LeetCode #3826 — HARD

Minimum Partition Score

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

The Problem

Problem Statement

You are given an integer array nums and an integer k.

Your task is to partition nums into exactly k subarrays and return an integer denoting the minimum possible score among all valid partitions.

The score of a partition is the sum of the values of all its subarrays.

The value of a subarray is defined as sumArr * (sumArr + 1) / 2, where sumArr is the sum of its elements.

Example 1:

Input: nums = [5,1,2,1], k = 2

Output: 25

Explanation:

We must partition the array into k = 2 subarrays. One optimal partition is [5] and [1, 2, 1].
The first subarray has sumArr = 5 and value = 5 × 6 / 2 = 15.
The second subarray has sumArr = 1 + 2 + 1 = 4 and value = 4 × 5 / 2 = 10.
The score of this partition is 15 + 10 = 25, which is the minimum possible score.

Example 2:

Input: nums = [1,2,3,4], k = 1

Output: 55

Explanation:

Since we must partition the array into k = 1 subarray, all elements belong to the same subarray: [1, 2, 3, 4].
This subarray has sumArr = 1 + 2 + 3 + 4 = 10 and value = 10 × 11 / 2 = 55.
The score of this partition is 55, which is the minimum possible score.

Example 3:

Input: nums = [1,1,1], k = 3

Output: 3

Explanation:

We must partition the array into k = 3 subarrays. The only valid partition is [1], [1], [1].
Each subarray has sumArr = 1 and value = 1 × 2 / 2 = 1.
The score of this partition is 1 + 1 + 1 = 3, which is the minimum possible score.

Constraints:

1 <= nums.length <= 1000
1 <= nums[i] <= 10⁴
1 <= k <= nums.length

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums and an integer k. Your task is to partition nums into exactly k subarrays and return an integer denoting the minimum possible score among all valid partitions. The score of a partition is the sum of the values of all its subarrays. The value of a subarray is defined as sumArr * (sumArr + 1) / 2, where sumArr is the sum of its elements.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming · Monotonic Queue

Example 1

[5,1,2,1]
2

Example 2

[1,2,3,4]
1

Example 3

[1,1,1]
3

Step 02

Core Insight

What unlocks the optimal approach

Let <code>dp[k][i]</code> be the minimum score to partition the first <code>i</code> elements into <code>k</code> subarrays. The transition is <code>dp[k][i] = min(dp[k-1][j] + value(subarray nums[j...i-1]))</code> for <code>j < i</code>.
The naive DP approach takes <code>O(K*N<sup>2</sup>)</code>, which is too slow. Look for optimizations applicable to partitioning problems.
The cost function is convex. This suggests that the optimal splitting point satisfies the Quadrangle Inequality, enabling Divide and Conquer Optimization to reduce the complexity.

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 #3826: Minimum Partition Score
// Auto-generated Java example from go.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (go):
// // Accepted solution for LeetCode #3826: Minimum Partition Score
// package main
// 
// import (
// 	"math"
// 	"math/big"
// )
// 
// // 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) dot(b vec) int { return a.x*b.x + a.y*b.y }
// func (a vec) det(b vec) int { return a.x*b.y - a.y*b.x } // 如果乘法会溢出，用 detCmp
// func (a vec) detCmp(b vec) int {
// 	v := new(big.Int).Mul(big.NewInt(int64(a.x)), big.NewInt(int64(b.y)))
// 	w := new(big.Int).Mul(big.NewInt(int64(a.y)), big.NewInt(int64(b.x)))
// 	return v.Cmp(w)
// }
// 
// func minPartitionScore(nums []int, k int) int64 {
// 	n := len(nums)
// 	sum := make([]int, n+1)
// 	for i, x := range nums {
// 		sum[i+1] = sum[i] + x
// 	}
// 
// 	f := make([]int, n+1)
// 	for i := 1; i <= n; i++ {
// 		f[i] = math.MaxInt / 2
// 	}
// 
// 	for K := 1; K <= k; K++ {
// 		s := sum[K-1]
// 		q := []vec{{s, f[K-1] + s*s - s}}
// 		for i := K; i <= n-(k-K); i++ {
// 			s = sum[i]
// 			p := vec{-2 * s, 1}
// 			for len(q) > 1 && p.dot(q[0]) >= p.dot(q[1]) {
// 				q = q[1:]
// 			}
// 
// 			v := vec{s, f[i] + s*s - s}
// 			f[i] = p.dot(q[0]) + s*s + s
// 
// 			// 读者可以把 detCmp 改成 det 感受下这个算法的效率
// 			// 目前 det 也能过，可以试试 hack 一下
// 			for len(q) > 1 && q[len(q)-1].sub(q[len(q)-2]).detCmp(v.sub(q[len(q)-1])) <= 0 {
// 				q = q[:len(q)-1]
// 			}
// 			q = append(q, v)
// 		}
// 	}
// 
// 	return int64(f[n] / 2)
// }
// 
// //func main() {
// //	a := make([]int, 1000)
// //	for i := range a {
// //		a[i] = 1e4
// //	}
// //	fmt.Println(minPartitionScore(a, 2))
// //}

