LeetCode 2483 店铺最小罚款（Minimum Penalty for a Shop）：前缀后缀统计与四类扫描解法精讲-北京尧图网络

LeetCode 2483 店铺最小罚款Minimum Penalty for a Shop前缀后缀统计与四类扫描解法精讲【免费下载链接】leetcodeLeetcode solutions项目地址: https://gitcode.com/GitHub_Trending/leetcode1/leetcode本指南围绕 LeetCode 2483「店铺最小罚款」Minimum Penalty for a Shop展开系统讲解如何根据店铺的顾客到访日志Y/N 字符串求出使罚款最小的打烊时刻。文章覆盖暴力枚举、前缀和与后缀和、两遍扫描、单遍扫描四类解法的完整推导、多语言实现与复杂度分析并结合本仓库 cpp/2483-minimum-penalty-for-a-shop.cpp、java/2483-minimum-penalty-for-a-shop.java、kotlin/2483-minimum-penalty-for-a-shop.kt、python/2483-minimum-penalty-for-a-shop.py 的源码实现进行印证。读完本文你将掌握将字符串计数问题转化为前缀/后缀累加问题、再进一步压缩为常数空间的完整套路。问题定义与罚则语义题目给出一条0索引字符串customers只包含字符Y与N第i个字符为Y表示第i小时有顾客到店第i个字符为N表示第i小时没有顾客到店。若店铺在第j小时0 j nn为字符串长度打烊罚款计算规则如下营业但无人到店店铺营业的每个小时里若没有顾客来罚款1打烊但顾客到店店铺打烊的每个小时里若有顾客来罚款1。要求返回罚款最小时最早的打烊时刻。仓库 C 源码 cpp/2483-minimum-penalty-for-a-shop.cpp 的头部注释完整描述了上述规则并给出了标准示例Input: customers YYNY Output: 2逐步核算来自 cpp/2483-minimum-penalty-for-a-shop.cpp第0小时打烊1101 3全程打烊错过 3 个 Y第1小时打烊0101 2第2小时打烊0001 1第3小时打烊0011 2第4小时打烊0010 1。第2与第4小时均取得最小罚款1按题意返回更早的时刻2。关键语义店铺在第j小时打烊表示它在第j小时起关闭——即营业时段为0到j-1关闭时段为j到n-1。这是所有解法构造罚则的基础一旦混淆极易产生 off-by-one 错误详见文末「常见陷阱」。在 README.md 的题解完成表中该题已收录 Python、Java、Kotlin、C 等多语言实现本文下述四种解法与其逐一对应。前置知识前缀和Prefix Sum预先统计每个位置之前的字符计数使任意前缀区间的计数查询降到 O(1)后缀和Suffix Sum预先统计从每个位置起往后的字符计数与前缀和对称字符串顺序迭代单次从左到右扫描同时维护运行中的累加值贪心式最优追踪遍历所有候选打烊时刻时只保留当前见过的最小罚款及其时刻。1. 暴力枚举Brute Force思路店铺可以在0到n含之间的任意时刻打烊。若在第i小时打烊则位置0到i-1中的每个N计1分营业却无人位置i到n-1中的每个Y计1分打烊却有人。枚举全部n1个候选打烊时刻取罚款最小者即可。算法步骤初始化res与minPenalty为n最坏情况下罚款不会超过n对每个候选打烊时刻i0到n统计位置0到i-1中N的个数统计位置i到n-1中Y的个数两者之和即为该时刻的罚款若小于minPenalty则更新minPenalty与res返回res。多语言实现class Solution: def bestClosingTime(self, customers: str) - int: n len(customers) res n minPenalty n for i in range(n 1): penalty 0 for j in range(i): if customers[j] N: penalty 1 for j in range(i, n): if customers[j] Y: penalty 1 if penalty minPenalty: minPenalty penalty res i return respublic class Solution { public int bestClosingTime(String customers) { int n customers.length(); int res n, minPenalty n; for (int i 0; i n; i) { int penalty 0; for (int j 0; j i; j) { if (customers.charAt(j) N) { penalty; } } for (int j i; j n; j) { if (customers.charAt(j) Y) { penalty; } } if (penalty minPenalty) { minPenalty penalty; res i; } } return res; } }class Solution { public: int bestClosingTime(string customers) { int n customers.size(); int res n, minPenalty n; for (int i 0; i n; i) { int penalty 0; for (int j 0; j i; j) { if (customers[j] N) { penalty; } } for (int j i; j n; j) { if (customers[j] Y) { penalty; } } if (penalty minPenalty) { minPenalty penalty; res i; } } return res; } };class Solution { /** * param {string} customers * return {number} */ bestClosingTime(customers) { const n customers.length; let res n, minPenalty n; for (let i 0; i n; i) { let penalty 0; for (let j 0; j i; j) { if (customers[j] N) { penalty; } } for (let j i; j n; j) { if (customers[j] Y) { penalty; } } if (penalty minPenalty) { minPenalty penalty; res