Maximum Profit in Job Scheduling in C++

Suppose we have n different tasks, where every task is scheduled to be done from startTime[i] to endTime[i], for that task we algo get profit of profit[i]. We know the startTime , endTime and profit lists, we have to find the maximum profit we can take such that there are no 2 tasks in the subset with overlapping time range. If we choose a task that ends at time X we will be able to start another task that starts at time X.
So, if the input is like startTime = [1,2,3,3], endTime = [3,4,5,6] profit = [500,100,400,700]
then the output will be 1200
To solve this, we will follow these steps −
Define one Data with start, end and cost values
make one array of Data j
make one array of Data j
n := size of s
n := size of s
for initialize i := 0, when i < n, update (increase i by 1), do −Create one Data temp(s[i], e[i], p[i])insert temp at the end of j
for initialize i := 0, when i < n, update (increase i by 1), do −
Create one Data temp(s[i], e[i], p[i])
Create one Data temp(s[i], e[i], p[i])
insert temp at the end of j
insert temp at the end of j
sort the array j based on the end time
sort the array j based on the end time
Define an array dp of size n
Define an array dp of size n
dp[0] := j[0].cost
dp[0] := j[0].cost
for initialize i := 1, when i < n, update (increase i by 1), do −temp := 0, low := 0, high := i - 1while low < high, do −mid := low + (high - low + 1) / 2if j[mid].end <= j[i].start, then −low := midOtherwisehigh := mid - 1dp[i] := j[i].costif j[low].end <= j[i].start, then −dp[i] := dp[i] + dp[low]dp[i] := maximum of dp[i] and dp[i - 1]
for initialize i := 1, when i < n, update (increase i by 1), do −
temp := 0, low := 0, high := i - 1
temp := 0, low := 0, high := i - 1
while low < high, do −mid := low + (high - low + 1) / 2if j[mid].end <= j[i].start, then −low := midOtherwisehigh := mid - 1dp[i] := j[i].costif j[low].end <= j[i].start, then −dp[i] := dp[i] + dp[low]
while low < high, do −
mid := low + (high - low + 1) / 2
mid := low + (high - low + 1) / 2
if j[mid].end <= j[i].start, then −low := mid
if j[mid].end <= j[i].start, then −
low := mid
low := mid
Otherwisehigh := mid - 1
Otherwise
high := mid - 1
high := mid - 1
dp[i] := j[i].cost
dp[i] := j[i].cost
if j[low].end <= j[i].start, then −dp[i] := dp[i] + dp[low]
if j[low].end <= j[i].start, then −
dp[i] := dp[i] + dp[low]
dp[i] := dp[i] + dp[low]
dp[i] := maximum of dp[i] and dp[i - 1]
dp[i] := maximum of dp[i] and dp[i - 1]
return dp[n - 1]
return dp[n - 1]
Let us see the following implementation to get better understanding −
 Live Demo
#include <bits/stdc++.h>
using namespace std;
struct Data{
   int s,e,c;
   Data(int x, int y, int z){
      s= x;
      e= y;
      c = z;
   }
};
bool cmp(Data a, Data b){
   return a.e<b.e;
}
class Solution {
   public:
   int jobScheduling(vector<int>& s, vector<int>& e, vector<int>& p){
      vector<Data> j;
      int n = s.size();
      for (int i = 0; i < n; i++) {
         Data temp(s[i], e[i], p[i]);
         j.push_back(temp);
      }
      sort(j.begin(), j.end(), cmp);
      vector<int> dp(n);
      dp[0] = j[0].c;
      for (int i = 1; i < n; i++) {
         int temp = 0;
         int low = 0;
         int high = i - 1;
         while (low < high) {
            int mid = low + (high - low + 1) / 2;
            if (j[mid].e <= j[i].s)
               low = mid;
            else
               high = mid - 1;
         }
         dp[i] = j[i].c;
         if (j[low].e <= j[i].s)
            dp[i] += dp[low];
         dp[i] = max(dp[i], dp[i - 1]);
      }
      return dp[n - 1];
   }
};
main(){
   Solution ob;
   vector<int> startTime = {1,2,3,3}, endTime = {3,4,5,6}, profit =
   {500,100,400,700};
   cout << (ob.jobScheduling(startTime, endTime, profit));
}
{1,2,3,3}, {3,4,5,6}, {500,100,400,700}
1200