// Accepted solution for LeetCode #3826: Minimum Partition Score
package main

import (
	"math"
	"math/big"
)

// 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) dot(b vec) int { return a.x*b.x + a.y*b.y }
func (a vec) det(b vec) int { return a.x*b.y - a.y*b.x } // 如果乘法会溢出，用 detCmp
func (a vec) detCmp(b vec) int {
	v := new(big.Int).Mul(big.NewInt(int64(a.x)), big.NewInt(int64(b.y)))
	w := new(big.Int).Mul(big.NewInt(int64(a.y)), big.NewInt(int64(b.x)))
	return v.Cmp(w)
}

func minPartitionScore(nums []int, k int) int64 {
	n := len(nums)
	sum := make([]int, n+1)
	for i, x := range nums {
		sum[i+1] = sum[i] + x
	}

	f := make([]int, n+1)
	for i := 1; i <= n; i++ {
		f[i] = math.MaxInt / 2
	}

	for K := 1; K <= k; K++ {
		s := sum[K-1]
		q := []vec{{s, f[K-1] + s*s - s}}
		for i := K; i <= n-(k-K); i++ {
			s = sum[i]
			p := vec{-2 * s, 1}
			for len(q) > 1 && p.dot(q[0]) >= p.dot(q[1]) {
				q = q[1:]
			}

			v := vec{s, f[i] + s*s - s}
			f[i] = p.dot(q[0]) + s*s + s

			// 读者可以把 detCmp 改成 det 感受下这个算法的效率
			// 目前 det 也能过，可以试试 hack 一下
			for len(q) > 1 && q[len(q)-1].sub(q[len(q)-2]).detCmp(v.sub(q[len(q)-1])) <= 0 {
				q = q[:len(q)-1]
			}
			q = append(q, v)
		}
	}

	return int64(f[n] / 2)
}

//func main() {
//	a := make([]int, 1000)
//	for i := range a {
//		a[i] = 1e4
//	}
//	fmt.Println(minPartitionScore(a, 2))
//}

# Accepted solution for LeetCode #3826: Minimum Partition Score
# Time:  O(n * log(n * r)) = O(nlogn + nlogr), r = max(nums)
# Space: O(n)

import collections


# prefix sum, dp, convex hull trick, wqs binary search, alien trick
class Solution(object):
    def minPartitionScore(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        def binary_search(left, right, check):
            while left <= right:
                mid = left+(right-left)//2
                if check(mid):
                    right = mid-1
                else:
                    left = mid+1
            return left

        def check(l1, l2, l3):
            return (l2[1]-l1[1])*(l2[0]-l3[0]) < (l3[1]-l2[1])*(l1[0]-l2[0])

        def max_lambda():
            mx, total = 0, prefix[-1]*(prefix[-1]+1)//2
            for i in xrange(1, len(nums)):
                c1, c2 = prefix[i], prefix[-1]-prefix[i]
                mx = max(mx, total-(c1*(c1+1)//2+c2*(c2+1)//2))
            return mx

        def f(l):
            dp = cnt = 0
            hull = collections.deque([(0, 0, 0)])
            for i in xrange(len(nums)):
                x = prefix[i+1]
                while len(hull) >= 2 and hull[0][0]*x+hull[0][1] >= hull[1][0]*x+hull[1][1]:
                    hull.popleft()
                dp, cnt = (hull[0][0]*x+hull[0][1])+(x*x+x)//2+l, hull[0][2]+1
                line = (-x, dp+(x*x-x)//2, cnt)
                while len(hull) >= 2 and not check(hull[-2], hull[-1], line):
                    hull.pop()
                hull.append(line)
            return dp, cnt

        prefix = [0]*(len(nums)+1)
        for i in xrange(len(nums)):
            prefix[i+1] = prefix[i]+nums[i]
        mx = max_lambda()
        assert(f(mx+1)[1] == 1)
        l = binary_search(0, mx, lambda x: f(x)[1] <= k)
        return f(l)[0]-k*l


# Time:  O(n * k)
# Space: O(n)
import collections


# prefix sum, dp, convex hull trick
class Solution2(object):
    def minPartitionScore(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        def check(l1, l2, l3):
            return (l2[1]-l1[1])*(l2[0]-l3[0]) < (l3[1]-l2[1])*(l1[0]-l2[0])

        INF = float("inf")
        prefix = [0]*(len(nums)+1)
        for i in xrange(len(nums)):
            prefix[i+1] = prefix[i]+nums[i]
        dp = [INF]*(len(nums)+1)
        dp[0] = 0
        for j in xrange(k):
            new_dp = [INF]*(len(nums)+1)
            hull = collections.deque()
            for i in xrange(j, len(nums)):
                if dp[i] is not INF:
                    x = prefix[i]
                    line = (-x, dp[i]+(x*x-x)//2)
                    while len(hull) >= 2 and not check(hull[-2], hull[-1], line):
                        hull.pop()
                    hull.append(line)
                x = prefix[i+1]
                while len(hull) >= 2 and hull[0][0]*x+hull[0][1] >= hull[1][0]*x+hull[1][1]:
                    hull.popleft()
                new_dp[i+1] = hull[0][0]*x+hull[0][1]+(x*x+x)//2
            dp = new_dp
        return dp[-1]

