How do you solve Minimum Cost to Divide Array Into Subarrays on LeetCode?

Minimum Cost to Divide Array Into Subarrays (LeetCode #3500) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Dynamic programming helps to break down the problem into manageable subproblems.

What is the time complexity of Minimum Cost to Divide Array Into Subarrays?

The optimal solution for Minimum Cost to Divide Array Into Subarrays runs in O(n²) time and uses O(n) space. The outer loop runs n times, and the inner loop can run up to n times, leading to O(n²) time complexity, while dp array uses O(n) space.

What is the brute force solution for Minimum Cost to Divide Array Into Subarrays?

The brute force approach for Minimum Cost to Divide Array Into Subarrays is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try every possible way to split the array into subarrays and calculate the total cost for each division. This method guarantees finding the minimum cost but is inefficient. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Minimum Cost to Divide Array Into Subarrays?

Minimum Cost to Divide Array Into Subarrays uses the following concepts: Array, Dynamic Programming, Prefix Sum. The recommended patterns to study are: Dynamic Programming, Prefix Sum.

What are the interview tips for Minimum Cost to Divide Array Into Subarrays?

Explain your thought process clearly. Discuss edge cases and how your solution handles them. Practice similar problems to build confidence.

What are common mistakes when solving Minimum Cost to Divide Array Into Subarrays?

Not considering the impact of the order of subarrays on cost. Overlooking the need to update the dp array correctly.

#3500

Minimum Cost to Divide Array Into Subarrays

Hard

ArrayDynamic ProgrammingPrefix SumDynamic ProgrammingPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n)

💡

Intuition

Time O(n²)Space O(n)

Using dynamic programming, we can build a solution by calculating the minimum cost for each suffix of the array, leveraging previously computed results to avoid redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Initialize a dp array where dp[i] represents the minimum cost for the suffix starting at index i.
2Step 2: Iterate backwards through the array, calculating the cost for each possible subarray ending at the current index.
3Step 3: Update dp[i] based on the minimum cost found for all subarrays starting from i.

solution.py12 lines

1def minCost(nums, cost, k):
2    n = len(nums)
3    dp = [float('inf')] * (n + 1)
4    dp[n] = 0
5    for i in range(n - 1, -1, -1):
6        subarray_sum = 0
7        subarray_cost = 0
8        for j in range(i, n):
9            subarray_sum += nums[j]
10            subarray_cost += cost[j]
11            dp[i] = min(dp[i], (subarray_sum + k * (n - j)) * subarray_cost + dp[j + 1])
12    return dp[0]

ℹ

Complexity note: The outer loop runs n times, and the inner loop can run up to n times, leading to O(n²) time complexity, while dp array uses O(n) space.

1Dynamic programming helps to break down the problem into manageable subproblems.
2Cost accumulation can be efficiently calculated using prefix sums.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.