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1052.grumpy-bookstore-owner.cpp
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63 lines (58 loc) · 2.13 KB
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// Tag: Array, Sliding Window
// Time: O(N)
// Space: O(1)
// Ref: -
// Note: -
// 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.
//
// 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.
//
//
class Solution {
public:
int maxSatisfied(vector<int>& customers, vector<int>& grumpy, int minutes) {
int n = grumpy.size();
int res = 0;
int count = 0;
for (int i = 0; i < n; i++) {
if (grumpy[i]) {
count += customers[i];
}
if (i >= minutes - 1) {
res = max(res, count);
if (grumpy[i - minutes + 1]) {
count -= customers[i - minutes + 1];
}
}
}
for (int i = 0; i < n; i++) {
if (!grumpy[i]) {
res += customers[i];
}
}
return res;
}
};