// Accepted solution for LeetCode #3826: Minimum Partition Score
// Rust example auto-generated from go 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 (go):
// // Accepted solution for LeetCode #3826: Minimum Partition Score
// package main
// 
// import (
// 	"math"
// 	"math/big"
// )
// 
// // 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) dot(b vec) int { return a.x*b.x + a.y*b.y }
// func (a vec) det(b vec) int { return a.x*b.y - a.y*b.x } // 如果乘法会溢出，用 detCmp
// func (a vec) detCmp(b vec) int {
// 	v := new(big.Int).Mul(big.NewInt(int64(a.x)), big.NewInt(int64(b.y)))
// 	w := new(big.Int).Mul(big.NewInt(int64(a.y)), big.NewInt(int64(b.x)))
// 	return v.Cmp(w)
// }
// 
// func minPartitionScore(nums []int, k int) int64 {
// 	n := len(nums)
// 	sum := make([]int, n+1)
// 	for i, x := range nums {
// 		sum[i+1] = sum[i] + x
// 	}
// 
// 	f := make([]int, n+1)
// 	for i := 1; i <= n; i++ {
// 		f[i] = math.MaxInt / 2
// 	}
// 
// 	for K := 1; K <= k; K++ {
// 		s := sum[K-1]
// 		q := []vec{{s, f[K-1] + s*s - s}}
// 		for i := K; i <= n-(k-K); i++ {
// 			s = sum[i]
// 			p := vec{-2 * s, 1}
// 			for len(q) > 1 && p.dot(q[0]) >= p.dot(q[1]) {
// 				q = q[1:]
// 			}
// 
// 			v := vec{s, f[i] + s*s - s}
// 			f[i] = p.dot(q[0]) + s*s + s
// 
// 			// 读者可以把 detCmp 改成 det 感受下这个算法的效率
// 			// 目前 det 也能过，可以试试 hack 一下
// 			for len(q) > 1 && q[len(q)-1].sub(q[len(q)-2]).detCmp(v.sub(q[len(q)-1])) <= 0 {
// 				q = q[:len(q)-1]
// 			}
// 			q = append(q, v)
// 		}
// 	}
// 
// 	return int64(f[n] / 2)
// }
// 
// //func main() {
// //	a := make([]int, 1000)
// //	for i := range a {
// //		a[i] = 1e4
// //	}
// //	fmt.Println(minPartitionScore(a, 2))
// //}

// Accepted solution for LeetCode #3826: Minimum Partition Score
// Auto-generated TypeScript example from go.
function exampleSolution(): void {
}
// Reference (go):
// // Accepted solution for LeetCode #3826: Minimum Partition Score
// package main
// 
// import (
// 	"math"
// 	"math/big"
// )
// 
// // 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) dot(b vec) int { return a.x*b.x + a.y*b.y }
// func (a vec) det(b vec) int { return a.x*b.y - a.y*b.x } // 如果乘法会溢出，用 detCmp
// func (a vec) detCmp(b vec) int {
// 	v := new(big.Int).Mul(big.NewInt(int64(a.x)), big.NewInt(int64(b.y)))
// 	w := new(big.Int).Mul(big.NewInt(int64(a.y)), big.NewInt(int64(b.x)))
// 	return v.Cmp(w)
// }
// 
// func minPartitionScore(nums []int, k int) int64 {
// 	n := len(nums)
// 	sum := make([]int, n+1)
// 	for i, x := range nums {
// 		sum[i+1] = sum[i] + x
// 	}
// 
// 	f := make([]int, n+1)
// 	for i := 1; i <= n; i++ {
// 		f[i] = math.MaxInt / 2
// 	}
// 
// 	for K := 1; K <= k; K++ {
// 		s := sum[K-1]
// 		q := []vec{{s, f[K-1] + s*s - s}}
// 		for i := K; i <= n-(k-K); i++ {
// 			s = sum[i]
// 			p := vec{-2 * s, 1}
// 			for len(q) > 1 && p.dot(q[0]) >= p.dot(q[1]) {
// 				q = q[1:]
// 			}
// 
// 			v := vec{s, f[i] + s*s - s}
// 			f[i] = p.dot(q[0]) + s*s + s
// 
// 			// 读者可以把 detCmp 改成 det 感受下这个算法的效率
// 			// 目前 det 也能过，可以试试 hack 一下
// 			for len(q) > 1 && q[len(q)-1].sub(q[len(q)-2]).detCmp(v.sub(q[len(q)-1])) <= 0 {
// 				q = q[:len(q)-1]
// 			}
// 			q = append(q, v)
// 		}
// 	}
// 
// 	return int64(f[n] / 2)
// }
// 
// //func main() {
// //	a := make([]int, 1000)
// //	for i := range a {
// //		a[i] = 1e4
// //	}
// //	fmt.Println(minPartitionScore(a, 2))
// //}

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.