Python Program to Find the Most Occurring Character and its Count

Given a string, the task is to find the character that appears the most number of times along with its count. For example:

Input: geeksforgeeks
Output: ('e', 4)

Let’s explore different methods in Python to solve this problem efficiently.

Using collections.Counter()

The Counter() class from the collections module counts how many times each character appears in a string. Then, we use the most_common(1) method to get the character with the highest frequency.

from collections import Counter
s = "geeksforgeeks"
count = Counter(s)
res = count.most_common(1)[0]
print(res)

('e', 4)

Explanation:

• Counter(s) returns a dictionary-like object with characters as keys and their counts as values.
• most_common(1) gives a list of the top 1 most frequent character.
• [0] extracts that tuple from the list.

Using Dictionary and max() Function

This method manually creates a frequency dictionary and finds the character with the maximum count using max() with a key argument.

s = "geeksforgeeks"
freq = {}
for ch in s:
    freq[ch] = freq.get(ch, 0) + 1

res = max(freq, key=freq.get)
print(res, freq[res])

e 4

Explanation:

• freq.get(ch, 0) + 1 increments the count of each character.
• max(freq, key=freq.get) finds the key (character) with the highest value (count).

Using Two Lists

In this method, one list stores unique characters, and the other stores their counts using the built-in count() method.

s = "geeksforgeeks"
a = []
b = []

for ch in s:
    if ch not in a:
        a.append(ch)
        b.append(s.count(ch))

res = max(b)
ele = a[b.index(res)]
print(ele, res)

e 4

Explanation:

• a keeps track of unique characters.
• s.count(ch) returns how many times each character appears.
• The index of the maximum value in b gives the most frequent character from a.

Using operator.countOf() Function

This method uses the operator.countOf() function to count occurrences of each character instead of the string method .count().

import operator
s = "geeksforgeeks"
a = []
b = []

for ch in s:
    if ch not in a:
        a.append(ch)
        b.append(operator.countOf(s, ch))

res = max(b)
ele = a[b.index(res)]
print(ele, res)

e 4

Explanation:

• operator.countOf(s, ch) counts occurrences of ch in s.
• Works similar to count() but comes from the operator module.