LeetCode #2483 — MEDIUM

Minimum Penalty for a Shop

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

The Problem

Problem Statement

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 i^th character is 'Y', it means that customers come at the i^th hour
whereas 'N' indicates that no customers come at the i^th hour.

If the shop closes at the j^th 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 j^th hour, it means the shop is closed at the hour j.

Example 1:

Input: customers = "YYNY"
Output: 2
Explanation: 
- Closing the shop at the 0^th hour incurs in 1+1+0+1 = 3 penalty.
- Closing the shop at the 1^st hour incurs in 0+1+0+1 = 2 penalty.
- Closing the shop at the 2^nd hour incurs in 0+0+0+1 = 1 penalty.
- Closing the shop at the 3^rd hour incurs in 0+0+1+1 = 2 penalty.
- Closing the shop at the 4^th hour incurs in 0+0+1+0 = 1 penalty.
Closing the shop at 2^nd or 4^th 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 0^th hour as no customers arrive.

Example 3:

Input: customers = "YYYY"
Output: 4
Explanation: It is best to close the shop at the 4^th hour as customers arrive at each hour.

Constraints:

1 <= customers.length <= 10⁵
customers consists only of characters 'Y' and 'N'.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

"YYNY"

Example 2

"NNNNN"

Example 3

"YYYY"

Core Insight

What unlocks the optimal approach

At any index, the penalty is the sum of prefix count of ‘N’ and suffix count of ‘Y’.
Enumerate all indices and find the minimum such value.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2483: Minimum Penalty for a Shop
class Solution {
    public int bestClosingTime(String customers) {
        int n = customers.length();
        int ans = 0, cost = 0;
        for (int i = 0; i < n; i++) {
            if (customers.charAt(i) == 'Y') {
                cost++;
            }
        }
        int mn = cost;
        for (int j = 1; j <= n; j++) {
            cost += customers.charAt(j - 1) == 'N' ? 1 : -1;
            if (cost < mn) {
                ans = j;
                mn = cost;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2483: Minimum Penalty for a Shop
func bestClosingTime(customers string) int {
	ans := 0
	cost := strings.Count(customers, "Y")
	mn := cost
	for j := 1; j <= len(customers); j++ {
		c := customers[j-1]
		if c == 'N' {
			cost++
		} else {
			cost--
		}
		if cost < mn {
			ans = j
			mn = cost
		}
	}
	return ans
}

# Accepted solution for LeetCode #2483: Minimum Penalty for a Shop
class Solution:
    def bestClosingTime(self, customers: str) -> int:
        ans = 0
        mn = cost = customers.count("Y")
        for j, c in enumerate(customers, 1):
            cost += 1 if c == "N" else -1
            if cost < mn:
                ans, mn = j, cost
        return ans

// Accepted solution for LeetCode #2483: Minimum Penalty for a Shop
impl Solution {
    pub fn best_closing_time(customers: String) -> i32 {
        let bytes = customers.as_bytes();

        let mut cost: i32 = bytes.iter().filter(|&&c| c == b'Y').count() as i32;
        let mut mn = cost;
        let mut ans: i32 = 0;

        for j in 1..=bytes.len() {
            let c = bytes[j - 1];
            if c == b'N' {
                cost += 1;
            } else {
                cost -= 1;
            }
            if cost < mn {
                mn = cost;
                ans = j as i32;
            }
        }
        ans
    }
}

// Accepted solution for LeetCode #2483: Minimum Penalty for a Shop
function bestClosingTime(customers: string): number {
    let ans = 0;
    let cost = 0;
    for (const ch of customers) {
        if (ch === 'Y') {
            cost++;
        }
    }
    let mn = cost;

    for (let j = 1; j <= customers.length; j++) {
        const c = customers[j - 1];
        cost += c === 'N' ? 1 : -1;
        if (cost < mn) {
            mn = cost;
            ans = j;
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.