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Formatted question description: https://leetcode.ca/all/2483.html
2483. Minimum Penalty for a Shop
 Difficulty: Medium.
 Related Topics: String, Prefix Sum.
 Similar Questions: Grid Game.
Problem
You are given the customer visit log of a shop represented by a 0indexed string customers
consisting only of characters 'N'
and 'Y'
:

if the
ith
character is'Y'
, it means that customers come at theith
hour 
whereas
'N'
indicates that no customers come at theith
hour.
If the shop closes at the jth
hour (0 <= j <= n
), the penalty is calculated as follows:

For every hour when the shop is open and no customers come, the penalty increases by
1
. 
For every hour when the shop is closed and customers come, the penalty increases by
1
.
Return** the earliest hour at which the shop must be closed to incur a minimum penalty.**
Note that if a shop closes at the jth
hour, it means the shop is closed at the hour j
.
Example 1:
Input: customers = "YYNY"
Output: 2
Explanation:
 Closing the shop at the 0th hour incurs in 1+1+0+1 = 3 penalty.
 Closing the shop at the 1st hour incurs in 0+1+0+1 = 2 penalty.
 Closing the shop at the 2nd hour incurs in 0+0+0+1 = 1 penalty.
 Closing the shop at the 3rd hour incurs in 0+0+1+1 = 2 penalty.
 Closing the shop at the 4th hour incurs in 0+0+1+0 = 1 penalty.
Closing the shop at 2nd or 4th hour gives a minimum penalty. Since 2 is earlier, the optimal closing time is 2.
Example 2:
Input: customers = "NNNNN"
Output: 0
Explanation: It is best to close the shop at the 0th hour as no customers arrive.
Example 3:
Input: customers = "YYYY"
Output: 4
Explanation: It is best to close the shop at the 4th hour as customers arrive at each hour.
Constraints:

1 <= customers.length <= 105

customers
consists only of characters'Y'
and'N'
.
Solution (Java, C++, Python)

class Solution { public int bestClosingTime(String customers) { int n = customers.length(); int[] s = new int[n + 1]; for (int i = 0; i < n; ++i) { s[i + 1] = s[i] + (customers.charAt(i) == 'Y' ? 1 : 0); } int ans = 0, cost = 1 << 30; for (int j = 0; j <= n; ++j) { int t = j  s[j] + s[n]  s[j]; if (cost > t) { ans = j; cost = t; } } return ans; } }

class Solution { public: int bestClosingTime(string customers) { int n = customers.size(); vector<int> s(n + 1); for (int i = 0; i < n; ++i) { s[i + 1] = s[i] + (customers[i] == 'Y'); } int ans = 0, cost = 1 << 30; for (int j = 0; j <= n; ++j) { int t = j  s[j] + s[n]  s[j]; if (cost > t) { ans = j; cost = t; } } return ans; } };

class Solution: def bestClosingTime(self, customers: str) > int: n = len(customers) s = [0] * (n + 1) for i, c in enumerate(customers): s[i + 1] = s[i] + int(c == 'Y') ans, cost = 0, inf for j in range(n + 1): t = j  s[j] + s[1]  s[j] if cost > t: ans, cost = j, t return ans

func bestClosingTime(customers string) (ans int) { n := len(customers) s := make([]int, n+1) for i, c := range customers { s[i+1] = s[i] if c == 'Y' { s[i+1]++ } } cost := 1 << 30 for j := 0; j <= n; j++ { t := j  s[j] + s[n]  s[j] if cost > t { ans, cost = j, t } } return }
Explain:
nope.
Complexity:
 Time complexity : O(n).
 Space complexity : O(n).