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# Using Counter() in Python to find minimum character removal to make two strings anagram

• Difficulty Level : Easy
• Last Updated : 23 Nov, 2020

Given two strings in lowercase, the task is to make them Anagram. The only allowed operation is to remove a character from any string. Find minimum number of characters to be deleted to make both the strings anagram?
If two strings contains same data set in any order then strings are called Anagrams.

Examples:

```Input : str1 = "bcadeh" str2 = "hea"
Output: 3
We need to remove b, c and d from str1.

Input : str1 = "cddgk" str2 = "gcd"
Output: 2

Input : str1 = "bca" str2 = "acb"
Output: 0
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

This problem has existing solution please refer Remove minimum number of characters so that two strings become anagram link. We will solve this problem in python quickly using Counter() and Dictionary Data Structure and intersection property of Set data structure. Approach is very simple,

1. Convert each string into a dictionary data structure using Counter(iterable) method.
2. Count number of keys in both dictionaries ( count1, count2) and count number of keys common in both dictionaries.
3. If no common keys found that means we need to remove count1 + count2 characters from both the strings.
4. Else (max(count1, count2) – countCommon) will be the number of characters to be removed
 `# Function remove minimum number of characters so that ``# two strings become anagram ``from` `collections ``import` `Counter ``def` `removeChars(str1, str2): `` ` `    ``# make dictionaries from both strings ``    ``dict1 ``=` `Counter(str1) ``    ``dict2 ``=` `Counter(str2) `` ` `    ``# extract keys from dict1 and dict2 ``    ``keys1 ``=` `dict1.keys() ``    ``keys2 ``=` `dict2.keys() `` ` `    ``# count number of keys in both lists of keys ``    ``count1 ``=` `len``(keys1) ``    ``count2 ``=` `len``(keys2) `` ` `    ``# convert list of keys in set to find common keys ``    ``set1 ``=` `set``(keys1) ``    ``commonKeys ``=` `len``(set1.intersection(keys2)) `` ` `    ``if` `(commonKeys ``=``=` `0``): ``        ``return` `count1 ``+` `count2 ``    ``else``: ``        ``return` `(``max``(count1, count2)``-``commonKeys) `` ` `# Driver program ``if` `__name__ ``=``=` `"__main__"``: ``    ``str1 ``=``'bcadeh'``    ``str2 ``=``'hea'``    ``print` `(removeChars(str1, str2)) `

Output:

```3
```

Alternate Solution :

1. Convert each string into a dictionary data structure using Counter(iterable) method.
2. Find the common elements from both dictonary
3. Add up the values from common dictionary in order to get the total number of common elements.
 `# Function remove minimum number of characters so that ``# two strings become anagram ``from` `collections ``import` `Counter ``def` `removeChars(a, b): `` ` `    ``# make dictionaries from both strings ``    ``c1 ``=` `Counter(a) ``    ``c2 ``=` `Counter(b) ``     ` `    ``# finding the common elements from both dictonary ``    ``common ``=` `c1&c2 ``    ``value ``=` `0`` ` `    ``# adding up the key from common dictionary in order ``    ``# to get the total number of common elements ``    ``for` `key ``in` `common: ``        ``value ``=` `value ``+` `common[key] ``         ` `    ``# returning the number of elements to be ``    ``# removed to form an anagram ``    ``return` `(``len``(a)``-``2``*``value``+` `len``(b))         `` ` `# Driver program ``if` `__name__ ``=``=` `"__main__"``: ``    ``str1 ``=``'bcadeh'``    ``str2 ``=``'hea'``    ``print` `(removeChars(str1, str2)) `

Output:

```3
```

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