# Check order of character in string using Ordered

Given an input string and a pattern, check if characters in the input string follows the same order as determined by characters present in the pattern. Assume there won’t be any duplicate characters in the pattern.

Examples –

```Input:
original_dict = {'a':1, 'b':2}
item to be inserted ('c', 3)

Output:  {'c':3, 'a':1, 'b':2}

Input:
original_dict = {'akshat':1, 'manjeet':2}
item to be inserted ('nikhil', 3)

```
``````Input:
string = "engineers rock"
pattern = "er";
Output: true
Explanation:
All 'e' in the input string are before all 'r'.

Input:
string = "engineers rock"
pattern = "egr";
Output: false
Explanation:
There are two 'e' after 'g' in the input string.

Input:
string = "engineers rock"
pattern = "gsr";
Output: false
Explanation:
There are one 'r' before 's' in the input string.``````

Recommended: Please try your approach on {IDE} first, before moving on to the solution.
We have existing solution for this problem, please refer Check if string follows order of characters defined by a pattern or not | Set 1. Here we solve this problem quickly in python using OrderedDict(). Approach is very simple,

Create an OrderedDict of input string which contains characters of input strings as Key only.
Now set a pointer at the start of pattern string.
Now traverse generated OrderedDict and match keys with individual character of pattern string, if key and character matches with each other then increment pointer by 1.
If pointer of pattern reaches it’s end that means string follows order of characters defined by a pattern otherwise not.

``````# Function to check if string follows order of
# characters defined by a pattern
from collections import OrderedDict

def checkOrder(input, pattern):

# create empty OrderedDict
# output will be like {'a': None,'b': None, 'c': None}
dict = OrderedDict.fromkeys(input)

# traverse generated OrderedDict parallel with
# pattern string to check if order of characters
# are same or not
ptrlen = 0
for key,value in dict.items():
if (key == pattern[ptrlen]):
ptrlen = ptrlen + 1

# check if we have traverse complete
# pattern string
if (ptrlen == (len(pattern))):
return 'true'

# if we come out from for loop that means
# order was mismatched
return 'false'

# Driver program
if __name__ == "__main__":
input = 'engineers rock'
pattern = 'egr'
print (checkOrder(input,pattern))
``````

Output:

`true`