1052. Grumpy Bookstore Owner.cpp

// 1052. Grumpy Bookstore Owner
// Solved
// Medium
// Topics
// Companies
// Hint
// There is a bookstore owner that has a store open for n minutes. Every minute, some number of customers enter the store. You are given an integer array customers of length n where customers[i] is the number of the customer that enters the store at the start of the ith minute and all those customers leave after the end of that minute.

// On some 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 of that minute are not satisfied, otherwise, they are satisfied.

// The bookstore owner knows a secret technique to keep themselves not grumpy for minutes consecutive minutes, but can only use it 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=customers.size();
        int currUns=0;
        int maxUns=0;
        for(int i=0;i<minutes;i++){
            currUns+=customers[i]*grumpy[i];
        }
        maxUns=currUns;
        int i=0;
        int j=minutes;
        while(j<n){
            currUns+=customers[j]*grumpy[j];
            currUns-=customers[i]*grumpy[i];
            maxUns=max(maxUns,currUns);
            i++;
            j++;
        }
        int totSat=maxUns;
        for(int i=0;i<n;i++){
            if(grumpy[i]==0){
                totSat+=customers[i];
            }
        }
        return totSat;
    }
};