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: 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.
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
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