Leetcode #2067: Number of Equal Count Substrings
In this guide, we solve Leetcode #2067 Number of Equal Count Substrings in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a 0-indexed string s consisting of only lowercase English letters, and an integer count. A substring of s is said to be an equal count substring if, for each unique letter in the substring, it appears exactly count times in the substring.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Hash Table, String, Counting, Sliding Window
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: s = "aaabcbbcc", count = 3
Output: 3
Explanation:
The substring that starts at index 0 and ends at index 2 is "aaa".
The letter 'a' in the substring appears exactly 3 times.
The substring that starts at index 3 and ends at index 8 is "bcbbcc".
The letters 'b' and 'c' in the substring appear exactly 3 times.
The substring that starts at index 0 and ends at index 8 is "aaabcbbcc".
The letters 'a', 'b', and 'c' in the substring appear exactly 3 times.
Python Solution
class Solution:
def equalCountSubstrings(self, s: str, count: int) -> int:
ans = 0
for i in range(1, 27):
k = i * count
if k > len(s):
break
cnt = Counter()
t = 0
for j, c in enumerate(s):
cnt[c] += 1
t += cnt[c] == count
t -= cnt[c] == count + 1
if j >= k:
cnt[s[j - k]] -= 1
t += cnt[s[j - k]] == count
t -= cnt[s[j - k]] == count - 1
ans += i == t
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.