Merge pull request neetcode-gh#2932 from Abe0770/main

Create 2483-minimum-penalty-for-a-shop.cpp

Author: felivalencia3

cpp/2483-minimum-penalty-for-a-shop.cpp

@@ -0,0 +1,43 @@
+/*
+  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.
+
+  Ex. Input: customers = "YYNY"
+      Output: 2
+      Explanation: 
+      - Closing the shop at the 0th hour incurs in 1+1+0+1 = 3 penalty.
+      - Closing the shop at the 1st hour incurs in 0+1+0+1 = 2 penalty.
+      - Closing the shop at the 2nd hour incurs in 0+0+0+1 = 1 penalty.
+      - Closing the shop at the 3rd hour incurs in 0+0+1+1 = 2 penalty.
+      - Closing the shop at the 4th hour incurs in 0+0+1+0 = 1 penalty.
+      Closing the shop at 2nd or 4th hour gives a minimum penalty. Since 2 is earlier, the optimal closing time is 2.
+
+  Time  : O(N)
+  Space : O(1)
+*/
+
+class Solution {
+public:
+    int bestClosingTime(string customers) {
+        int res = -1, maxi = 0, pen = 0;
+        for(int i = 0 ; i < customers.size() ; ++i) {
+            if(customers[i] == 'Y')
+                ++pen;
+            else 
+                --pen;
+            if(pen > maxi) {
+                maxi = pen;
+                res = i;
+            }
+        }
+        return ++res;
+    }
+};