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Minimum Difficulty of a Job Schedule

Hard
100%
Updated 6/1/2025

Minimum Difficulty of a Job Schedule

1. What is this problem about?

Minimum Difficulty of a Job Schedule is a classic partitioning problem. You are given an array of jobs, where each job has a difficulty level, and a number of days d. You must schedule all jobs over exactly d days. The rules are:

  • Jobs must be done in the order given.
  • Each day, you must finish at least one job.
  • The difficulty of a day is the maximum difficulty of the jobs done that day.
  • The total difficulty is the sum of the difficulties of each day. The goal is to find the minimum possible total difficulty.

2. Why is this asked in interviews?

This "Hard" problem is a favorite at Amazon, Apple, and Salesforce. The Minimum Difficulty of a Job Schedule interview question is a perfect test of Multi-stage Dynamic Programming. It assesses your ability to define a state that captures both the progress through the jobs and the number of days used. It also tests optimization, as the naive DP can be improved using a monotonic stack.

3. Algorithmic pattern used

The algorithmic pattern is Dynamic Programming. Let dp[i][j]dp[i][j] be the minimum difficulty to complete the first jj jobs in ii days. To compute dp[i][j]dp[i][j], we consider all possible split points kk for the last day: dp[i][j]=mini1k<j(dp[i1][k]+max(jobs[k...j1]))dp[i][j] = \min_{i-1 \leq k < j} (dp[i-1][k] + \max(jobs[k...j-1])). This "Array, Dynamic Programming interview pattern" typically runs in O(D×N2)O(D \times N^2). With N300N \leq 300, this approach is well within the time limits.

4. Example explanation

Jobs: [6, 5, 4, 3, 2, 1], d=2d = 2.

  • Split after 6: Day 1 (6), Day 2 (5,4,3,2,1). Total = 6+max(5,4,3,2,1)=6+5=116 + \max(5,4,3,2,1) = 6 + 5 = 11.
  • Split after 6,5,4,3,2: Day 1 (6,5,4,3,2), Day 2 (1). Total = 6+1=76 + 1 = 7. The minimum difficulty is 7.

5. Common mistakes candidates make

The most frequent error in the Minimum Difficulty of a Job Schedule coding problem is not handling the base cases correctly (e.g., when number of jobs < dd). Another mistake is failing to correctly find the maximum difficulty within a subarray efficiently, often recalculating it unnecessarily. Some candidates also struggle with the order of loops in their DP, leading to results that use the same day's data twice.

6. Interview preparation tip

Whenever you need to partition an array into KK segments, think of DP[k][n]DP[k][n]. This "Partitioning DP interview pattern" is a core skill. Practice identifying when the cost of a segment is a "max" or "sum," as this dictates how you optimize the transition. For "max" segments, look into using a monotonic stack for O(D×N)O(D \times N) optimization.

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