i; } } return res; } }public class Solution { public int BestClosingTime(string customers) { int n customers.Length; int res n, minPenalty n; for (int i 0; i n; i) { int penalty 0; for (int j 0; j i; j) { if (customers[j] N) { penalty; } } for (int j i; j n; j) { if (customers[j] Y) { penalty; } } if (penalty minPenalty) { minPenalty penalty; res i; } } return res; } }func bestClosingTime(customers string) int { n : len(customers) res, minPenalty : n, n for i : 0; i n; i { penalty : 0 for j : 0; j i; j { if customers[j] N { penalty } } for j : i; j n; j { if customers[j] Y { penalty } } if penalty minPenalty { minPenalty penalty res i } } return res }class Solution { fun bestClosingTime(customers: String): Int { val n customers.length var res n var minPenalty n for (i in 0..n) { var penalty 0 for (j in 0 until i) { if (customers[j] N) { penalty } } for (j in i until n) { if (customers[j] Y) { penalty } } if (penalty minPenalty) { minPenalty penalty res i } } return res } }class Solution { func bestClosingTime(_ customers: String) - Int { let n customers.count let chars Array(customers) var res n var minPenalty n for i in 0...n { var penalty 0 for j in 0..i { if chars[j] N { penalty 1 } } for j in i..n { if chars[j] Y { penalty 1 } } if penalty minPenalty { minPenalty penalty res i } } return res } }impl Solution { pub fn best_closing_time(customers: String) - i32 { let s customers.as_bytes(); let n s.len(); let mut res n as i32; let mut min_penalty n as i32; for i in 0..n { let mut penalty 0; for j in 0..i { if s[j] bN { penalty 1; } } for j in i..n { if s[j] bY { penalty 1; } } if penalty min_penalty { min_penalty penalty; res i as i32; } } res } }复杂度时间复杂度$O(n^2)$——每个候选时刻i都要重新扫描两侧区间空间复杂度$O(1)$——只使用常数个变量。该解法思路直观、不易出错适合作为正确性基准但性能不足以应对大输入。2. 前缀和与后缀和Prefix Suffix思路暴力解在每个i处重复计数造成平方复杂度。改进办法预先算出每个位置之前的N个数前缀以及从每个位置往后的Y个数后缀则任意打烊时刻的罚款就是两个预计算值的和查询降到 O(1)。prefixN[i] 位置0到i-1中N的个数suffixY[i] 位置i到n-1中Y的个数。算法步骤构建prefixN从左到右扫描先记录当前累计N数遇到N再自增最后补上末尾一项数组长度为n1构建suffixY从右到左扫描先继承右侧累计值遇到Y再自增对每个打烊时刻i0到n罚款 prefixN[i] suffixY[i]追踪最小罚款及其对应时刻返回最小罚款对应的时刻。仓库中的 Java 实现 java/2483-minimum-penalty-for-a-shop.java 正是这一思路先构造pre_n前缀N计数与post_y后缀Y计数再以pre_n[i] post_y[i]扫描取最小Kotlin 实现 kotlin/2483-minimum-penalty-for-a-shop.kt 同样使用prefix与postfix两个数组完成。多语言实现class Solution: def bestClosingTime(self, customers: str) - int: n len(customers) cnt 0 prefixN [] for c in customers: prefixN.append(cnt) if c N: cnt 1 prefixN.append(cnt) suffixY [0] * (n 1) for i in range(n - 1, -1, -1): suffixY[i] suffixY[i 1] if customers[i] Y: suffixY[i] 1 res n minPenalty n for i in range(n 1): penalty prefixN[i] suffixY[i] if penalty minPenalty: minPenalty penalty res i return respublic class Solution { public int bestClosingTime(String customers) { int n customers.length(); int cnt 0; int[] prefixN new int[n 1]; for (int i 0; i n; i) { prefixN[i] cnt; if (customers.charAt(i) N) { cnt; } } prefixN[n] cnt; int[] suffixY new int[n 1]; for (int i n - 1; i 0; i--) { suffixY[i] suffixY[i 1]; if (customers.charAt(i) Y) { suffixY[i]; } } int res n, minPenalty n; for (int i 0; i n; i) { int penalty prefixN[i] suffixY[i]; if (penalty minPenalty) { minPenalty penalty; res i; } } return res; } }class Solution { public: int bestClosingTime(string customers) { int n customers.size(), cnt 0; vectorint prefixN(n 1); for (int i 0; i n; i) { prefixN[i] cnt; if (customers[i] N) { cnt; } } prefixN[n] cnt; vectorint suffixY(n 1, 0); for (int i n - 1; i 0; i--) { suffixY[i] suffixY[i 1]; if (customers[i] Y) { suffixY[i]; } } int res n, minPenalty n; for (int i 0; i n; i) { int penalty prefixN[i] suffixY[i]; if (penalty minPenalty) { minPenalty penalty; res i; } } return res; } };class Solution { /** * param {string} customers * return {number} */ bestClosingTime(customers) { const n customers.length; let cnt 0; const prefixN []; for (const c of customers) { prefixN.push(cnt); if (c N) { cnt; } } prefixN.push(cnt); const suffixY new Array(n 1).fill(0); for (let i n - 1; i 0; i--) { suffixY[i] suffixY[i 1]; if (customers[i] Y) { suffixY[i]; } } let res n, minPenalty n; for (let i 0; i n; i) { const penalty prefixN[i] suffixY[i]; if (penalty minPenalty) { minPenalty penalty; res i; } } return res; } }public class Solution { public int BestClosingTime(string customers) { int n customers.Length; int cnt 0; int[] prefixN new int[n 1]; for (int i 0; i n; i) { prefixN[i] cnt; if (customers[i] N) { cnt; } } prefixN[n] cnt; int[] suffixY new int[n 1]; for (int i n - 1; i 0; i--) { suffixY[i] suffixY[i 1]; if (customers[i] Y) { suffixY[i]; } } int res n, minPenalty n; for (int i 0; i n; i) { int penalty prefixN[i] suffixY[i]; if (penalty minPenalty) { minPenalty penalty; res i; } } return res; } }func bestClosingTime(customers string) int { n : len(customers) cnt : 0 prefixN : make([]int, n1) for i : 0; i n; i { prefixN[i] cnt if customers[i] N { cnt } } prefixN[n] cnt suffixY : make([]int, n1) for i : n - 1; i 0; i-- { suffixY[i] suffixY[i1] if customers[i] Y { suffixY[i] } } res, minPenalty : n, n for i : 0; i n; i { penalty : prefixN[i] suffixY[i] if penalty minPenalty { minPenalty penalty res i } } return res }class Solution { fun bestClosingTime(customers: String): Int { val n customers.length var cnt 0 val prefixN IntArray(n 1) for (i in 0 until n) { prefixN[i] cnt if (customers[i] N) { cnt } } prefixN[n] cnt val suffixY IntArray(n 1) for (i in n - 1 downTo 0) { suffixY[i] suffixY[i 1] if (customers[i] Y) { suffixY[i] } } var res n var minPenalty n for (i in 0..n) { val penalty prefixN[i] suffixY[i] if (penalty minPenalty) { minPenalty penalty res i } } return res } }class Solution { func bestClosingTime(_ customers: String) - Int { let n customers.count let chars Array(customers) var cnt 0 var prefixN Int for i in 0..n { prefixN[i] cnt if chars[i] N { cnt 1 } } prefixN[n] cnt var suffixY Int for i in stride(from: n - 1, through: 