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Replace missing white spaces in a string with the least frequent character using Pandas

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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" 
'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" 
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: 

Now, Let’s see the implementation:


# importing pandas library
import pandas as pd
# taking string with white spaces
newstr1 = 'akash loves gfg'
# printing the original string
print("Original String given by user:",
# converting string into
# list of characters
ser = pd.Series(list(newstr1))
# counting the frequency
# of characters
element_freq = ser.value_counts()
# printing character and their
# respective frequency
current_freq = element_freq.dropna().index[-1]
# function element_freq.dropna()
# will  Return a new Series with
# missing values removed
result = "".join(ser.replace(' ',



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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Last Updated : 17 Apr, 2023
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