Asked in: Amazon
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All Testcases Passed ✔
Solution
from functools import lru_cache
from sys import setrecursionlimit
setrecursionlimit(10**6)
def findPartitionCost(cost, k):
// ... rest of solution available after purchaseExplanation
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To approach this problem, it is important to first thoroughly understand the definition of partitioning and how the cost is computed for each partition. We have an array representing costs associated with each server in a linear chain, and the goal is to split this chain into exactly k contiguous segments (partitions). The cost for each segment is calculated as the sum of the cost at the start of the segment plus the cost at the end of the segment. The total cost is the sum of these individual partition costs, and we need to find both the minimum and maximum possible total costs achievable by choosing different partitions.
Step 1: Understand the partitioning concept
- We have an array of n elements, cost[1..n].
- We want to partition it into k contiguous segments. For example, if k=3, the array is split into 3 parts, such as [cost[1..i]], [cost[i+1..j]], [cost[j+1..n]].
- Each partition covers a contiguous range of servers; no overlap or omission is allowed.
- The cost of a partition covering indices from i to j is cost[i] + cost[j].
- The total cost is sum of the costs of all k partitions.
Step 2: Identify the critical components affecting the total cost
- The cost for each partition depends only on the cost of the first and last element in that partition.
- When we partition into k segments, there will be k segments and k starting costs and k ending costs.
- Notice that for the entire array partitioned into k segments, the total cost can be thought of as the sum of:
- The cost of the very first server (cost[1]) and
- The cost of the very last server (cost[n]), since these will always be included as starts or ends of partitions.
- Plus, the costs at the boundaries between partitions (where one partition ends and another begins).
Step 3: Reframe the problem to identifying partition boundaries
- Since partitions are contiguous and cover the entire array, the k-1 partition boundaries are where we split the array.
- Each boundary defines the end of one partition and the start of the next partition.
- The cost contribution at each partition boundary is cost of the server at that boundary position from the left partition’s end and from the right partition’s start. But the problem states that the partition cost is cost[i] + cost[j] where i and j are the first and last servers of that partition. So the boundary servers appear as end of one partition and start of the next.
- Therefore, the total cost can be thought of as:
- cost[1] + cost[n] + sum of cost values at the partition boundaries chosen (because the boundary server’s cost is counted twice—once as end of one partition, once as start of the next).
Step 4: Transform the problem into a selection problem over boundaries
- The key insight is that the total cost depends on the chosen partition boundaries.
- There are n-1 possible partition boundaries between consecutive servers.
- We must choose exactly k-1 boundaries to split the chain into k partitions.
- The total cost = cost[1] + cost[n] + sum of costs at these k-1 boundaries (each boundary corresponds to the cost at the server that separates partitions).
- But since the cost at boundaries are counted twice (once for left partition end, once for right partition start), the cost added per boundary is cost of the boundary server itself (i.e., cost at the position where partition is made).
- Therefore, minimizing or maximizing the total cost boils down to selecting k-1 boundaries with the smallest or largest possible sum of cost values at these boundaries.
Step 5: Extract the relevant costs for boundaries
- Construct an array of costs corresponding to the boundaries.
- The boundaries are between servers 1 and 2, 2 and 3, ..., n-1 and n.
- For each boundary between servers i and i+1, the contribution to total cost is cost[i] + cost[i+1].
- But careful reading suggests the problem states that partition cost is cost[start] + cost[end].
- At boundary i, the last server of the left partition is i, and the first server of the right partition is i+1.
- So for the boundary itself, the contribution is cost[i] + cost[i+1].
- When summing across all partitions, cost[1] and cost[n] will be counted once, but intermediate boundaries are counted twice (once at left partition end, once at right partition start).
- To avoid double counting, it helps to think in terms of the sum of cost[i] + cost[i+1] for each boundary i, because this value directly relates to how “expensive” a boundary is.
- Thus, create a list of boundary costs where each element is cost[i] + cost[i+1] for i in 1 to n-1.
Step 6: Find minimum and maximum total cost
- Since total cost includes cost[1] and cost[n], and the sum of boundary costs for the chosen k-1 boundaries, the problem reduces to:
- Minimum total cost = cost[1] + cost[n] + sum of (k-1) smallest boundary costs.
- Maximum total cost = cost[1] + cost[n] + sum of (k-1) largest boundary costs.
Step 7: Algorithm outline
- Compute the boundary costs array of length n-1.
- Sort this array.
- Sum up the first k-1 elements for the minimum total cost.
- Sum up the last k-1 elements for the maximum total cost.
- Add cost[1] and cost[n] to both sums to get final answers.
Step 8: Edge cases and constraints
- For k=1, the whole array is one partition, so total cost is cost[1] + cost[n] only, no boundaries selected.
- If k equals n, every server is its own partition, so total cost includes cost of every server twice (except cost[1] and cost[n]), calculated via boundary costs.
- Make sure to handle large inputs efficiently by using sorting algorithms with appropriate complexity (O(n log n)).
Step 9: Summarize insights
- The complex problem of partitioning and summing costs simplifies to selecting boundaries with appropriate cost sums.
- The careful observation about how costs add up in partitions and boundaries reveals an efficient solution.
- This understanding enables computing minimum and maximum total costs without enumerating all partition combinations.
By following these steps, one can confidently solve the problem by focusing on the boundary costs array and choosing k-1 boundaries that minimize or maximize the sum of boundary costs, then adding the fixed costs at the start and end servers.
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