0, by: -1) { suffixY[i] suffixY[i 1] if chars[i] Y { suffixY[i] 1 } } var res n var minPenalty n for i in 0...n { let penalty prefixN[i] suffixY[i] if penalty minPenalty { minPenalty penalty res i } } return res } }impl Solution { pub fn best_closing_time(customers: String) - i32 { let s customers.as_bytes(); let n s.len(); let mut cnt 0i32; let mut prefix_n vec![0i32; n 1]; for i in 0..n { prefix_n[i] cnt; if s[i] bN { cnt 1; } } prefix_n[n] cnt; let mut suffix_y vec![0i32; n 1]; for i in (0..n).rev() { suffix_y[i] suffix_y[i 1]; if s[i] bY { suffix_y[i] 1; } } let mut res n as i32; let mut min_penalty n as i32; for i in 0..n { let penalty prefix_n[i] suffix_y[i]; if penalty min_penalty { min_penalty penalty; res i as i32; } } res } }复杂度时间复杂度$O(n)$——两轮预计算各 O(n)一轮扫描 O(n)空间复杂度$O(n)$——两个长度为n1的辅助数组。这是「以空间换时间」的典型模式也是许多区间查询问题的通用前置步骤。3. 两遍扫描Iteration, Two Pass思路前缀/后缀解法的两个辅助数组本质上是冗余的我们完全可以在一遍计数一遍扫描中动态维护左右两侧的罚款贡献从而把空间压到 O(1)。先统计全部Y的个数cntY。若第0小时就打烊店铺全程关闭会错过所有顾客此时罚款恰为cntY。随后从左到右推进打烊时刻每越过一个Y说明这个顾客在打烊前被服务到罚项减少1cntY--每越过一个N说明这段时间营业却无人罚项增加1cntN。扫描过程中任意时刻的罚款恒等于cntN cntY只需记录其最小值。算法步骤统计cntY为Y总数初始化minPenalty cntY、res 0、cntN 0遍历下标i处的字符若是YcntY--少一个被错过的顾客若是NcntN多一个白营业的小时计算当前罚款cntN cntY若小于minPenalty更新minPenalty并把res置为i 1返回res。多语言实现class Solution: def bestClosingTime(self, customers: str) - int: cntY sum(c Y for c in customers) minPenalty cntY res cntN 0 for i, c in enumerate(customers): if c Y: cntY - 1 else: cntN 1 penalty cntN cntY if penalty minPenalty: res i 1 minPenalty penalty return respublic class Solution { public int bestClosingTime(String customers) { int cntY 0; for (char c : customers.toCharArray()) { if (c Y) cntY; } int minPenalty cntY, res 0, cntN 0; for (int i 0; i customers.length(); i) { if (customers.charAt(i) Y) { cntY--; } else { cntN; } int penalty cntN cntY; if (penalty minPenalty) { res i 1; minPenalty penalty; } } return res; } }class Solution { public: int bestClosingTime(string customers) { int cntY count(customers.begin(), customers.end(), Y); int minPenalty cntY, res 0, cntN 0; for (int i 0; i customers.size(); i) { if (customers[i] Y) { cntY--; } else { cntN; } int penalty cntN cntY; if (penalty minPenalty) { res i 1; minPenalty penalty; } } return res; } };class Solution { /** * param {string} customers * return {number} */ bestClosingTime(customers) { let cntY 0; for (let c of customers) { if (c Y) cntY; } let minPenalty cntY, res 0, cntN 0; for (let i 0; i customers.length; i) { if (customers[i] Y) { cntY--; } else { cntN; } const penalty cntN cntY; if (penalty minPenalty) { res i 1; minPenalty penalty; } } return res; } }public class Solution { public int BestClosingTime(string customers) { int cntY 0; foreach (char c in customers) { if (c Y) cntY; } int minPenalty cntY, res 0, cntN 0; for (int i 0; i customers.Length; i) { if (customers[i] Y) { cntY--; } else { cntN; } int penalty cntN cntY; if (penalty minPenalty) { res i 1; minPenalty penalty; } } return