The rarest method accepts a map with string keys and integer values, representing people's names mapped to their ages. It returns the integer value (age) that occurs least frequently in the map. If there is a tie, it returns the younger age. If the map is empty, it throws an exception. The example map shows three people aged 20, two aged 22, and four aged 25. So the method would return 22. If another person aged 22 was added, creating a tie between ages 20 and 22, it would return 20 since that is the younger age.

1. RAREST Method
Write a method rarest that accepts a map whose keys are strings and whose values are integers as
a parameter and returns the integer value that occurs the fewest times in the map. If there is a tie,
return the smaller integer value. If the map is empty, throw an exception.
For example, suppose the map contains mappings from students' names (strings) to their ages
(integers). Your method would return the least frequently occurring age. Consider a map variable
m containing the following key/value pairs:
Three people are age 20 (Jeff, Kasey, and Kim), two people are age 22 (Alyssa and Stef), and
four people are age 25 (Char, Dan, Mogran, and Ryan). So a call of rarest(m) returns 22 because
only two people are that age.
If there is a tie (two or more rarest ages that occur the same number of times), return the
youngest age among them. For example, if we added another pair of Kelly=22 to the map above,
there would now be a tie of three people of age 20 (Jeff, Kasey, Kim) and three people of age 22
(Alyssa, Kelly, Stef). So a call of rarest(m) would now return 20 because 20 is the smaller of the
rarest values.
Solution
Rarest Method:
Algorithm:
Input: Key value pairs as a dictionary
Expected output: Least frequent value from key value pairs
We see that '22' has occurred twice, and '25' four times. Now we have to find it using out
algorithm:
Now, we have the following values in 'min_age' and 'frequency' variables:
min_age = 22
frequency = 2
If we add kelly to the initial key value pair, we will get the min_age as 20 and frequency as 3 as
expected.
The following python code can be referred to understand the algorithm:
def rarest(d):
age_count = {}
for key,value in d.items():

2. if value in age_count:
age_count[value]+=1
else:
age_count[value]=1
min_age=0
freq=0
for key,value in age_count.items():
if freq==0:
freq=value
min_age=key
elif freq>value:
freq=value
min_age=key
elif freq==value:
if min_age>key:
min_age=key
return min_age,freq,frd
d={'Alyssa':22, 'Char':25, 'Dan':25, 'Jeff':20, 'Kasey':20, 'Kim':20, 'Mogran':25,
'Ryan':25, 'Stef':22}
r = rarest(d)
print(r)