Welcome to Subscribe On Youtube

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 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 (Java, C++, Python)

• Java
• C++
• Python
• Go
• RenderScript

• 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
  }
  

• impl Solution {
      #[allow(dead_code)]
      pub fn best_closing_time(customers: String) -> i32 {
          let n = customers.len();
          let mut penalty = i32::MAX;
          let mut ret = -1;
          let mut prefix_sum = vec![0; n + 1];
  
          // Initialize the vector
          for (i, c) in customers.chars().enumerate() {
              prefix_sum[i + 1] = prefix_sum[i] + if c == 'Y' { 1 } else { 0 };
          }
  
          // Calculate the answer
          for i in 0..=n {
              if penalty > (prefix_sum[n] - prefix_sum[i]) as i32 + (i - prefix_sum[i]) as i32 {
                  penalty = (prefix_sum[n] - prefix_sum[i]) as i32 + (i - prefix_sum[i]) as i32;
                  ret = i as i32;
              }
          }
  
          ret
      }
  }
  

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).

All Problems

All Solutions