Solve sliding-window and disjoint-string-pairs tasks
NetflixSoftware Engineer · Technical ScreenLast updated
This question tests the ability to combine sliding window techniques with bit manipulation or frequency counting to identify pairs of substrings that share no common characters.
Problem Statement
Given a string s, find the maximum product of the lengths of two disjoint substrings s[i...j] and s[k...m] such that they share no common characters. Two substrings are disjoint if their index ranges do not overlap. If it is impossible to find such a pair, return 0.
Function Signature
def max_product_disjoint_substrings(s: str) -> int:
Constraints
2 <= len(s) <= 100- The string contains only lowercase English letters.
Examples
Example 1:
Input: s = "abcde"
Output: 4
Explanation: We can pick "ab" (length 2) and "cd" (length 2). Their product is 4. They are disjoint and share no common characters.
Example 2:
Input: s = "aaaaa"
Output: 0
Explanation: Any two disjoint substrings will contain the character 'a', so no pair can satisfy the condition.
Considerations
- Focus on efficiently checking the 'no common characters' condition.
- Ensure the index ranges
[i, j]and[k, m]are strictly separated (i.e.,j < korm < i).
Hints
Hint 1
Represent the set of characters present in any substring using a bitmask of length 26.
Hint 2
Precompute all possible substrings and their associated bitmasks and lengths, storing them in a list or hash map.
Hint 3
Iterate through all pairs of stored substrings, checking if their bitmasks have a bitwise AND equal to zero and ensuring their indices do not overlap.
Follow-up questions
- How would your approach change if the string length was increased to 10,000?
- Can you optimize the storage if many substrings share the same bitmask?
- How would you modify the algorithm if we wanted to find the maximum product of three disjoint substrings with no common characters?
Solution walkthrough is for members
Get the full walkthrough with reference code and complexity analysis, plus Kitty Coach at 300 messages a day. Hints, the code runner and community answers stay free.