mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-11 02:58:13 +08:00
174 lines
18 KiB
JSON
174 lines
18 KiB
JSON
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"questionId": "1000221",
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"questionFrontendId": "LCP 33",
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"categoryTitle": "Algorithms",
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"boundTopicId": 711347,
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"title": "蓄水",
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"titleSlug": "o8SXZn",
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"content": "",
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"translatedTitle": "蓄水",
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"translatedContent": "给定 N 个无限容量且初始均空的水缸，每个水缸配有一个水桶用来打水，第 `i` 个水缸配备的水桶容量记作 `bucket[i]`。小扣有以下两种操作：\n- 升级水桶：选择任意一个水桶，使其容量增加为 `bucket[i]+1`\n- 蓄水：将全部水桶接满水，倒入各自对应的水缸\n\n每个水缸对应最低蓄水量记作 `vat[i]`，返回小扣至少需要多少次操作可以完成所有水缸蓄水要求。\n\n注意：实际蓄水量 **达到或超过** 最低蓄水量，即完成蓄水要求。\n\n**示例 1：**\n>输入：`bucket = [1,3], vat = [6,8]`\n>\n>输出：`4`\n>\n>解释：\n>第 1 次操作升级 bucket[0]；\n>第 2 ~ 4 次操作均选择蓄水，即可完成蓄水要求。\n\n\n\n\n**示例 2：**\n>输入：`bucket = [9,0,1], vat = [0,2,2]`\n>\n>输出：`3`\n>\n>解释：\n>第 1 次操作均选择升级 bucket[1]\n>第 2~3 次操作选择蓄水，即可完成蓄水要求。\n\n**提示：**\n- `1 <= bucket.length == vat.length <= 100`\n- `0 <= bucket[i], vat[i] <= 10^4`\n",
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"difficulty": "Easy",
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"code": "class Solution {\npublic:\n int storeWater(vector<int>& bucket, vector<int>& vat) {\n\n }\n};",
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"code": "class Solution(object):\n def storeWater(self, bucket, vat):\n \"\"\"\n :type bucket: List[int]\n :type vat: List[int]\n :rtype: int\n \"\"\"",
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"code": "# @param {Integer[]} bucket\n# @param {Integer[]} vat\n# @return {Integer}\ndef store_water(bucket, vat)\n\nend",
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