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Maximum Number of Subsequences After One Inserting
Maximize the number of "LCT" subsequences in a string after one insertion of an uppercase letter.
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Medium · Maximum Number of Subsequences After One Inserting core interview pattern
Answer-first summary
Maximize the number of "LCT" subsequences in a string after one insertion of an uppercase letter.
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To solve this problem, we need to insert one uppercase letter at any position in the string and maximize the number of "LCT" subsequences. The solution can be efficiently achieved by precomputing arrays such as preL, preLC, sufT, and sufCT to count subsequences from the left and right of each position, which can help determine the optimal insertion point for the highest subsequence count.
Problem Statement
You are given a string s consisting of uppercase English letters. You are allowed to insert at most one uppercase English letter at any position in the string, including the beginning or the end. Your task is to find the maximum number of "LCT" subsequences that can be formed after performing this insertion.
A subsequence is a sequence derived by deleting some or no elements from a string without changing the order of the remaining elements. In this problem, you are interested in the subsequences that form the sequence "LCT". You need to determine the best place to insert the letter to maximize the count of such subsequences.
Examples
Example 1
Input: s = "LMCT"
Output: 2
We can insert a "L" at the beginning of the string s to make "LLMCT" , which has 2 subsequences, at indices [0, 3, 4] and [1, 3, 4].
Example 2
Input: s = "LCCT"
Output: 4
We can insert a "L" at the beginning of the string s to make "LLCCT" , which has 4 subsequences, at indices [0, 2, 4], [0, 3, 4], [1, 2, 4] and [1, 3, 4].
Example 3
Input: s = "L"
Output: 0
Since it is not possible to obtain the subsequence "LCT" by inserting a single letter, the result is 0.
Constraints
- 1 <= s.length <= 105
- s consists of uppercase English letters.
Solution Approach
Precompute Key Arrays
First, precompute arrays to track the number of 'L', 'LC', 'T', and 'CT' subsequences at each position. These arrays, namely preL, preLC, sufT, and sufCT, will allow us to count subsequences efficiently from the left and right of each insertion point.
Evaluate Insertion Points
Next, simulate the insertion of a letter at each possible position in the string and compute the resulting subsequence count. For each position, use the precomputed arrays to determine how many subsequences of "LCT" can be formed by adding the letter.
Return Maximum Result
Finally, return the maximum number of "LCT" subsequences formed after inserting a letter at any position. This will be the optimal solution.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
The time complexity depends on the method used for precomputing the arrays and evaluating each insertion point. Precomputing the arrays takes linear time, O(n), and evaluating each insertion also takes O(n), so the overall time complexity is O(n). The space complexity is O(n) due to the storage of precomputed arrays.
What Interviewers Usually Probe
- The candidate uses precomputation effectively to reduce the complexity of subsequence counting.
- The solution demonstrates efficient handling of string manipulation and counting.
- The candidate optimizes the problem using dynamic programming or a similar approach to minimize redundant calculations.
Common Pitfalls or Variants
Common pitfalls
- Failing to precompute the necessary arrays efficiently may lead to a brute-force solution with higher time complexity.
- Incorrect handling of subsequence counting, especially at the insertion points, can result in incorrect answers.
- Overlooking edge cases like strings that already contain no valid subsequences or strings where inserting a letter doesn't change the result.
Follow-up variants
- Consider the problem where more than one insertion is allowed, which would require modifying the approach to account for multiple insertion points.
- Modify the problem to include case-insensitive subsequences, where the letter insertion might be any alphabet character.
- Allow insertions at positions that may also change the string's original order, introducing more complexity in subsequence counting.
FAQ
What is the core pattern of the "Maximum Number of Subsequences After One Inserting" problem?
The core pattern involves optimizing subsequence counting through precomputation of helper arrays and determining the optimal insertion point for maximizing subsequences of "LCT".
How can I improve my approach to handling large strings in this problem?
To handle large strings efficiently, focus on precomputing subsequences and avoid recalculating values unnecessarily. Use dynamic programming or similar techniques to reduce time complexity.
What is the significance of precomputing arrays in this problem?
Precomputing arrays like preL, preLC, sufT, and sufCT allows you to quickly calculate subsequences from both the left and right sides of the string without redundant operations, speeding up the solution.
Are there any edge cases I should be aware of for this problem?
Yes, edge cases include strings that already have no valid subsequences, and strings where inserting a letter does not increase the number of subsequences, such as when there is no 'L' or 'C' in the string.
How does GhostInterview help with this problem's interview preparation?
GhostInterview helps by breaking down the problem into digestible steps, providing optimized solutions, and guiding users through common pitfalls and efficient algorithmic strategies for the maximum subsequence problem.
Solution
Solution 1: Enumeration
We can first calculate the number of "LCT" subsequences in the original string, then consider the case of inserting one letter.
class Solution:
def numOfSubsequences(self, s: str) -> int:
def calc(t: str) -> int:
cnt = a = 0
for c in s:
if c == t[1]:
cnt += a
a += int(c == t[0])
return cnt
l, r = 0, s.count("T")
ans = mx = 0
for c in s:
r -= int(c == "T")
if c == "C":
ans += l * r
l += int(c == "L")
mx = max(mx, l * r)
mx = max(mx, calc("LC"), calc("CT"))
ans += mx
return ans