Python | Check order of character in string using OrderedDict( )

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 wonâ€™t be any duplicate characters in the pattern. Examples:

`Input: string = "engineers rock"pattern = "er";Output: trueExplanation: All 'e' in the input string are before all 'r'.Input: string = "engineers rock"pattern = "gsr";Output: falseExplanation:There are one 'r' before 's' in the input string.`

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.

Python3

 `# 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 ``=` `'er'``    ``print` `(checkOrder(``input``,pattern)) `

Output:

`true`

Check order of character in string Using two pointers

This Approach checks if the characters in the given pattern appear in the same order in the given string. It uses two pointers to keep track of the positions of the characters being compared. If all the characters in the pattern are found in the same order in the string, the function returns True, otherwise False.

Algorithm

1. Initialize two pointers, one for the string and one for the pattern, both starting at 0.
2. Traverse through the string character by character.
3. If the current character in the string matches the current character in the pattern, increment the pattern pointer.
4. Increment the string pointer after processing each character.
5. If the pattern pointer is equal to the length of the pattern, return True.
6. If the end of the string is reached and the pattern pointer is less than the length of the pattern, return False.

Python3

 `def` `check_order(string, pattern):``    ``i, j ``=` `0``, ``0``    ``for` `char ``in` `string:``        ``if` `char ``=``=` `pattern[j]:``            ``j ``+``=` `1``        ``if` `j ``=``=` `len``(pattern):``            ``return` `True``        ``i ``+``=` `1` `    ``return` `False``string ``=` `'engineers rock'``pattern ``=` `'er'``print``(check_order(string, pattern))`

Output
```True
```

Time Complexity: O(n), where n is length of string
Space Complexity: O(1)

Previous
Next