From 33685798a840ee22b51a2336a77b5e75da4a307b Mon Sep 17 00:00:00 2001
From: Banseedhar Gondaliya <56584368+Banseedhar01@users.noreply.github.com>
Date: Mon, 11 Oct 2021 09:48:25 +0530
Subject: [PATCH] Create Max_ProfitinJobScheduling.cpp

---
 .../cpp/Max_ProfitinJobScheduling.cpp         | 31 +++++++++++++++++++
 1 file changed, 31 insertions(+)
 create mode 100644 Dynamic Programming/WeightedJobScheduling/cpp/Max_ProfitinJobScheduling.cpp

diff --git a/Dynamic Programming/WeightedJobScheduling/cpp/Max_ProfitinJobScheduling.cpp b/Dynamic Programming/WeightedJobScheduling/cpp/Max_ProfitinJobScheduling.cpp
new file mode 100644
index 000000000..e3f182b0a
--- /dev/null
+++ b/Dynamic Programming/WeightedJobScheduling/cpp/Max_ProfitinJobScheduling.cpp	
@@ -0,0 +1,31 @@
+/*
+We have n jobs, where every job is scheduled to be done from startTime[i] to endTime[i], obtaining a profit of profit[i].
+
+You're given the startTime, endTime and profit arrays, return the maximum profit you can take such that there are no two jobs in the subset with overlapping time range.*/
+
+
+//Explanation of the code below:
+/*
+1)sorting jobs according to end time.
+2)declare dp which stores {end time, total profit till that time}.
+3)go through sorted jobs, find the best-fit end time(previous of upper bound of current starting time) from dp, and add current job profit to that one if it is greater than the maximum profit that has been present in dp, then add that calculated profit to the dp.
+4)maximum profit has been stored at the end of dp, thus return (last element)->second.
+*/
+class Solution {
+public:
+    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
+    int n=startTime.size();
+	vector<vector<int>> jobs;
+        for (int i = 0; i < n; ++i) {
+            jobs.push_back({endTime[i], startTime[i], profit[i]});
+        }
+        sort(jobs.begin(), jobs.end()); // sort jobs according to end time
+        map<int, int> dp = {{0, 0}}; 
+        for (auto& job : jobs) {
+            int cur = prev(dp.upper_bound(job[1]))->second + job[2]; 
+            if (cur > dp.rbegin()->second)
+                dp[job[0]] = cur;
+        }
+        return dp.rbegin()->second;
+    }
+};