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#2.2 Implement an algorithm to find the kth to last element of a singly linked list.
# supose that the length of the link is unknown

from LinkedList import Node, LinkedList

def nthtolast(head, k):
    
    if head is None:
        return  
    p1 = head

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class Node:

    def __init__(self, data):
        self.data = data
        self.next = None
    


class LinkedList:

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#cc 1.8 Assume you have a method isSubstring which checks if one word is a substring of another. Given two strings, s1 and s2, write code to check if s2 is a rotation of s1 using only one call to isSubstring (e.g.,"waterbottle"is a rotation of"erbottlewat").


def isRotation(s1, s2):

    if (len(s1) == len(s2) and len(s1) > 0):
        s = s1 + s1
        return isSubstring(s, s2)
    else:
        return False 

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1.7  Write an algorithm such that if an element in an MxN matrix is 0, its entire row and column are set to 0.

# deal matrix as a list


def listIndex(x, y, n):
    return y + x * n

def toList(matrix, m, n):
    global listm

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#cc 1.6 Given an image represented by an N*N matrix, where each pixel in the image is 4 bytes. write a method to rotate the image by 90 degrees
#Assume clockwise rotation

def rotate(matrix):

    n = len(matrix)
    for layer in range(0, n/2):
        first = layer
        last = n-layer-1
        for i in range(first, last):

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# cc1.5 Implement a method to perform string compression using counts of repeated characters. If the compressed string is not smaller than the ooriginal string, return the original one.


def compression(s):
    
    list1 = list(s)
    list2 = []
    c = 1
    j = 0
    for i in range(1, len(list1)):

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def replaceS(s):

    replist = list(s)
    for i in range (0, len(s)):
        if ord(replist[i]) == 32:
            replist[i] = '20%'
    newlist = ''.join(replist)
    return newlist

       

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#1.3 Given two strings, write a method to decide is one is a permutation of the other.
    
def permutation(s1, s2):
#write string into list 
    list1 = list(s1)
    list2 = list(s2)
#compare the length, if not the same re turn false
    if len(list1) != len(list2):
        return False
    else:

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#1.1 Implement an algorithm to determine if a string has all unique characters. Whatif you cannot use additional data structures? 

test =[0 for i in range(255)]
re = 0
s= raw_input('Input:')

if s.strip() =='':
    print'This is empty string'
else: