{ "data": { "question": { "questionId": "3948", "questionFrontendId": "3628", "categoryTitle": "Algorithms", "boundTopicId": 3732707, "title": "Maximum Number of Subsequences After One Inserting", "titleSlug": "maximum-number-of-subsequences-after-one-inserting", "content": "
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.
\n\nReturn the maximum number of "LCT"
subsequences that can be formed in the resulting string after at most one insertion.
\n
Example 1:
\n\nInput: s = "LMCT"
\n\nOutput: 2
\n\nExplanation:
\n\nWe 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:
\n\nInput: s = "LCCT"
\n\nOutput: 4
\n\nExplanation:
\n\nWe 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:
\n\nInput: s = "L"
\n\nOutput: 0
\n\nExplanation:
\n\nSince it is not possible to obtain the subsequence "LCT"
by inserting a single letter, the result is 0.
\n
Constraints:
\n\n1 <= s.length <= 105
s
consists of uppercase English letters.给你一个由大写英文字母组成的字符串 s
。
你可以在字符串的 任意 位置(包括字符串的开头或结尾)最多插入一个 大写英文字母。
\n\n返回在 最多插入一个字母 后,字符串中可以形成的 \"LCT\"
子序列的 最大 数量。
子序列 是从另一个字符串中删除某些字符(可以不删除)且不改变剩余字符顺序后得到的一个 非空 字符串。
\n\n\n\n
示例 1:
\n\n输入: s = \"LMCT\"
\n\n输出: 2
\n\n解释:
\n\n可以在字符串 s
的开头插入一个 \"L\"
,变为 \"LLMCT\"
,其中包含 2 个子序列,分别位于下标 [0, 3, 4] 和 [1, 3, 4]。
示例 2:
\n\n输入: s = \"LCCT\"
\n\n输出: 4
\n\n解释:
\n\n可以在字符串 s
的开头插入一个 \"L\"
,变为 \"LLCCT\"
,其中包含 4 个子序列,分别位于下标 [0, 2, 4]、[0, 3, 4]、[1, 2, 4] 和 [1, 3, 4]。
示例 3:
\n\n输入: s = \"L\"
\n\n输出: 0
\n\n解释:
\n\n插入一个字母无法获得子序列 \"LCT\"
,结果为 0。
\n\n
提示:
\n\n1 <= s.length <= 105
s
仅由大写英文字母组成。preL
, preLC
, sufT
, and sufCT
arrays to count L’s, LC’s, T’s, and CT’s at each position.",
"Compute base
as the sum over all i of preLC[i] * sufT[i]
.",
"For each insert position i, compute gains sufCT[i]
for ‘L’, preL[i] * sufT[i]
for ‘C’, and preLC[i]
for ‘T’, and take the maximum of base
and base + gain
."
],
"solution": null,
"status": null,
"sampleTestCase": "\"LMCT\"",
"metaData": "{\n \"name\": \"numOfSubsequences\",\n \"params\": [\n {\n \"name\": \"s\",\n \"type\": \"string\"\n }\n ],\n \"return\": {\n \"type\": \"long\"\n }\n}",
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