Today I have done leetcode's August LeetCoding Challenge with cpp.

August LeetCoding Challenge 29

Description

Minimum Penalty for a Shop

You are given the customer visit log of a shop represented by a 0-indexed string customers consisting only of characters 'N' and 'Y':

• if the ith character is 'Y', it means that customers come at the ith hour
• whereas 'N' indicates that no customers come at the ith 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

class Solution {
public:
  int bestClosingTime(string customers) {
    int Y = 0;
    for(auto c : customers) {
      Y += 'Y' == c;
    }

    int answer = 0;
    int result = Y;
    int currentY = 0;
    int currentN = 0;
    for(int i = 0; i < customers.length(); ++i) {
      currentY += customers[i] == 'Y';
      currentN += customers[i] == 'N';
      if(result > (Y - currentY) + currentN) {
        result = Y - currentY + currentN;
        answer = i + 1;
      }
    }
    return answer;
  }
};

// Accepted
// 42/42 cases passed (11 ms)
// Your runtime beats 100 % of cpp submissions
// Your memory usage beats 86.38 % of cpp submissions (10.6 MB)