How do you solve Minimum Partition Score on LeetCode?

Minimum Partition Score (LeetCode #3826) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(k * n²) time complexity and O(n) space complexity. Key insight: Dynamic programming is effective for partitioning problems.

What is the brute force solution for Minimum Partition Score?

The brute force approach for Minimum Partition Score is Brute Force, with O(n²) time complexity and O(1) space complexity. We can explore all possible partitions of the array into k subarrays and calculate their scores. This method guarantees finding the minimum score but is inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(k * n²).

What are the interview tips for Minimum Partition Score?

Explain your thought process clearly. Discuss edge cases and constraints. Practice similar partitioning problems.

What are common mistakes when solving Minimum Partition Score?

Not using prefix sums for efficiency. Overlooking the base cases in DP initialization.

#3826

Minimum Partition Score

Hard

ArrayDivide and ConquerDynamic ProgrammingQueuePrefix SumMonotonic QueueDynamic ProgrammingPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(k * n²)Space O(n)

Using dynamic programming, we can build up the solution by calculating the minimum score for smaller subproblems, leveraging previously computed results to avoid redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Initialize a DP table where dp[k][i] represents the minimum score for partitioning the first i elements into k subarrays.
2Step 2: Calculate prefix sums to efficiently compute subarray sums.
3Step 3: Use a nested loop to fill the DP table, minimizing the score by considering all possible previous partitions.

solution.py12 lines

1def minPartitionScore(nums, k):
2    n = len(nums)
3    prefix = [0] * (n + 1)
4    for i in range(n): prefix[i + 1] = prefix[i] + nums[i]
5    dp = [[float('inf')] * (n + 1) for _ in range(k + 1)]
6    dp[0][0] = 0
7    for i in range(1, n + 1):
8        for j in range(1, k + 1):
9            for p in range(j - 1, i):
10                sumArr = prefix[i] - prefix[p]
11                dp[j][i] = min(dp[j][i], dp[j - 1][p] + sumArr * (sumArr + 1) // 2)
12    return dp[k][n]

ℹ

Complexity note: The complexity arises from filling the DP table, where each entry requires iterating over previous partitions, leading to a quadratic factor.

1Dynamic programming is effective for partitioning problems.
2Prefix sums optimize subarray sum calculations.

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