# Python Dictionary to find mirror characters in a string

Given a string and a number N, we need to mirror the characters from the N-th position up to the length of the string in alphabetical order. In mirror operation, we change â€˜aâ€™ to â€˜zâ€™, â€˜bâ€™ to â€˜yâ€™, and so on.

Examples:

```Input : N = 3
Output : paizwlc
We mirror characters from position 3 to end.

Input : N = 6
pneumonia
Output : pneumlmrz```

We have an existing solution for this problem please refer to Mirror characters of a string link. We can solve this problem in Python using Dictionary Data Structure. The mirror value of ‘a’ is ‘z’,’b’ is ‘y’, etc, so we create a dictionary data structure and one-to-one map reverse sequence of alphabets onto the original sequence of alphabets. Now traverse characters from length k in given string and change characters into their mirror value using a dictionary.

Implementation:

## Python3

 `# function to mirror characters of a string`   `def` `mirrorChars(``input``,k):`   `    ``# create dictionary` `    ``original ``=` `'abcdefghijklmnopqrstuvwxyz'` `    ``reverse ``=` `'zyxwvutsrqponmlkjihgfedcba'` `    ``dictChars ``=` `dict``(``zip``(original,reverse))`   `    ``# separate out string after length k to change` `    ``# characters in mirror` `    ``prefix ``=` `input``[``0``:k``-``1``]` `    ``suffix ``=` `input``[k``-``1``:]` `    ``mirror ``=` `''`   `    ``# change into mirror` `    ``for` `i ``in` `range``(``0``,``len``(suffix)):` `         ``mirror ``=` `mirror ``+` `dictChars[suffix[i]]`   `    ``# concat prefix and mirrored part` `    ``print` `(prefix``+``mirror)` `         `  `# Driver program` `if` `__name__ ``=``=` `"__main__"``:` `    ``input` `=` `'paradox'` `    ``k ``=` `3` `    ``mirrorChars(``input``,k)`

Output

`paizwlc`

Time complexity: O(n)
Auxiliary Space: O(n)

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