LeetCode #3500 — HARD

Minimum Cost to Divide Array Into Subarrays

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

You are given two integer arrays, nums and cost, of the same size, and an integer k.

You can divide nums into subarrays. The cost of the i^th subarray consisting of elements nums[l..r] is:

(nums[0] + nums[1] + ... + nums[r] + k * i) * (cost[l] + cost[l + 1] + ... + cost[r]).

Note that i represents the order of the subarray: 1 for the first subarray, 2 for the second, and so on.

Return the minimum total cost possible from any valid division.

Example 1:

Input: nums = [3,1,4], cost = [4,6,6], k = 1

Output: 110

Explanation:

The minimum total cost possible can be achieved by dividing nums into subarrays [3, 1] and [4].

The cost of the first subarray [3,1] is (3 + 1 + 1 * 1) * (4 + 6) = 50.
The cost of the second subarray [4] is (3 + 1 + 4 + 1 * 2) * 6 = 60.

Example 2:

Input: nums = [4,8,5,1,14,2,2,12,1], cost = [7,2,8,4,2,2,1,1,2], k = 7

Output: 985

Explanation:

The minimum total cost possible can be achieved by dividing nums into subarrays [4, 8, 5, 1], [14, 2, 2], and [12, 1].

The cost of the first subarray [4, 8, 5, 1] is (4 + 8 + 5 + 1 + 7 * 1) * (7 + 2 + 8 + 4) = 525.
The cost of the second subarray [14, 2, 2] is (4 + 8 + 5 + 1 + 14 + 2 + 2 + 7 * 2) * (2 + 2 + 1) = 250.
The cost of the third subarray [12, 1] is (4 + 8 + 5 + 1 + 14 + 2 + 2 + 12 + 1 + 7 * 3) * (1 + 2) = 210.

Constraints:

1 <= nums.length <= 1000
cost.length == nums.length
1 <= nums[i], cost[i] <= 1000
1 <= k <= 1000

Step 01

Brute Force Baseline

Problem summary: You are given two integer arrays, nums and cost, of the same size, and an integer k. You can divide nums into subarrays. The cost of the ith subarray consisting of elements nums[l..r] is: (nums[0] + nums[1] + ... + nums[r] + k * i) * (cost[l] + cost[l + 1] + ... + cost[r]). Note that i represents the order of the subarray: 1 for the first subarray, 2 for the second, and so on. Return the minimum total cost possible from any valid division.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming

Example 1

[3,1,4]
[4,6,6]
1

Example 2

[4,8,5,1,14,2,2,12,1]
[7,2,8,4,2,2,1,1,2]
7

Core Insight

What unlocks the optimal approach

<code>dp[i]</code> is the minimum cost to split the array suffix starting at <code>i</code>.
Observe that no matter how many subarrays we have, if we have the first subarray on the left, the total cost of the previous subarrays increases by <code>k * total_cost_of_the_subarray</code>. This is because when we increase <code>i</code> to <code>(i + 1)</code>, the cost increase is just the suffix sum of the cost array.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3500: Minimum Cost to Divide Array Into Subarrays
class Solution {
  public long minimumCost(int[] nums, int[] cost, int k) {
    final int n = nums.length;
    long[] prefixNums = new long[n + 1];
    long[] prefixCost = new long[n + 1];
    // dp[i] := the minimum cost to divide nums[i..n - 1] into subarrays
    long[] dp = new long[n + 1];

    for (int i = 0; i < n; ++i) {
      prefixNums[i + 1] = prefixNums[i] + nums[i];
      prefixCost[i + 1] = prefixCost[i] + cost[i];
    }

    Arrays.fill(dp, Long.MAX_VALUE);
    dp[n] = 0;

    for (int i = n - 1; i >= 0; --i)
      for (int j = i; j < n; ++j)
        dp[i] = Math.min(dp[i], prefixNums[j + 1] * (prefixCost[j + 1] - prefixCost[i]) +
                                    k * (prefixCost[n] - prefixCost[i]) + dp[j + 1]);

    return dp[0];
  }
}

// Accepted solution for LeetCode #3500: Minimum Cost to Divide Array Into Subarrays
package main

import (
	"math"
)

