Move from brute-force thinking to an efficient approach using dynamic programming strategy.
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 (including the beginning or end) of the string.
Return the maximum number of "LCT" subsequences that can be formed in the resulting string after at most one insertion.
Example 1:
Input: s = "LMCT"
Output: 2
Explanation:
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
Explanation:
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
Explanation:
Since it is not possible to obtain the subsequence "LCT" by inserting a single letter, the result is 0.
Constraints:
1 <= s.length <= 105s consists of uppercase English letters.Problem summary: 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 (including the beginning or end) of the string. Return the maximum number of "LCT" subsequences that can be formed in the resulting string after at most one insertion.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Dynamic Programming · Greedy
"LMCT"
"LCCT"
"L"
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3628: Maximum Number of Subsequences After One Inserting
class Solution {
private char[] s;
public long numOfSubsequences(String S) {
s = S.toCharArray();
int l = 0, r = 0;
for (char c : s) {
if (c == 'T') {
++r;
}
}
long ans = 0, mx = 0;
for (char c : s) {
r -= c == 'T' ? 1 : 0;
if (c == 'C') {
ans += 1L * l * r;
}
l += c == 'L' ? 1 : 0;
mx = Math.max(mx, 1L * l * r);
}
mx = Math.max(mx, Math.max(calc("LC"), calc("CT")));
ans += mx;
return ans;
}
private long calc(String t) {
long cnt = 0;
int a = 0;
for (char c : s) {
if (c == t.charAt(1)) {
cnt += a;
}
a += c == t.charAt(0) ? 1 : 0;
}
return cnt;
}
}
// Accepted solution for LeetCode #3628: Maximum Number of Subsequences After One Inserting
func numOfSubsequences(s string) int64 {
calc := func(t string) int64 {
cnt, a := int64(0), int64(0)
for _, c := range s {
if c == rune(t[1]) {
cnt += a
}
if c == rune(t[0]) {
a++
}
}
return cnt
}
l, r := int64(0), int64(0)
for _, c := range s {
if c == 'T' {
r++
}
}
ans, mx := int64(0), int64(0)
for _, c := range s {
if c == 'T' {
r--
}
if c == 'C' {
ans += l * r
}
if c == 'L' {
l++
}
mx = max(mx, l*r)
}
mx = max(mx, calc("LC"), calc("CT"))
ans += mx
return ans
}
# Accepted solution for LeetCode #3628: Maximum Number of Subsequences After One Inserting
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
// Accepted solution for LeetCode #3628: Maximum Number of Subsequences After One Inserting
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
pub fn rust_example() {
// Port the logic from the reference block below.
}
}
// Reference (java):
// // Accepted solution for LeetCode #3628: Maximum Number of Subsequences After One Inserting
// class Solution {
// private char[] s;
//
// public long numOfSubsequences(String S) {
// s = S.toCharArray();
// int l = 0, r = 0;
// for (char c : s) {
// if (c == 'T') {
// ++r;
// }
// }
// long ans = 0, mx = 0;
// for (char c : s) {
// r -= c == 'T' ? 1 : 0;
// if (c == 'C') {
// ans += 1L * l * r;
// }
// l += c == 'L' ? 1 : 0;
// mx = Math.max(mx, 1L * l * r);
// }
// mx = Math.max(mx, Math.max(calc("LC"), calc("CT")));
// ans += mx;
// return ans;
// }
//
// private long calc(String t) {
// long cnt = 0;
// int a = 0;
// for (char c : s) {
// if (c == t.charAt(1)) {
// cnt += a;
// }
// a += c == t.charAt(0) ? 1 : 0;
// }
// return cnt;
// }
// }
// Accepted solution for LeetCode #3628: Maximum Number of Subsequences After One Inserting
function numOfSubsequences(s: string): number {
const calc = (t: string): number => {
let [cnt, a] = [0, 0];
for (const c of s) {
if (c === t[1]) cnt += a;
if (c === t[0]) a++;
}
return cnt;
};
let [l, r] = [0, 0];
for (const c of s) {
if (c === 'T') r++;
}
let [ans, mx] = [0, 0];
for (const c of s) {
if (c === 'T') r--;
if (c === 'C') ans += l * r;
if (c === 'L') l++;
mx = Math.max(mx, l * r);
}
mx = Math.max(mx, calc('LC'));
mx = Math.max(mx, calc('CT'));
ans += mx;
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.
Wrong move: Locally optimal choices may fail globally.
Usually fails on: Counterexamples appear on crafted input orderings.
Fix: Verify with exchange argument or monotonic objective before committing.