Is Count Substrings With K-Frequency Characters II easy or hard?

Count Substrings With K-Frequency Characters II is categorized as a hard problem because the counting logic inside the sliding window can be tricky. Recognizing when a window becomes valid and how many substrings it contributes requires careful reasoning about frequency updates and pointer movement.

Count Substrings With K-Frequency Characters II Python/Java solution

Most implementations follow the same sliding window pattern regardless of language. Maintain two pointers, update a frequency array or map for each character, and count valid substrings once a character appears k times. The logic translates directly to Python, Java, C++, Go, and TypeScript.

How to solve Count Substrings With K-Frequency Characters II in O(n)?

Traverse the string with a sliding window using two pointers. Maintain a frequency array for characters in the current window. When any character reaches frequency k, count all valid substrings ending at the current index and shrink the window from the left while maintaining the condition. Each index is processed once, giving linear complexity.

What is the best approach for Count Substrings With K-Frequency Characters II?

The most efficient solution uses a sliding window with a frequency counter. Two pointers maintain a dynamic substring while tracking character counts in a hash table or fixed array. When a character reaches frequency k, the window becomes valid and contributes multiple substrings at once. This approach runs in O(n) time and O(1) space.

Is Count Substrings With K-Frequency Characters II asked at Google/Amazon/Meta?

Substring counting and sliding window frequency problems appear frequently in interviews at companies like Google, Amazon, and Meta. Variants that involve counting substrings with specific frequency constraints are common because they test understanding of two‑pointer techniques and hash maps.

What data structure is used in Count Substrings With K-Frequency Characters II?

The main data structure is a frequency map, typically implemented as a fixed-size array or hash table. It tracks how many times each character appears in the current sliding window. Combined with two pointers, this structure allows constant‑time updates while scanning the string.

What is the time complexity of Count Substrings With K-Frequency Characters II?

The optimal sliding window algorithm runs in O(n) time because each character enters and leaves the window at most once. Frequency updates and checks are constant time operations. Space complexity is O(1) since the frequency array stores counts for a fixed alphabet size.

Description Hints 2 Solution Video 0:48

3329. Count Substrings With K-Frequency Characters II

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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.

Example 1:

Input: s = "abacb", k = 2

Output: 4

Explanation:

The valid substrings are:

"aba" (character 'a' appears 2 times).
"abac" (character 'a' appears 2 times).
"abacb" (character 'a' appears 2 times).
"bacb" (character 'b' appears 2 times).

Example 2:

Input: s = "abcde", k = 1

Output: 15

Explanation:

All substrings are valid because every character appears at least once.

Constraints:

1 <= s.length <= 3 * 10⁵
1 <= k <= s.length
s consists only of lowercase English letters.

Topics

Hints open one at a time

Hint 1

If substring [i, j] is valid, then for all k >= j, [i, k] is valid too.

Hint 2

For each i, find j with a sliding window or binary search.

0:48

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Examples

Case 1Case 2

Input

"abacb"
2

Input

"abcde"
1

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