Minimum Difficulty of a Job Schedule in C++

Suppose we want to schedule a list of tasks in d days. The tasks are dependent so, to work on the i-th task, we have to finish all the tasks j where 0 <= j < i.

We have to finish at least one task in each day. The difficulty of a task schedule is actually the sum of difficulties of each day of the d number of days. The difficulty of a day is the maximum difficulty of a task that done in that day.

So we have an array of integers called taskDifficulty and an integer d. The difficulty of the i-th job is taskDifficulty[i]. We have to find the minimum difficulty of a task schedule. If we cannot find a schedule for the tasks, then return -1.

So, if the input is like taskDifficulty = [6,5,4,3,2,1], d = 2,

then the output will be 7, as on the day 1 we can finish the first 5 jobs, the total difficulty is 6. Now on day 2 we can finish the last job, total difficulty is 1, so the difficulty of the schedule will be 6 + 1 = 7.

To solve this, we will follow these steps −

• Define a function solve(), this will take an array v, idx, k, one 2D dp,

• if idx is same as size of v and k is same as 0, then −

  • return 0

• if k < 0 or idx is same as size of v and k > 0, then −

  • return 1^6

• if dp[idx, k] is not equal to -1, then −

  • return dp[idx, k]

• maxVal := 0

• ret := inf

• for initialize i := idx, when i < size of v, update (increase i by 1), do −

  • maxVal := maximum of v[i] and maxVal

  • ret := minimum of ret and maxVal + solve(v, i + 1, k - 1, dp)

• dp[idx, k] := ret

• return ret

• From the main method do the following −

• n := size of j

• if d > n, then −

  • return -1

• Define one 2D array dp of size n x (d + 1) and fill this with -1

• return solve(j, 0, d, dp)

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
   public:
   int solve(vector<int>& v, int idx, int k, vector<vector<int> >&
   dp){
      if (idx == v.size() && k == 0)
      return 0;
      if (k < 0 || idx == v.size() && k > 0)
      return 1e6;
      if (dp[idx][k] != -1)
      return dp[idx][k];
      int maxVal = 0;
      int ret = INT_MAX;
      for (int i = idx; i < v.size(); i++) {
         maxVal = max(v[i], maxVal);
         ret = min(ret, maxVal + solve(v, i + 1, k - 1, dp));
      }
      return dp[idx][k] = ret;
   }
   int minDifficulty(vector<int>& j, int d){
      int n = j.size();
      if (d > n)
      return -1;
      vector<vector<int> > dp(n, vector<int>(d + 1, -1));
      return solve(j, 0, d, dp);
   }
};
main(){
   Solution ob;
   vector<int> v = {6,5,4,3,2,1};
   cout << (ob.minDifficulty(v, 2));
}

Input

{6,5,4,3,2,1}, 2

Output

7

Arnab Chakraborty

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Updated on: 08-Jun-2020

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