res; } }func bestClosingTime(customers string) int { cntY : 0 for _, c : range customers { if c Y { cntY } } minPenalty, res, cntN : cntY, 0, 0 for i : 0; i len(customers); i { if customers[i] Y { cntY-- } else { cntN } penalty : cntN cntY if penalty minPenalty { res i 1 minPenalty penalty } } return res }class Solution { fun bestClosingTime(customers: String): Int { var cntY customers.count { it Y } var minPenalty cntY var res 0 var cntN 0 for (i in customers.indices) { if (customers[i] Y) { cntY-- } else { cntN } val penalty cntN cntY if (penalty minPenalty) { res i 1 minPenalty penalty } } return res } }class Solution { func bestClosingTime(_ customers: String) - Int { let chars Array(customers) var cntY chars.filter { $0 Y }.count var minPenalty cntY var res 0 var cntN 0 for i in 0..chars.count { if chars[i] Y { cntY - 1 } else { cntN 1 } let penalty cntN cntY if penalty minPenalty { res i 1 minPenalty penalty } } return res } }impl Solution { pub fn best_closing_time(customers: String) - i32 { let s customers.as_bytes(); let mut cnt_y s.iter().filter(|c| c bY).count() as i32; let mut min_penalty cnt_y; let mut res 0i32; let mut cnt_n 0i32; for i in 0..s.len() { if s[i] bY { cnt_y - 1; } else { cnt_n 1; } let penalty cnt_n cnt_y; if penalty min_penalty { res i as i32 1; min_penalty penalty; } } res } }复杂度时间复杂度$O(n)$——统计一遍、扫描一遍各 O(n)空间复杂度$O(1)$——仅使用三个计数器。仓库中的 Python 实现 python/2483-minimum-penalty-for-a-shop.py 采用了与之同构但符号方向相反的变体以Y减分、N加分维护curPenalty并在curPenalty创新低时记录i1同样只需 O(1) 空间。4. 单遍扫描Iteration, One Pass思路两遍扫描还能再进一步不直接维护绝对罚款而是维护一个相对分值。把Y视为1营业带来的收益N视为-1营业付出的成本从左到右累加。直觉上累计分值最高的点意味着「营业带来的净收益最大」因此最优打烊时刻就是该最高点之后的那个小时。数学上等价于若在时刻i打烊罚款 i之前N数i之后Y数。令T为总Y数则罚款 i前N数 T - i前Y数 T (i 前 N 数 - i 前 Y 数)。括号内正是以N-1, Y1计分时的前缀和取负。因此最小化罚款等价于最大化该前缀和最优时刻落在前缀和首次达到最大值的边界处。仓库中 C 实现 cpp/2483-minimum-penalty-for-a-shop.cpp 正是这种写法Y使pen自增、N使pen自减当pen刷新最大值时记录i最终返回res即最高点后的小时。Kotlin 文件 kotlin/2483-minimum-penalty-for-a-shop.kt 也给出了同样的「Kadane 风格」单遍版本。算法步骤初始化res 0、minPenalty 0、penalty 0遍历下标i处的字符若是Ypenalty 1否则penalty - 1若penalty minPenalty更新minPenalty penalty并令res i 1返回res。注意更新条件是严格大于且res记录的是累计分值创新高之后的位置即i1初始时刻0penalty 0天然是候选无需特判。多语言实现class Solution: def bestClosingTime(self, customers: str) - int: res minPenalty 0 penalty 0 for i, c in enumerate(customers): penalty 1 if c Y else -1 if penalty minPenalty: minPenalty penalty res i 1 return respublic class Solution { public int bestClosingTime(String customers) { int res 0, minPenalty 0, penalty 0; for (int i 0; i customers.length(); i) { penalty customers.charAt(i) Y ? 1 : -1; if (penalty minPenalty) { minPenalty penalty; res i 1; } } return res; } }class Solution { public: int bestClosingTime(string customers) { int res 0, minPenalty 0, penalty 0; for (int i 0; i customers.size(); i) { penalty customers[i] Y ? 1 : -1; if (penalty minPenalty) { minPenalty penalty; res i 1; } } return res; } };class Solution { /** * param {string} customers * return {number} */ bestClosingTime(customers) { let res 0, minPenalty 0, penalty 0; for (let i 0; i customers.length; i) { penalty customers[i] Y ? 