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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:

Flowchart: Count number of substrings from a given string of lowercase alphabets with exactly k distinct characters

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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