LeetCode #1052 — MEDIUM

Grumpy Bookstore Owner

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

There is a bookstore owner that has a store open for n minutes. You are given an integer array customers of length n where customers[i] is the number of the customers that enter the store at the start of the i^th minute and all those customers leave after the end of that minute.

During certain minutes, the bookstore owner is grumpy. You are given a binary array grumpy where grumpy[i] is 1 if the bookstore owner is grumpy during the i^th minute, and is 0 otherwise.

When the bookstore owner is grumpy, the customers entering during that minute are not satisfied. Otherwise, they are satisfied.

The bookstore owner knows a secret technique to remain not grumpy for minutes consecutive minutes, but this technique can only be used once.

Return the maximum number of customers that can be satisfied throughout the day.

Example 1:

Input: customers = [1,0,1,2,1,1,7,5], grumpy = [0,1,0,1,0,1,0,1], minutes = 3

Output: 16

Explanation:

The bookstore owner keeps themselves not grumpy for the last 3 minutes.

The maximum number of customers that can be satisfied = 1 + 1 + 1 + 1 + 7 + 5 = 16.

Example 2:

Input: customers = [1], grumpy = [0], minutes = 1

Output: 1

Constraints:

n == customers.length == grumpy.length
1 <= minutes <= n <= 2 * 10⁴
0 <= customers[i] <= 1000
grumpy[i] is either 0 or 1.

Step 01

Brute Force Baseline

Problem summary: There is a bookstore owner that has a store open for n minutes. You are given an integer array customers of length n where customers[i] is the number of the customers that enter the store at the start of the ith minute and all those customers leave after the end of that minute. During certain minutes, the bookstore owner is grumpy. You are given a binary array grumpy where grumpy[i] is 1 if the bookstore owner is grumpy during the ith minute, and is 0 otherwise. When the bookstore owner is grumpy, the customers entering during that minute are not satisfied. Otherwise, they are satisfied. The bookstore owner knows a secret technique to remain not grumpy for minutes consecutive minutes, but this technique can only be used once. Return the maximum number of customers that can be satisfied throughout the day.

Baseline thinking

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

Pattern signal: Array · Sliding Window

Example 1

[1,0,1,2,1,1,7,5]
[0,1,0,1,0,1,0,1]
3

Example 2

[1]
[0]
1

Step 02

Core Insight

What unlocks the optimal approach

Say the store owner uses their power in minute 1 to X and we have some answer A. If they instead use their power from minute 2 to X+1, we only have to use data from minutes 1, 2, X and X+1 to update our answer A.

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 #1052: Grumpy Bookstore Owner
class Solution {
    public int maxSatisfied(int[] customers, int[] grumpy, int minutes) {
        int cnt = 0;
        int tot = 0;
        for (int i = 0; i < minutes; ++i) {
            cnt += customers[i] * grumpy[i];
            tot += customers[i] * (grumpy[i] ^ 1);
        }
        int mx = cnt;
        int n = customers.length;
        for (int i = minutes; i < n; ++i) {
            cnt += customers[i] * grumpy[i];
            cnt -= customers[i - minutes] * grumpy[i - minutes];
            mx = Math.max(mx, cnt);
            tot += customers[i] * (grumpy[i] ^ 1);
        }
        return tot + mx;
    }
}

// Accepted solution for LeetCode #1052: Grumpy Bookstore Owner
func maxSatisfied(customers []int, grumpy []int, minutes int) int {
	var cnt, tot int
	for i, c := range customers[:minutes] {
		cnt += c * grumpy[i]
		tot += c * (grumpy[i] ^ 1)
	}
	mx := cnt
	for i := minutes; i < len(customers); i++ {
		cnt += customers[i] * grumpy[i]
		cnt -= customers[i-minutes] * grumpy[i-minutes]
		mx = max(mx, cnt)
		tot += customers[i] * (grumpy[i] ^ 1)
	}
	return tot + mx
}

# Accepted solution for LeetCode #1052: Grumpy Bookstore Owner
class Solution:
    def maxSatisfied(
        self, customers: List[int], grumpy: List[int], minutes: int
    ) -> int:
        mx = cnt = sum(c * g for c, g in zip(customers[:minutes], grumpy))
        for i in range(minutes, len(customers)):
            cnt += customers[i] * grumpy[i]
            cnt -= customers[i - minutes] * grumpy[i - minutes]
            mx = max(mx, cnt)
        return sum(c * (g ^ 1) for c, g in zip(customers, grumpy)) + mx

// Accepted solution for LeetCode #1052: Grumpy Bookstore Owner
impl Solution {
    pub fn max_satisfied(customers: Vec<i32>, grumpy: Vec<i32>, minutes: i32) -> i32 {
        let mut cnt = 0;
        let mut tot = 0;
        let minutes = minutes as usize;
        for i in 0..minutes {
            cnt += customers[i] * grumpy[i];
            tot += customers[i] * (1 - grumpy[i]);
        }
        let mut mx = cnt;
        let n = customers.len();
        for i in minutes..n {
            cnt += customers[i] * grumpy[i];
            cnt -= customers[i - minutes] * grumpy[i - minutes];
            mx = mx.max(cnt);
            tot += customers[i] * (1 - grumpy[i]);
        }
        tot + mx
    }
}

// Accepted solution for LeetCode #1052: Grumpy Bookstore Owner
function maxSatisfied(customers: number[], grumpy: number[], minutes: number): number {
    let [cnt, tot] = [0, 0];
    for (let i = 0; i < minutes; ++i) {
        cnt += customers[i] * grumpy[i];
        tot += customers[i] * (grumpy[i] ^ 1);
    }
    let mx = cnt;
    for (let i = minutes; i < customers.length; ++i) {
        cnt += customers[i] * grumpy[i];
        cnt -= customers[i - minutes] * grumpy[i - minutes];
        mx = Math.max(mx, cnt);
        tot += customers[i] * (grumpy[i] ^ 1);
    }
    return tot + mx;
}

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(k)

Approach Breakdown

BRUTE FORCE

O(n × k) time

O(1) space

For each starting index, scan the next k elements to compute the window aggregate. There are n−k+1 starting positions, each requiring O(k) work, giving O(n × k) total. No extra space since we recompute from scratch each time.

SLIDING WINDOW

O(n) time

O(k) space

The window expands and contracts as we scan left to right. Each element enters the window at most once and leaves at most once, giving 2n total operations = O(n). Space depends on what we track inside the window (a hash map of at most k distinct elements, or O(1) for a fixed-size window).

Shortcut: Each element enters and exits the window once → O(n) amortized, regardless of window size.

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.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.