// https://space.bilibili.com/206214
type vec struct{ x, y int }

func (a vec) sub(b vec) vec { return vec{a.x - b.x, a.y - b.y} }
func (a vec) det(b vec) int { return a.x*b.y - a.y*b.x }
func (a vec) dot(b vec) int { return a.x*b.x + a.y*b.y }

func minimumCost(nums, cost []int, k int) int64 {
	totalCost := 0
	for _, c := range cost {
		totalCost += c
	}

	q := []vec{{}}
	sumNum, sumCost := 0, 0
	for i, x := range nums {
		sumNum += x
		sumCost += cost[i]

		p := vec{-sumNum - k, 1}
		for len(q) > 1 && p.dot(q[0]) >= p.dot(q[1]) {
			q = q[1:]
		}

		// 一边算 DP 一边构建下凸包
		p = vec{sumCost, p.dot(q[0]) + sumNum*sumCost + k*totalCost}
		for len(q) > 1 && q[len(q)-1].sub(q[len(q)-2]).det(p.sub(q[len(q)-1])) <= 0 {
			q = q[:len(q)-1]
		}
		q = append(q, p)
	}
	return int64(q[len(q)-1].y)
}

func minimumCost1(nums, cost []int, k int) int64 {
	n := len(nums)
	s := make([]int, n+1)
	for i, c := range cost {
		s[i+1] = s[i] + c
	}

	f := make([]int, n+1)
	sumNum := 0
	for i, x := range nums {
		i++
		sumNum += x
		res := math.MaxInt
		for j := range i {
			res = min(res, f[j]+sumNum*(s[i]-s[j])+k*(s[n]-s[j]))
		}
		f[i] = res
	}
	return int64(f[n])
}

# Accepted solution for LeetCode #3500: Minimum Cost to Divide Array Into Subarrays
class Solution:
  def minimumCost(self, nums: list[int], cost: list[int], k: int) -> int:
    n = len(nums)
    prefixNums = list(itertools.accumulate(nums, initial=0))
    prefixCost = list(itertools.accumulate(cost, initial=0))
    # dp[i] := the minimum cost to divide nums[i..n - 1] into subarrays
    dp = [math.inf] * n + [0]

    for i in range(n - 1, -1, -1):
      for j in range(i, n):
        dp[i] = min(dp[i],
                    prefixNums[j + 1] * (prefixCost[j + 1] - prefixCost[i]) +
                    k * (prefixCost[n] - prefixCost[i]) + dp[j + 1])

    return dp[0]

// Accepted solution for LeetCode #3500: Minimum Cost to Divide Array Into Subarrays
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3500: Minimum Cost to Divide Array Into Subarrays
// class Solution {
//   public long minimumCost(int[] nums, int[] cost, int k) {
//     final int n = nums.length;
//     long[] prefixNums = new long[n + 1];
//     long[] prefixCost = new long[n + 1];
//     // dp[i] := the minimum cost to divide nums[i..n - 1] into subarrays
//     long[] dp = new long[n + 1];
// 
//     for (int i = 0; i < n; ++i) {
//       prefixNums[i + 1] = prefixNums[i] + nums[i];
//       prefixCost[i + 1] = prefixCost[i] + cost[i];
//     }
// 
//     Arrays.fill(dp, Long.MAX_VALUE);
//     dp[n] = 0;
// 
//     for (int i = n - 1; i >= 0; --i)
//       for (int j = i; j < n; ++j)
//         dp[i] = Math.min(dp[i], prefixNums[j + 1] * (prefixCost[j + 1] - prefixCost[i]) +
//                                     k * (prefixCost[n] - prefixCost[i]) + dp[j + 1]);
// 
//     return dp[0];
//   }
// }

// Accepted solution for LeetCode #3500: Minimum Cost to Divide Array Into Subarrays
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #3500: Minimum Cost to Divide Array Into Subarrays
// class Solution {
//   public long minimumCost(int[] nums, int[] cost, int k) {
//     final int n = nums.length;
//     long[] prefixNums = new long[n + 1];
//     long[] prefixCost = new long[n + 1];
//     // dp[i] := the minimum cost to divide nums[i..n - 1] into subarrays
//     long[] dp = new long[n + 1];
// 
//     for (int i = 0; i < n; ++i) {
//       prefixNums[i + 1] = prefixNums[i] + nums[i];
//       prefixCost[i + 1] = prefixCost[i] + cost[i];
//     }
// 
//     Arrays.fill(dp, Long.MAX_VALUE);
//     dp[n] = 0;
// 
//     for (int i = n - 1; i >= 0; --i)
//       for (int j = i; j < n; ++j)
//         dp[i] = Math.min(dp[i], prefixNums[j + 1] * (prefixCost[j + 1] - prefixCost[i]) +
//                                     k * (prefixCost[n] - prefixCost[i]) + dp[j + 1]);
// 
//     return dp[0];
//   }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.