Python - Test substring order
Given two strings, check if substring characters occur in correct order in string.
Input : test_str = 'geeksforgeeks', K = 'sees' Output : True Explanation : "s" after that "ee" and then "s" is present in order in string 1.
Input : test_str = 'geeksforgeeks', K = 'seef' Output : False Explanation : Unordered String.
Method #1 : Using join() + generator expression + in operator
In this, we check we join all the characters which occur in the substring using join(), post that check if substring is present using in operator.
# Python3 code to demonstrate working of
# Test substring order
# Using join() + in operator + generator expression
# initializing string
test_str = 'geeksforgeeks'
# printing original string
print("The original string is : " + str(test_str))
# initializing substring
K = 'seek'
# concatenating required characters
def temp(sub): return ''.join(chr for chr in sub if chr in set(K))
# checking in order
res = K in temp(test_str)
# printing result
print("Is substring in order : " + str(res))
Output
The original string is : geeksforgeeks Is substring in order : True
Time Complexity: O(n)
Space Auxiliary: O(n)
Method #2 : Using all() + next() + generator expression
In this, we get the string with just substring characters using next() and generator expression, to check for order, all() operation is used for each character in substring.
# Python3 code to demonstrate working of
# Test substring order
# Using all() + next() + generator expression
# initializing string
test_str = 'geeksforgeeks'
# printing original string
print("The original string is : " + str(test_str))
# initializing substring
K = 'seek'
# concatenating required characters using next()
# all() used to test order
test_str = iter(test_str)
res = all(next((ele for ele in test_str if ele == chr), None) is not None for chr in K)
# printing result
print("Is substring in order : " + str(res))
Output
The original string is : geeksforgeeks Is substring in order : True
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #3 : Using find() method
# Python3 code to demonstrate working of
# Test substring order
# initializing string
test_str = 'geeksforgeeks'
# printing original string
print("The original string is : " + str(test_str))
# initializing substring
K = 'seek'
ns = ""
for i in test_str:
if i in K:
ns += i
res = False
if(ns.find(K) != -1):
res = True
# printing result
print("Is substring in order : " + str(res))
Output
The original string is : geeksforgeeks Is substring in order : True
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #4 : Using replace() and find() methods
# Python3 code to demonstrate working of
# Test substring order
# initializing string
test_str = 'geeksforgeeks'
# printing original string
print("The original string is : " + str(test_str))
# initializing substring
K = 'seek'
for i in test_str:
if i not in K:
test_str=test_str.replace(i,"")
res=False
if(test_str.find(K)!=-1):
res=True
# printing result
print("Is substring in order : " + str(res))
Output
The original string is : geeksforgeeks Is substring in order : True
Method #5 : Using operator.contains() method
Approach
- Replace all characters of test_str which are not present in K using replace() method
- Now check whether K is present in test_str using operator.contains() method
- Display the return value of operator.contains()
# Python3 code to demonstrate working of
# Test substring order
# initializing string
test_str = 'geeksforgeeks'
# printing original string
print("The original string is : " + str(test_str))
# initializing substring
K = 'seek'
for i in test_str:
if i not in K:
test_str=test_str.replace(i,"")
import operator
res=operator.contains(test_str,K)
# printing result
print("Is substring in order : " + str(res))
Output
The original string is : geeksforgeeks Is substring in order : True
Time Complexity : O(N) N- length of string
Auxiliary Space - O(1) because res is a single variable
Method 6 : using list comprehension and the all() function
# Python3 code to demonstrate working of
# Test substring order
# initializing string
test_str = 'geeksforgeeks'
# printing original string
print("The original string is : " + str(test_str))
# initializing substring
K = 'seek'
# using list comprehension and all()
res = all(K[i] in test_str[test_str.find(K[i-1])+1:] for i in range(1, len(K)))
# printing result
print("Is substring in order : " + str(res))
Output
The original string is : geeksforgeeks Is substring in order : True
The time complexity of this approach is O(n^2), where n is the length of the substring K, because for each character in K, we need to search through the remaining part of test_str to find its position.
The auxiliary space complexity is O(1), because we only use a few extra variables to store the current position in test_str, the previous position found by find(), and the result of the all() function. No additional data structures are created.
