Determine if a string is a substring of another string

substring.py

#NOTE: #### denotes additions to the code between questions

#Find the length of a string
#Nerf toss question to warm them up



#Write a function that takes two strings, and find if the first string contains the second string

def findSubstring( main, sub ):
  lengthMain = len(main)
  if( len(sub) > len(main) ):
    return False
  for i in range(0, lengthMain):
    if main[i] == sub[0]:
      matchFlag = True
      for j in range(0,len(sub)):
        if( (i+j) > lengthMain-1 ):
          matchFlag = False
          break
        if( sub[j] != main[i+j] ):
          matchFlag = False
          break
      if matchFlag:
        return True
        break
  return False


if( findSubstring( "abbbcd", "te\st" ) ):
  print "YES"





#Add in functionality for the escape character , where any character after the \ is ignored

def findSubstring2( main, sub ):
  lengthMain = len(main)
  print main
  print sub
  if( len(sub) > len(main) ):
    return False
  for i in range(0, lengthMain):
    if main[i] == sub[0]:
      matchFlag = True
      continueFlag = False      #####
      newIterator = i           #####
      for j in range(0,len(sub)):
        if( continueFlag ):     #####
          continueFlag = False  #####
          continue              #####
        if( sub[j] == "\\" ):   #####
          continueFlag = True   #####
          newIterator-=2        #####
          continue              #####
        if( (newIterator+j) > lengthMain-1 ): #####
          matchFlag = False
          break
        if( sub[j] != main[newIterator+j] ): #####
          matchFlag = False
          break
      if matchFlag:
        return True
        break
  return False

if( findSubstring2( "te\st", "te\st" ) ):
  print "YES7"
if( findSubstring2( "te\st", "te\st\\" ) ):
  print "YES8"
if( findSubstring2( "te\st", "te\st" ) ):
  print "YES9"

#Add functionality for a single wildcard "?" in the second argument, where the "?" can be replaced with any single character


def findSubstring3( main, sub ):
  lengthMain = len(main)
  if( len(sub) > len(main) ):
    return False
  for i in range(0, lengthMain):
    if main[i] == sub[0] or sub[0] == "?": #####
      matchFlag = True
      for j in range(0,len(sub)):
        if( (i+j) > lengthMain-1 ):
          matchFlag = False
          break
        if( sub[j] == "?" ):                #####important to go second
          continue                         #####
        if( sub[j] != main[i+j] ):
          matchFlag = False
          break
      if matchFlag:
        return True
        break
  return False

if( findSubstring3( "test", "??" ) ):
  print "YES3"
if( findSubstring3( "test", "t?st" ) ):
  print "YES4"
if( findSubstring3( "test", "st?" ) ):
  print "NO5"
if( findSubstring3( "test", "?????" ) ):
  print "NO6"




if( findSubstring3( "test", "t*" ) ):
  print "YES6"
if( findSubstring3( "test", "**" ) ):
  print "YES7"
if( findSubstring3( "test", "****" ) ):
  print "YES8"
if( findSubstring3( "testing", "t*g" ) ):
  print "YES9"
if( findSubstring3( "testing", "t*t*g" ) ):
  print "YES9"
if( findSubstring3( "testing", "t*g*g" ) ):
  print "YES9"
#Implement the multiple character wildcard "*", where if found in the second argument, it could represent any number of any possible characters