1 : -1; if (penalty minPenalty) { minPenalty penalty; res i 1; } } return res; } }public class Solution { public int BestClosingTime(string customers) { int res 0, minPenalty 0, penalty 0; for (int i 0; i customers.Length; i) { penalty customers[i] Y ? 1 : -1; if (penalty minPenalty) { minPenalty penalty; res i 1; } } return res; } }func bestClosingTime(customers string) int { res, minPenalty, penalty : 0, 0, 0 for i : 0; i len(customers); i { if customers[i] Y { penalty } else { penalty-- } if penalty minPenalty { minPenalty penalty res i 1 } } return res }class Solution { fun bestClosingTime(customers: String): Int { var res 0 var minPenalty 0 var penalty 0 for (i in customers.indices) { penalty if (customers[i] Y) 1 else -1 if (penalty minPenalty) { minPenalty penalty res i 1 } } return res } }class Solution { func bestClosingTime(_ customers: String) - Int { let chars Array(customers) var res 0 var minPenalty 0 var penalty 0 for i in 0..chars.count { penalty chars[i] Y ? 1 : -1 if penalty minPenalty { minPenalty penalty res i 1 } } return res } }impl Solution { pub fn best_closing_time(customers: String) - i32 { let s customers.as_bytes(); let mut res 0i32; let mut min_penalty 0i32; let mut penalty 0i32; for i in 0..s.len() { penalty if s[i] bY { 1 } else { -1 }; if penalty min_penalty { min_penalty penalty; res i as i32 1; } } res } }复杂度时间复杂度$O(n)$——仅一轮扫描空间复杂度$O(1)$——只使用三个变量。这是本题的最优解一次遍历、常数空间、代码最短适合作为面试中的最终答案。四解法复杂度对比解法思路时间复杂度空间复杂度适用场景1. 暴力枚举每个候选时刻重扫两侧区间$O(n^2)$$O(1)$小规模输入、正确性基准2. 前缀和后缀和预计算prefixN与suffixY$O(n)$$O(n)$需要理解前缀/后缀思想的入门题3. 两遍扫描计数cntY后单次推进维护cntN cntY$O(n)$$O(1)$常规最优解直观易写4. 单遍扫描Y1, N-1维护相对分值最高点$O(n)$$O(1)$面试展示代码最精简四种解法最终都返回罚款最小时最早的打烊时刻输出一致区别只在于计算罚款的方式与时空开销。常见陷阱1. 混淆打烊时刻的语义店铺在第i小时打烊表示营业时段为0到i-1、关闭时段为i到n-1。很多实现错误地把它理解成「第i小时仍在营业」导致罚则计算整体错位产生 off-by-one 错误。建议先在纸上用YYNY逐步核算见文首示例再编码。2. 漏掉第0小时或第n小时合法打烊时刻覆盖0全天不营业到n全天营业。若循环范围写成1..n-1之类的区间就会漏掉「最优策略是干脆不开门」或「干脆开满全天」这两种情况得到错误答案。三种线性解法均通过循环i从0到n含规避此问题。3. 平局时返回了错误的索引当多个打烊时刻罚款相同且都是最小时题目要求返回最早的一个。实现上若用而非更新结果或扫描顺序不正确就会把答案覆盖成更晚的时刻。因此务必使用严格小于才更新res保证只保留首次遇到的最优解。仓库源码印证本仓库为本题提供了多语言、多思路的实现可在 README.md 的完成状态表中确认收录情况具体包括cpp/2483-minimum-penalty-for-a-shop.cpp头部注释完整给出题目罚则与YYNY手算示例正文采用解法 4单遍扫描O(N)时间、O(1)空间java/2483-minimum-penalty-for-a-shop.java采用解法 2前缀pre_n 后缀post_ykotlin/2483-minimum-penalty-for-a-shop.kt同时给出解法 2前缀/后缀数组与解法 4Kadane 风格单遍扫描两版实现python/2483-minimum-penalty-for-a-shop.py单遍扫描的符号反转变体以curPenalty创新低处记录i1。对照阅读可以发现四种解法的核心差异仅在于「如何得到某个时刻的罚款」暴力重扫、前缀后缀查表、两计数器动态维护、单变量相对分值。掌握这条从 O(n²) 到 O(n) 再到 O(1) 空间的优化主线就理解了本题的全部考查点也能将其迁移到其他「前缀/后缀统计最优分割点」类问题如分割数组、平衡括号区间等中。【免费下载链接】leetcodeLeetcode solutions项目地址: https://gitcode.com/GitHub_Trending/leetcode1/leetcode创作声明：本文部分内容由AI辅助生成（AIGC），仅供参考

LeetCode 2483 店铺最小罚款（Minimum Penalty for a Shop）：前缀后缀统计与四类扫描解法精讲

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