Added Leetcode-Reducing-Dishes

C++/ReducingDishes.cpp

@@ -0,0 +1,81 @@
+/*
+ #Hacktoberfest2021
+A chef has collected data on the satisfaction level of his n dishes. Chef can cook any dish in 1 unit of time.
+
+Like-time coefficient of a dish is defined as the time taken to cook that dish including previous dishes multiplied by its satisfaction level i.e. time[i] * satisfaction[i].
+
+Return the maximum sum of like-time coefficient that the chef can obtain after dishes preparation.
+
+Dishes can be prepared in any order and the chef can discard some dishes to get this maximum value.
+
+
+
+Example 1:
+
+Input: satisfaction = [-1,-8,0,5,-9]
+Output: 14
+Explanation: After Removing the second and last dish, the maximum total like-time coefficient will be equal to (-1*1 + 0*2 + 5*3 = 14).
+Each dish is prepared in one unit of time.
+
+
+Example 2:
+
+Input: satisfaction = [4,3,2]
+Output: 20
+Explanation: Dishes can be prepared in any order, (2*1 + 3*2 + 4*3 = 20)
+
+
+Example 3:
+
+Input: satisfaction = [-1,-4,-5]
+Output: 0
+Explanation: People don't like the dishes. No dish is prepared.
+
+
+Example 4:
+
+Input: satisfaction = [-2,5,-1,0,3,-3]
+Output: 35
+
+
+Constraints:
+
+n == satisfaction.length
+1 <= n <= 500
+-1000 <= satisfaction[i] <= 1000
+*/
+class Solution {
+public:
+ int maxSatisfaction(vector<int>& st) {
+
+ sort(st.begin(),st.end());
+
+ int n = st.size();
+ int sum = 0;
+
+ for(int i=0;i<n;i++)
+ sum+=st[i];
+
+ int i=0;
+
+ if(sum<0) {
+ //int i=0;
+ while(sum<0) {
+ sum = sum-st[i++];
+ }
+
+ }
+
+ int t = 1,ans = 0;
+
+ for(int k=i;k<n;k++) {
+ ans+=(st[k]*t);
+ t++;
+ }
+
+ return ans;
+
+
+
+ }
+};