How do you solve Minimum Penalty for a Shop on LeetCode?

Minimum Penalty for a Shop (LeetCode #2483) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding how to calculate penalties efficiently is crucial.

What is the brute force solution for Minimum Penalty for a Shop?

The brute force approach for Minimum Penalty for a Shop is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach calculates the penalty for each possible closing hour by iterating through the customer log. For each closing hour, we count how many hours the shop was open with no customers and how many hours it was closed with customers, summing these to get the penalty. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Penalty for a Shop?

Minimum Penalty for a Shop uses the following concepts: String, Prefix Sum. The recommended patterns to study are: Prefix Sum, Suffix Sum.

What are the interview tips for Minimum Penalty for a Shop?

Always explain your thought process while coding. Consider edge cases before finalizing your solution. Practice writing both brute-force and optimal solutions to strengthen your understanding.

What are common mistakes when solving Minimum Penalty for a Shop?

Not considering the edge case when all customers are 'N'. Failing to correctly update the minimum penalty and closing hour.

#2483

Minimum Penalty for a Shop

Medium

StringPrefix SumPrefix SumSuffix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses prefix sums to efficiently calculate penalties without nested loops. We can precompute how many 'N's are before any hour and how many 'Y's are after it, allowing us to compute the penalty in constant time for each closing hour.

⚙️

Algorithm

5 steps

1Step 1: Create a prefix sum array to count 'N's up to each hour.
2Step 2: Create a suffix sum array to count 'Y's from each hour to the end.
3Step 3: For each hour, calculate the penalty using the prefix and suffix sums.
4Step 4: Track the minimum penalty and the corresponding hour.
5Step 5: Return the hour with the minimum penalty.

solution.py16 lines

1def bestClosingTime(customers):
2    n = len(customers)
3    prefixN = [0] * (n + 1)
4    suffixY = [0] * (n + 1)
5    for i in range(n):
6        prefixN[i + 1] = prefixN[i] + (1 if customers[i] == 'N' else 0)
7    for i in range(n - 1, -1, -1):
8        suffixY[i] = suffixY[i + 1] + (1 if customers[i] == 'Y' else 0)
9    min_penalty = float('inf')
10    best_hour = 0
11    for j in range(n + 1):
12        penalty = prefixN[j] + suffixY[j]
13        if penalty < min_penalty:
14            min_penalty = penalty
15            best_hour = j
16    return best_hour

ℹ

Complexity note: This complexity is due to the linear scans needed to create the prefix and suffix arrays, allowing us to compute penalties in constant time thereafter.

1Understanding how to calculate penalties efficiently is crucial.
2Using prefix and suffix sums can significantly reduce time complexity.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.