DESHAW Coding Question – Solved

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To solve the problem of rearranging tasks across groups so that the resulting array is non-increasing from left to right using the minimum number of moves (where a move consists of shifting a single task between adjacent groups), you need to think carefully about the constraints and the nature of the allowed operations.

Step 1: Understand the Problem Requirements and Constraints
- You have an array representing the number of tasks in each group arranged linearly.
- The target is to make the tasks array non-increasing, i.e., tasks[0] ≥ tasks[1] ≥ tasks[2] ≥ ... ≥ tasks[n-1].
- You can move one task at a time between adjacent groups either left or right. Each such move increments the move count by 1.
- The goal is to find the minimum number of moves needed to achieve the non-increasing order.
- Important constraints include that the total tasks across all groups is limited (≤ 250), and the number of groups is relatively small (≤ 250), which hints that solutions involving dynamic programming or state enumeration might be feasible.

Step 2: Key Insights About The Target Configuration
Since you want a non-increasing sequence, the values must decrease or stay the same as you move from left to right. But, unlike a sorting problem, you are restricted to moving tasks only between adjacent groups. This means you cannot just shuffle tasks arbitrarily; you must count the cost of moving tasks along the array step by step.

Step 3: Observe That The Total Number of Tasks is Fixed
The total number of tasks in all groups combined does not change, so your final non-increasing sequence must sum to the same total as the original array. The problem is to distribute this fixed total in a non-increasing way that minimizes the task movement cost.

Step 4: Characterize The Final Array of Tasks
You want to find a sequence of integers F = [f0, f1, ..., fn-1] such that:
- f0 ≥ f1 ≥ f2 ≥ ... ≥ fn-1 ≥ 0
- sum of f_i = sum of tasks[i] (constant)
- The cost to transform tasks to F, where cost is the sum of tasks moved between adjacent groups, is minimized.

Step 5: Think About How To Compute The Cost For A Given Target F
For any candidate final sequence F, you can calculate the cost to rearrange tasks into F by considering the differences at each group. If tasks[i] > f_i, you must move out (tasks[i] - f_i) tasks from group i to adjacent groups; if tasks[i] < f_i, you must receive (f_i - tasks[i]) tasks from adjacent groups. But since movement is allowed only between adjacent groups, the cost depends on how far tasks have to travel.

Step 6: Understand The Movement Cost Pattern
When tasks are moved between adjacent groups, moving one task from group i to group j costs |i - j| moves (since one move moves it one step). Therefore, the minimal cost rearrangement will involve moving tasks as little as possible to match the target final configuration F.

Step 7: Consider Using Prefix Sums for Movement Costs
Define prefix sums of the original tasks and prefix sums of the final target F. The difference in prefix sums at each index represents how many tasks need to be moved across that boundary.

Specifically:
- Let prefixTasks[i] = sum(tasks[0..i])
- Let prefixF[i] = sum(F[0..i])
- The absolute difference between prefixTasks[i] and prefixF[i] gives the number of tasks that must cross between group i and group i+1 in the transformation.
- The total movement cost is then the sum over all i of |prefixTasks[i] - prefixF[i]|.

This insight allows you to quickly compute the cost to transform from tasks to F given the prefix sums.

Step 8: Use This Cost Function to Find the Optimal F
Your task reduces to finding a non-increasing array F of length n summing to total tasks, that minimizes the sum of |prefixTasks[i] - prefixF[i]|.

Step 9: How To Find Such an F Efficiently
Since n and the total sum are small (both ≤ 250), you can explore dynamic programming over positions and possible values of tasks for each group under the non-increasing constraint.

One way is to use dynamic programming where for each group i you decide the final number of tasks f_i, ensuring that f_i ≤ f_(i-1) (to maintain non-increasing order), and cumulatively build prefix sums for F.

At each step, calculate the incremental cost and choose the best option to minimize total cost.

Step 10: Practical Considerations
- Enumerate possible values for f_i from 0 up to the minimum of the previous group's value and remaining tasks.
- At each step, update DP with minimal cost considering prefix sums.
- Finally, DP[n-1][...] gives the minimal cost to achieve a valid non-increasing array F.

Step 11: Summary
- The problem is a constrained redistribution problem with adjacency constraints and cost proportional to distance moved.
- Key insight: cost can be expressed in terms of absolute differences in prefix sums of original and final arrays.
- The main challenge is to find the final non-increasing sequence F summing to total tasks that minimizes total cost.
- Use dynamic programming to explore valid sequences F efficiently, leveraging the prefix sums cost formula for quick cost evaluation.
- Since constraints are small, this method is computationally feasible and elegant.

This approach balances understanding of the problem’s structure with efficient algorithmic techniques to find the minimal number of moves required.
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