Python: Count number of substrings from a given string of lowercase alphabets with exactly k distinct characters
Python String: Exercise-76 with Solution
Write a Python program to count number of substrings from a given string of lowercase alphabets with exactly k distinct (given) characters.
Sample Solution:-
Python Code:
def count_k_dist(str1, k):
str_len = len(str1)
result = 0
ctr = [0] * 27
for i in range(0, str_len):
dist_ctr = 0
ctr = [0] * 27
for j in range(i, str_len):
if(ctr[ord(str1[j]) - 97] == 0):
dist_ctr += 1
ctr[ord(str1[j]) - 97] += 1
if(dist_ctr == k):
result += 1
if(dist_ctr > k):
break
return result
str1 = input("Input a string (lowercase alphabets):")
k = int(input("Input k: "))
print("Number of substrings with exactly", k, "distinct characters : ", end = "")
print(count_k_dist(str1, k))
Sample Output:
Input a string (lowercase alphabets): wolf Input k: 4 Number of substrings with exactly 4 distinct characters : 1
Flowchart:

Visualize Python code execution:
The following tool visualize what the computer is doing step-by-step as it executes the said program:
Python Code Editor:
Have another way to solve this solution? Contribute your code (and comments) through Disqus.
Previous: Write a Python program to find smallest window that contains all characters of a given string.
Next: Write a Python program to count number of non-empty substrings of a given string.
What is the difficulty level of this exercise?
Test your Programming skills with w3resource's quiz.
Python: Tips of the Day
Check if a given key already exists in a dictionary:
In is the intended way to test for the existence of a key in a dict.
d = {"key1": 10, "key2": 23} if "key1" in d: print("this will execute") if "nonexistent key" in d: print("this will not")
If you wanted a default, you can always use dict.get():
d = dict() for i in range(100): key = i % 10 d[key] = d.get(key, 0) + 1
and if you wanted to always ensure a default value for any key you can either use dict.setdefault() repeatedly or defaultdict from the collections module, like so:
from collections import defaultdict d = defaultdict(int) for i in range(100): d[i % 10] += 1
but in general, the in keyword is the best way to do it.
Ref: https://bit.ly/2XPMRyz
- New Content published on w3resource:
- HTML-CSS Practical: Exercises, Practice, Solution
- Java Regular Expression: Exercises, Practice, Solution
- Scala Programming Exercises, Practice, Solution
- Python Itertools exercises
- Python Numpy exercises
- Python GeoPy Package exercises
- Python Pandas exercises
- Python nltk exercises
- Python BeautifulSoup exercises
- Form Template
- Composer - PHP Package Manager
- PHPUnit - PHP Testing
- Laravel - PHP Framework
- Angular - JavaScript Framework
- Vue - JavaScript Framework
- Jest - JavaScript Testing Framework