# Minimum Difficulty of a Job Schedule Amazon OA 2023

## Minimum Difficulty of a Job Schedule Solution

You want to schedule a list of jobs in d days. Jobs are dependent (i.e To work on the i-th job, you have to finish all the jobs j where 0 <= j < i).

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

Given an array of integers jobDifficulty and an integer d. The difficulty of the i-th job is jobDifficulty[i].

Return the minimum difficulty of a job schedule. If you cannot find a schedule for the jobs return -1.

Example 1:

Input: jobDifficulty = [6,5,4,3,2,1], d = 2

Output: 7

Explanation: First day you can finish the first 5 jobs, total difficulty = 6. Second day you can finish the last job, total difficulty = 1. The difficulty of the schedule = 6 + 1 = 7

Example 2:

Input: jobDifficulty = [9,9,9], d = 4

Output: -1

Explanation: If you finish a job per day you will still have a free day. you cannot find a schedule for the given jobs.

Example 3:

Input: jobDifficulty = [1,1,1], d = 3

Output: 3

Explanation: The schedule is one job per day. total difficulty will be 3.

Example 4:

Input: jobDifficulty = [7,1,7,1,7,1], d = 3

Output: 15

Example 5:

Input: jobDifficulty = [11,111,22,222,33,333,44,444], d = 6

Output: 843

Constraints:

• 1 <= jobDifficulty.length <= 300
• 0 <= jobDifficulty[i] <= 1000
• 1 <= d <= 10

### SOLUTION

Program: Minimum Difficulty of a Job Schedule Solution in C++

Consider dp[i][j] as the min difficulty for first i jobs in first j days. Since every day should be assinged atleast one job, ith job is done on jth day. By the same logic atleast first (j-1) jobs should be left for first (j-1) days. Now let us schedule jobs from k to i on the jth day, k varies from j to i.

``````class Solution {
public:
int minDifficulty(vector<int>& a, int d) {
int n=a.size();
if(n<d) return -1;
int dp[n+1][d+1];
fill_n(&dp,(n+1)*(d+1),1e9);
dp=0;
for(int i=1;i<=n;i++){
for(int j=1;j<=d;j++){
int mx=0;
for(int k=i;k>=j;k--){
mx=max(mx,a[k-1]);
if(dp[k-1][j-1]<1e9)
dp[i][j]=min(dp[i][j],mx+dp[k-1][j-1]);
}
}
}
return dp[n][d];
}
};``````

Program: Minimum Difficulty of a Job Schedule Solution in Java

``````class Solution {
public int minDiff(int[] jobs, int start, int n, int d, int[][] dp){
if (n > d)
return 0;

//if calculated before
if (dp[start][n] != -1)
return dp[start][n];

//int the last step, we need to take the maximum of the rest of elements
if (n == d){
int currentMax = Integer.MIN_VALUE;
for (int i=start;i<jobs.length;i++){
currentMax = Math.max(currentMax, jobs[i]);
}
dp[start][n] = currentMax;
return currentMax;
}

int min = Integer.MAX_VALUE;
int currentMax = Integer.MIN_VALUE;
//loop until before jobs.length-(d-n) to make sure we have enough jobs for the rest of the days
for (int i=start;i<jobs.length-(d-n);i++){
currentMax = Math.max(currentMax, jobs[i]);
min = Math.min(min, minDiff(jobs, i+1, n+1, d, dp)+currentMax);
}
dp[start][n] = min;
return min;
}

public int minDifficulty(int[] jobDifficulty, int d) {
//if days is more than the jobs, then we can't proceed
if (d > jobDifficulty.length)
return -1;

//init our dynamic programming array
int[][] dp = new int[jobDifficulty.length][d+1];
for (int i=0;i<dp.length;i++){
for (int j=0;j<dp.length;j++)
dp[i][j] = -1;
}
return minDiff(jobDifficulty,0, 1, d, dp);
}
}``````

Program: Minimum Difficulty of a Job Schedule Solution in Python

``````"""
dp[i][k] := minimum difficulty of a k days job schedule, D[:i].
maxInRange[i][j] := max(D[i:j+1])
"""
class Solution(object):
def minDifficulty(self, D, K):
if not D or not K or K>len(D): return -1
N = len(D)
maxInRange = [[0 for _ in xrange(N)] for _ in xrange(N)]
for i in xrange(N): maxInRange[i][i] = D[i]
for l in xrange(2, N+1):
for i in xrange(N):
j = i+l-1
if j>=N: continue
maxInRange[i][j] = max(maxInRange[i+1][j-1], D[i], D[j])
dp = [[float('inf') for _ in xrange(K+1)] for _ in xrange(N+1)]
dp = 0
for i in xrange(1, N+1):
for k in xrange(1, min(i, K)+1):
for j in xrange(k, i+1):
dp[i][k] = min(dp[i][k], dp[j-1][k-1]+maxInRange[j-1][i-1])

#if you don't pre-calculate maxInRange
#dp[i][k] = min(dp[i][k], dp[j-1][k-1]+max(D[j-1:i]))
return dp[N][K]``````