{ "data": { "question": { "questionId": "2599", "questionFrontendId": "2516", "categoryTitle": "Algorithms", "boundTopicId": 2029759, "title": "Take K of Each Character From Left and Right", "titleSlug": "take-k-of-each-character-from-left-and-right", "content": "
You are given a string s
consisting of the characters 'a'
, 'b'
, and 'c'
and a non-negative integer k
. Each minute, you may take either the leftmost character of s
, or the rightmost character of s
.
Return the minimum number of minutes needed for you to take at least k
of each character, or return -1
if it is not possible to take k
of each character.
\n
Example 1:
\n\n\nInput: s = "aabaaaacaabc", k = 2\nOutput: 8\nExplanation: \nTake three characters from the left of s. You now have two 'a' characters, and one 'b' character.\nTake five characters from the right of s. You now have four 'a' characters, two 'b' characters, and two 'c' characters.\nA total of 3 + 5 = 8 minutes is needed.\nIt can be proven that 8 is the minimum number of minutes needed.\n\n\n
Example 2:
\n\n\nInput: s = "a", k = 1\nOutput: -1\nExplanation: It is not possible to take one 'b' or 'c' so return -1.\n\n\n
\n
Constraints:
\n\n1 <= s.length <= 105
s
consists of only the letters 'a'
, 'b'
, and 'c'
.0 <= k <= s.length
给你一个由字符 'a'
、'b'
、'c'
组成的字符串 s
和一个非负整数 k
。每分钟,你可以选择取走 s
最左侧 还是 最右侧 的那个字符。
你必须取走每种字符 至少 k
个,返回需要的 最少 分钟数;如果无法取到,则返回 -1
。
\n\n
示例 1:
\n\n\n输入:s = \"aabaaaacaabc\", k = 2\n输出:8\n解释:\n从 s 的左侧取三个字符,现在共取到两个字符 'a' 、一个字符 'b' 。\n从 s 的右侧取五个字符,现在共取到四个字符 'a' 、两个字符 'b' 和两个字符 'c' 。\n共需要 3 + 5 = 8 分钟。\n可以证明需要的最少分钟数是 8 。\n\n\n
示例 2:
\n\n\n输入:s = \"a\", k = 1\n输出:-1\n解释:无法取到一个字符 'b' 或者 'c',所以返回 -1 。\n\n\n
\n\n
提示:
\n\n1 <= s.length <= 105
s
仅由字母 'a'
、'b'
、'c'
组成0 <= k <= s.length
\\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\"],\"java\":[\"Java\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u88ab\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5305\\u542b Pair \\u7c7b: https:\\/\\/docs.oracle.com\\/javase\\/8\\/javafx\\/api\\/javafx\\/util\\/Pair.html <\\/p>\"],\"python\":[\"Python\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982\\uff1aarray<\\/a>, bisect<\\/a>, 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out-of-bounds<\\/code>\\u548c
use-after-free<\\/code>\\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\n
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Python 2.7.12<\\/code><\\/p>\\r\\n\\r\\n
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out-of-bounds<\\/code>\\u548c
use-after-free<\\/code>\\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\n
\\r\\nstruct hash_entry {\\r\\n int id; \\/* we'll use this field as the key *\\/\\r\\n char name[10];\\r\\n UT_hash_handle hh; \\/* makes this structure hashable *\\/\\r\\n};\\r\\n\\r\\nstruct hash_entry *users = NULL;\\r\\n\\r\\nvoid add_user(struct hash_entry *s) {\\r\\n HASH_ADD_INT(users, id, s);\\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\\r\\n\\r\\n
\\r\\nstruct hash_entry *find_user(int user_id) {\\r\\n struct hash_entry *s;\\r\\n HASH_FIND_INT(users, &user_id, s);\\r\\n return s;\\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\\r\\n\\r\\n
\\r\\nvoid delete_user(struct hash_entry *user) {\\r\\n HASH_DEL(users, user); \\r\\n}\\r\\n<\\/pre>\\r\\n<\\/p>\"],\"csharp\":[\"C#\",\"
Node.js 16.13.2<\\/code><\\/p>\\r\\n\\r\\n
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Ruby 3.1<\\/code>\\u6267\\u884c<\\/p>\\r\\n\\r\\n
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Go 1.21<\\/code><\\/p>\\r\\n\\r\\n
Python 3.10<\\/code><\\/p>\\r\\n\\r\\n
Scala 2.13<\\/code><\\/p>\"],\"kotlin\":[\"Kotlin\",\"
Kotlin 1.9.0<\\/code><\\/p>\\r\\n\\r\\n
rust 1.58.1<\\/code><\\/p>\\r\\n\\r\\n
PHP 8.1<\\/code>.<\\/p>\\r\\n\\r\\n