Minimum Penalty for a Shop

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 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 s) {
      
        s.push_back('N');
        vector<int>yes;
        vector<int>no;
        int size = s.length();
        int sum_yes =0;
        int sum_no =0;
        for(int i  =0 ; i < size ; i++)
        {
            if(s[i] == 'Y')
            {
                sum_yes += 1;
            }
            else
            {
                sum_no += 1;
            }
            yes.push_back(sum_yes);
            no.push_back(sum_no);
        }
        int palenty = yes[size -1];
        int index = 0;
        for(int  i =1 ; i < size ; i++)
        {
            int temp_panelty = yes[size-1] - yes[i-1]+no[i-1];
            if(temp_panelty < palenty)
            { 
                palenty = temp_panelty;
                index = i;
            }
        }
        return index;
      
    }
};


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