Suppose we have a number k and another string s, we have to find the size of the longest substring that contains at most k distinct characters.
So, if the input is like k = 3 s = "kolkata", then the output will be 4, as there are two longest substrings with 3 distinct characters these are "kolk" and "kata", which have length 4.
To solve this, we will follow these steps −
ans := 0, left := 0
table := a new map
for right is in range 0 to size of s − 1, do
table[s[right]] := 1 + (s[right] if exists otherwise 0)
if size of table <= k, then
ans := maximum of ans and (right − left + 1)
otherwise,
while size of table > k, do
left_char := s[left]
if table[left_char] is same as 1, then
delete left_char−th element from table
otherwise,
table[left_char] := table[left_char] − 1
left := left + 1
return ans
Let us see the following implementation to get better understanding −
Example
class Solution:
def solve(self, k, s):
ans = 0
left = 0
table = {}
for right in range(0, len(s)):
table[s[right]] = table.get(s[right], 0) + 1
if len(table) <= k:
ans = max(ans, right − left + 1)
else:
while len(table) > k:
left_char = s[left]
if table[left_char] == 1:
table.pop(left_char)
else:
table[left_char] −= 1
left += 1
return ans
ob = Solution()
k = 3
s = "kolkata"
print(ob.solve(k, s))Input
"anewcoffeepot"
Output
4