Smallest alphabet with maximum occurences in python string

There is a python string for example “programming”
I have to find the smallest letter with highest occurrences.

That is for given input string = “programming”
the output should be g 2

I tried this , but not able to achieve the solution

str1 = 'programming'


max_freq = {}
for i in str1:
    if i in max_freq:
        max_freq[i] += 1
    else:
        max_freq[i] = 1
res = max(max_freq, key = max_freq.get) 

Any help would be really appreciated

You can use Counter and achieve this.

• Count the frequency of each letter using Counter. This will give you a dict
• Sort the dict first by the values in descending order and then by the keys value in ascending order
• The First item is your answer

from collections import Counter
str1 = 'programming'
d = Counter(str1)

d = sorted(d.items(), key= lambda x: (-x[1], x[0]))
print(d[0])

('g', 2)

For your code to work, replace the last line with this

res = sorted(max_freq.items(), key=lambda x: (-x[1], x[0]))[0]

res will have the smallest letter with maximum occurrences. i.e, ('g', 2)

You are close, you just aren’t getting the max correctly. If all you care about is the number, then you could modify your example slightly:

str1 = 'programming'


max_freq = {}
for i in str1:
    if i in max_freq:
        max_freq[i] += 1
    else:
        max_freq[i] = 1
res = max(max_freq.values())