Minimum Difficulty of a Job Schedule Solution Amazon OA 2021

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].

See Also : Amazon Online Assessment Questions 2021 Preparation

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

Example 1:

Minimum Difficulty of a Job Schedule Solution Amazon OA 2021
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


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


Program 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 {
    int minDifficulty(vector<int>& a, int d) {
        int n=a.size();
        if(n<d) return -1;
        int dp[n+1][d+1];
        for(int i=1;i<=n;i++){
            for(int j=1;j<=d;j++){
                int mx=0;
                for(int k=i;k>=j;k--){
        return dp[n][d];

Program 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[0].length;j++)
                dp[i][j] = -1;

        return minDiff(jobDifficulty,0, 1, d, dp);

Program 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][0] = 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]


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