Replace missing white spaces in a string with the least frequent character using Pandas (original) (raw)
Last Updated : 17 Apr, 2023
Let’s create a program in python which will replace the white spaces
in a string with the character that occurs in the string very least
using the Pandas library.
Example 1:
String S = "akash loves gfg" here: 'g' comes: 2 times 's' comes: 2 times 'a' comes: 2 times 'h' comes: 1 time 'o' comes: 1 time 'k' comes: 1 time 'v' comes: 1 time 'e' comes: 1 time 'f' comes: 1 time 'l' comes: 1 time
In this example, there are 7 characters with least frequency 1 so, there can be 7 valid outputs One of the possible output is given below: So, the Output String will be: "akashlloveslgfg".
Example 2:
string ="goodd noon" here: g comes: 1 time o comes: 4 times d comes: 2 times n comes: 2 times So the character with the least frequency 1 is g So here white spaces will be replaced by the character g and the output will be: "gooddgnoon"
Now, Let’s see the implementation:
Python3
import
pandas as pd
newstr1
=
'akash loves gfg'
print
(
"Original String given by user:"
,
`` newstr1)
ser
=
pd.Series(
list
(newstr1))
element_freq
=
ser.value_counts()
print
(element_freq)
current_freq
=
element_freq.dropna().index[
-
1
]
result
=
"".join(ser.replace(
' '
,
`` current_freq))
print
(result)
Output:
The overall time complexity is O(n log n).
The auxiliary space complexity is O(n), where n is the length of the input string.
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