mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-10 18:48:13 +08:00
199 lines
21 KiB
JSON
199 lines
21 KiB
JSON
{
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"questionId": "1000484",
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"questionFrontendId": "LCP 69",
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"title": "Hello LeetCode!",
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"titleSlug": "rMeRt2",
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"content": "English description is not available for the problem. Please switch to Chinese.",
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"translatedTitle": "Hello LeetCode!",
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"translatedContent": "力扣嘉年华同样准备了纪念品展位，参观者只需要集齐 `helloleetcode` 的 `13` 张字母卡片即可获得力扣纪念章。\n\n在展位上有一些由字母卡片拼成的单词，`words[i][j]` 表示第 `i` 个单词的第 `j` 个字母。\n\n你可以从这些单词中取出一些卡片，但每次拿取卡片都需要消耗游戏代币，规则如下：\n\n- 从一个单词中取一个字母所需要的代币数量，为该字母左边和右边字母数量之积\n\n- 可以从一个单词中多次取字母，每个字母仅可被取一次\n\n> 例如：从 `example` 中取出字母 `a`，需要消耗代币 `2*4=8`，字母取出后单词变为 `exmple`；\n再从中取出字母 `m`，需要消耗代币 `2*3=6`，字母取出后单词变为 `exple`；\n\n请返回取得 `helloleetcode` 这些字母需要消耗代币的 **最少** 数量。如果无法取得，返回 `-1`。\n\n**注意：**\n- 取出字母的顺序没有要求\n- 取出的所有字母恰好可以拼成 `helloleetcode` \n\n**示例 1：**\n>输入：`words = [\"hold\",\"engineer\",\"cost\",\"level\"]`\n>\n>输出：`5`\n>\n>解释：最优方法为：\n>从 `hold` 依次取出 `h`、`o`、`l`、`d`， 代价均为 `0`\n>从 `engineer` 依次取出第 `1` 个 `e` 与最后一个 `e`， 代价为 `0` 和 `5*1=5`\n>从 `cost` 取出 `c`、`o`、`t`， 代价均为 `0`\n>从 `level` 依次取出 `l`、`l`、`e`、`e`， 代价均为 `0`\n>所有字母恰好可以拼成 `helloleetcode`，因此最小的代价为 `5`\n\n**示例 2：**\n>输入：`words = [\"hello\",\"leetcode\"]`\n>\n>输出：`0`\n\n**提示：**\n+ `n == words.length`\n+ `m == words[i].length`\n+ `1 <= n <= 24`\n+ `1 <= m <= 8`\n+ `words[i][j]` 仅为小写字母",
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