Program to find minimum sum of difficulties to complete jobs over k days in C++

Suppose we have a list of numbers called jobs and another value k. Now we want to finish all jobs in k different days. The jobs must be performed in the given order and in each day we have to complete one task. The difficulty of job i is stored at jobs[i] and the difficulty of completing a list of jobs on a day will be the maximum difficulty job performed on that day. So we have to find the minimum sum of the difficulties to perform the jobs over k different days.

So, if the input is like jobs = [2, 3, 4, 6, 3] k = 2, then the output will be 8, at first we do [2] then do [3, 4, 6, 3]. so the difficulty is 2 + maximum of (3, 4, 6, 3) = 8.

To solve this, we will follow these steps −

Define an array dp of size: 505 x 15.
Define a function dfs(), this will take start, k, and an array v,
if start >= size of v, then −
- return (if k is same as 0, then 0, otherwise inf)
if k < 0, then −
- return inf
if dp[start, k] is not equal to -1, then −
- return dp[start, k]
ret := inf
val := 0
for initialize i := start, when i < size of v, update (increase i by 1), do −
- val := maximum of val and v[i]
- ret := minimum of ret and (val + dfs(i + 1, k - 1, v))
dp[start, k] = ret
return ret
From the main method do the following −
fill dp with -1
return dfs(0, k, jobs)

Example (C++)

Let us see the following implementation to get better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
const int inf = 1e6;
int dp[505][15];
int dfs(int start, int k, vector <int>& v){
   if(start >= v.size()){
      return k == 0 ? 0 : inf;
   }
   if(k < 0)
      return inf;
   if(dp[start][k] != -1)
      return dp[start][k];
   int ret = inf;
   int val = 0;
   for(int i = start; i < v.size(); i++){
      val = max(val, v[i]);
      ret = min(ret, val + dfs(i + 1, k - 1, v));
   }
   return dp[start][k] = ret;
}
int solve(vector<int>& jobs, int k) {
   memset(dp ,-1, sizeof dp);
   return dfs(0, k, jobs);
}
int main(){
   vector<int> v = {2, 3, 4, 6, 3};
   int k = 2;
   cout << solve(v, k);
}