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# Python counter and dictionary intersection example (Make a string using deletion and rearrangement)

Given two strings, find if we can make first string from second by deleting some characters from second and rearranging remaining characters.

Examples:

```Input : s1 = ABHISHEKsinGH
: s2 = gfhfBHkooIHnfndSHEKsiAnG
Output : Possible

Input : s1 = Hello
: s2 = dnaKfhelddf
Output : Not Possible

Input : s1 = GeeksforGeeks
: s2 = rteksfoGrdsskGeggehes
Output : Possible```

We have existing solution for this problem please refer Make a string from another by deletion and rearrangement of characters link. We will this problem quickly in python. Approach is very simple,

1. Convert both string into dictionary using Counter(iterable) method, each dictionary contains characters within string as Key and their frequencies as Value.
2. Now take intersection of two dictionaries and compare resultant output with dictionary of first string, if both are equal that means it is possible to convert string otherwise not.

Implementation:

## Python3

 `# Python code to find if we can make first string``# from second by deleting some characters from``# second and rearranging remaining characters.``from` `collections ``import` `Counter` `def` `makeString(str1,str2):` `    ``# convert both strings into dictionaries``    ``# output will be like str1="aabbcc",``    ``# dict1={'a':2,'b':2,'c':2}``    ``# str2 = 'abbbcc', dict2={'a':1,'b':3,'c':2}``    ``dict1 ``=` `Counter(str1)``    ``dict2 ``=` `Counter(str2)` `    ``# take intersection of two dictionaries``    ``# output will be result = {'a':1,'b':2,'c':2}``    ``result ``=` `dict1 & dict2` `    ``# compare resultant dictionary with first``    ``# dictionary comparison first compares keys``    ``# and then compares their corresponding values``    ``return` `result ``=``=` `dict1` `# Driver program``if` `__name__ ``=``=` `"__main__"``:``    ``str1 ``=` `'ABHISHEKsinGH'``    ``str2 ``=` `'gfhfBHkooIHnfndSHEKsiAnG'``    ``if` `(makeString(str1,str2)``=``=``True``):``        ``print``(``"Possible"``)``    ``else``:``        ``print``(``"Not Possible"``)`

Output

`Possible`

Time Complexity: O(n)

Auxiliary Space: O(1)