Job Scheduling to Minimize Difficulty - Amazon Top Interview Questions


Problem Statement :


You are given a list of integers jobs and an integer k. You want to finish all jobs in k days. The jobs must be done in order and a job must be done each day.

The difficulty of job i is jobs[i] and the difficulty of doing a list of jobs on a day is defined to be the maximum difficulty job performed on that day.

Return the minimum sum of the difficulties to perform the jobs over k days.

Constraints

n ≤ 500 where n is the length of jobs

k ≤ 10

Example 1

Input

jobs = [1, 2, 3, 5, 2]

k = 2

Output

6

Explanation

We do [1] the first day and then do [2, 3, 5, 2]. The total difficulty is 1 + max(2, 3, 5, 2) = 6.



Solution :



title-img




                        Solution in C++ :

int solve(vector<int>& nums, int k) {
    int n = nums.size();
    vector<int> dp(n + 1);
    fill(dp.begin() + 1, dp.end(), INT_MAX);
    dp[0] = 0;
    while (k--) {
        vector<int> ndp(n + 1, INT_MAX);
        for (int i = 0; i < nums.size(); i++) {
            if (dp[i] == INT_MAX) continue;
            int maxd = 0;
            for (int j = i; j < nums.size(); j++) {
                maxd = max(maxd, nums[j]);
                ndp[j + 1] = min(ndp[j + 1], dp[i] + maxd);
            }
        }
        dp.swap(ndp);
    }
    return dp[n];
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    /**
    public int solve(int[] jobs, int k) {
        if (jobs.length < k) {
            return -1;
        }

        return helper(jobs, k, 0);
    }
    private int helper(int[] jobs, int k, int start) {

        int max = -1;
        int ans = Integer.MAX_VALUE;
        int end = jobs.length - k;

        if (k == 1) {
            for (int i = start; i <= end; i++) {
                max = Math.max(max, jobs[i]);
            }

            return max;
        } else {
            for (int i = start; i <= end; i++) {
                max = Math.max(max, jobs[i]);
                ans = Math.min(ans, max + helper(jobs, k - 1, i + 1));
            }

            return ans;
        }
    }
    */
    int[][] m;
    public int solve(int[] jobs, int k) {
        if (jobs.length < k) {
            return -1;
        }
        m = new int[k + 1][jobs.length];
        return helper(jobs, k, 0);
    }
    private int helper(int[] jobs, int k, int start) {
        if (m[k][start] > 0) {
            return m[k][start];
        }
        int max = -1;
        int ans = Integer.MAX_VALUE;
        int end = jobs.length - k;

        if (k == 1) {
            for (int i = start; i <= end; i++) {
                max = Math.max(max, jobs[i]);
            }
            m[k][start] = max;
            return max;
        } else {
            for (int i = start; i <= end; i++) {
                max = Math.max(max, jobs[i]);
                ans = Math.min(ans, max + helper(jobs, k - 1, i + 1));
            }
            m[k][start] = ans;
            return ans;
        }
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, jobs, k):
        N = len(jobs)

        @lru_cache(None)
        def traverse(idx, day, current_max):

            if day > k:
                return math.inf

            if idx == N:
                if k == day:
                    return current_max if current_max != -math.inf else 0
                return math.inf

            # choice-01 -> Stop the day with the current task and chill. (Lazy :D)
            # choice-02: do the next task the same day so that you can chill later. (Eager :|)

            val = jobs[idx]
            choice_01 = max(current_max, val) + traverse(idx + 1, day + 1, -math.inf)
            choice_02 = traverse(idx + 1, day, max(current_max, val))

            return min(choice_01, choice_02)

        return traverse(0, 0, -math.inf)
                    


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