Given a string s and an integer k, return the total number of substrings of s where at least one character appears at least k times.
Note: A substring is a contiguous sequence of characters within a string. For example, "edu" is a substring of "educative".
Constraints:
s.length
k s.length
s consists only of lowercase English letters.
A brute-force approach to this problem would involve checking all possible substrings and counting character frequencies to see if any character appears exactly k times. However, this becomes inefficient for longer strings due to the large number of substrings and repeated computations.
To avoid this, a sliding window technique can be used. The idea is to move through the string with two pointers that define a window, while maintaining a frequency count of characters within it. The right pointer extends the window forward one character at a time, while the left pointer adjusts the starting point of the window when necessary. As the window expands, a frequency table keeps track of how many times each character appears inside the current window. Alongside this, a separate counter tracks how many characters have reached a frequency of exactly k.
When a character in the current sliding window reaches a ...
Given a string s and an integer k, return the total number of substrings of s where at least one character appears at least k times.
Note: A substring is a contiguous sequence of characters within a string. For example, "edu" is a substring of "educative".
Constraints:
s.length
k s.length
s consists only of lowercase English letters.
A brute-force approach to this problem would involve checking all possible substrings and counting character frequencies to see if any character appears exactly k times. However, this becomes inefficient for longer strings due to the large number of substrings and repeated computations.
To avoid this, a sliding window technique can be used. The idea is to move through the string with two pointers that define a window, while maintaining a frequency count of characters within it. The right pointer extends the window forward one character at a time, while the left pointer adjusts the starting point of the window when necessary. As the window expands, a frequency table keeps track of how many times each character appears inside the current window. Alongside this, a separate counter tracks how many characters have reached a frequency of exactly k.
When a character in the current sliding window reaches a ...