1,106 @@
+"""
+Return Kth to Last: Implement an algorithm to find the kth to last element of
+a singly linked list.
+Hints:#8, #25, #41, #67, #126
+"""
+
+class Node:
+    #Singly link list
+    def __init__(self,data):
+        self.data = data
+        self.next = None
+
+class linklist:
+    #linkList class
+    def __init__(self):
+        self.head = None
+        self.size = 0
+
+    def push(self,data):
+        node = Node(data)# create a new node
+
+        if self.head == None: #check to see if the head node is empty
+            self.head = node # If its empty add it to the new node
+            self.size = 1
+            return
+        #if the head of the linklist is filled
+
+        current = self.head
+        while current.next is not None:#Check the current postion is empty
+        #Move to the next line if nothing is there
+            current = current.next
+
+
+        current.next = node #point self.head to a new node
+        self.size+=1
+
+
+
+
+    def lenght(self):
+        #note the count doesn't start at zero
+        cur = self.head
+        counter = 0
+        while cur is not None:
+            counter+=1
+            cur = cur.next
+        print('Linklist len: '+str(counter))
+        return counter
+
+    def printList(self):
+        curr = self.head
+        elem = []
+
+        while(curr != None):
+            elem.append(curr.data)
+            curr = curr.next
+        print(elem)
+    #1->2->3
+    def remove_node(self,data):
+        #1->2->3
+        curr = self.head
+        if curr is not None and curr.data == data:
+            self.head = curr.next
+            curr = None
+        #look for the node that has the data we are looking for
+        while curr is not None:
+            if curr.data == data:
+                break
+            prev = curr
+            curr = curr.next
+
+        #if data isn't found just reutrn
+        if(curr == None):
+            return
+
+        #allow us to unlink the nodes
+        prev.next = curr.next
+        curr = None
+
+    def Kth_to_Last(self,num):
+        counter = 1 # We need to keep track where we are
+        curr = self.head
+        goal = self.size - num# kth to last element
+        while curr is not None:
+            if goal == counter:
+                return curr.data
+            counter+=1
+            curr = curr.next
+
+
+
+
+
+
+llist = linklist()
+llist.push(1)
+llist.push(2)
+llist.push(3)
+llist.push(4)
+llist.push(5)
+llist.printList()
+
+llist.lenght()
+print(llist.Kth_to_Last(1))# should return 4
+print(llist.Kth_to_Last(2))# should return 4
+print(llist.Kth_to_Last(3))# should return 4