LeetCode #2547 — HARD

Minimum Cost to Split an Array

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.

Split the array into some number of non-empty subarrays. The cost of a split is the sum of the importance value of each subarray in the split.

Let trimmed(subarray) be the version of the subarray where all numbers which appear only once are removed.

For example, trimmed([3,1,2,4,3,4]) = [3,4,3,4].

The importance value of a subarray is k + trimmed(subarray).length.

For example, if a subarray is [1,2,3,3,3,4,4], then trimmed([1,2,3,3,3,4,4]) = [3,3,3,4,4].The importance value of this subarray will be k + 5.

Return the minimum possible cost of a split of nums.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [1,2,1,2,1,3,3], k = 2
Output: 8
Explanation: We split nums to have two subarrays: [1,2], [1,2,1,3,3].
The importance value of [1,2] is 2 + (0) = 2.
The importance value of [1,2,1,3,3] is 2 + (2 + 2) = 6.
The cost of the split is 2 + 6 = 8. It can be shown that this is the minimum possible cost among all the possible splits.

Example 2:

Input: nums = [1,2,1,2,1], k = 2
Output: 6
Explanation: We split nums to have two subarrays: [1,2], [1,2,1].
The importance value of [1,2] is 2 + (0) = 2.
The importance value of [1,2,1] is 2 + (2) = 4.
The cost of the split is 2 + 4 = 6. It can be shown that this is the minimum possible cost among all the possible splits.

Example 3:

Input: nums = [1,2,1,2,1], k = 5
Output: 10
Explanation: We split nums to have one subarray: [1,2,1,2,1].
The importance value of [1,2,1,2,1] is 5 + (3 + 2) = 10.
The cost of the split is 10. It can be shown that this is the minimum possible cost among all the possible splits.

Constraints:

1 <= nums.length <= 1000
0 <= nums[i] < nums.length
1 <= k <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums and an integer k. Split the array into some number of non-empty subarrays. The cost of a split is the sum of the importance value of each subarray in the split. Let trimmed(subarray) be the version of the subarray where all numbers which appear only once are removed. For example, trimmed([3,1,2,4,3,4]) = [3,4,3,4]. The importance value of a subarray is k + trimmed(subarray).length. For example, if a subarray is [1,2,3,3,3,4,4], then trimmed([1,2,3,3,3,4,4]) = [3,3,3,4,4].The importance value of this subarray will be k + 5. Return the minimum possible cost of a split of nums. A subarray is a contiguous non-empty sequence of elements within an array.

Baseline thinking

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

Pattern signal: Array · Hash Map · Dynamic Programming

Example 1

[1,2,1,2,1,3,3]
2

Example 2

[1,2,1,2,1]
2

Example 3

[1,2,1,2,1]
5

Core Insight

What unlocks the optimal approach

Let's denote dp[r] = minimum cost to partition the first r elements of nums. What would be the transitions of such dynamic programming?
dp[r] = min(dp[l] + importance(nums[l..r])) over all 0 <= l < r. This already gives us an O(n^3) approach, as importance can be calculated in linear time, and there are a total of O(n^2) transitions.
Can you think of a way to compute multiple importance values of related subarrays faster?
importance(nums[l-1..r]) is either importance(nums[l..r]) if a new unique element is added, importance(nums[l..r]) + 1 if an old element that appeared at least twice is added, or importance(nums[l..r]) + 2, if a previously unique element is duplicated. This allows us to compute importance(nums[l..r]) for all 0 <= l < r in O(n) by keeping a frequency table and decreasing l from r-1 down to 0.

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 #2547: Minimum Cost to Split an Array
class Solution {
    private Integer[] f;
    private int[] nums;
    private int n, k;

    public int minCost(int[] nums, int k) {
        n = nums.length;
        this.k = k;
        this.nums = nums;
        f = new Integer[n];
        return dfs(0);
    }

    private int dfs(int i) {
        if (i >= n) {
            return 0;
        }
        if (f[i] != null) {
            return f[i];
        }
        int[] cnt = new int[n];
        int one = 0;
        int ans = 1 << 30;
        for (int j = i; j < n; ++j) {
            int x = ++cnt[nums[j]];
            if (x == 1) {
                ++one;
            } else if (x == 2) {
                --one;
            }
            ans = Math.min(ans, k + j - i + 1 - one + dfs(j + 1));
        }
        return f[i] = ans;
    }
}

// Accepted solution for LeetCode #2547: Minimum Cost to Split an Array
func minCost(nums []int, k int) int {
	n := len(nums)
	f := make([]int, n)
	var dfs func(int) int
	dfs = func(i int) int {
		if i >= n {
			return 0
		}
		if f[i] > 0 {
			return f[i]
		}
		ans, one := 1<<30, 0
		cnt := make([]int, n)
		for j := i; j < n; j++ {
			cnt[nums[j]]++
			x := cnt[nums[j]]
			if x == 1 {
				one++
			} else if x == 2 {
				one--
			}
			ans = min(ans, k+j-i+1-one+dfs(j+1))
		}
		f[i] = ans
		return ans
	}
	return dfs(0)
}

# Accepted solution for LeetCode #2547: Minimum Cost to Split an Array
class Solution:
    def minCost(self, nums: List[int], k: int) -> int:
        @cache
        def dfs(i):
            if i >= n:
                return 0
            cnt = Counter()
            one = 0
            ans = inf
            for j in range(i, n):
                cnt[nums[j]] += 1
                if cnt[nums[j]] == 1:
                    one += 1
                elif cnt[nums[j]] == 2:
                    one -= 1
                ans = min(ans, k + j - i + 1 - one + dfs(j + 1))
            return ans

        n = len(nums)
        return dfs(0)

// Accepted solution for LeetCode #2547: Minimum Cost to Split an Array
// 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 #2547: Minimum Cost to Split an Array
// class Solution {
//     private Integer[] f;
//     private int[] nums;
//     private int n, k;
// 
//     public int minCost(int[] nums, int k) {
//         n = nums.length;
//         this.k = k;
//         this.nums = nums;
//         f = new Integer[n];
//         return dfs(0);
//     }
// 
//     private int dfs(int i) {
//         if (i >= n) {
//             return 0;
//         }
//         if (f[i] != null) {
//             return f[i];
//         }
//         int[] cnt = new int[n];
//         int one = 0;
//         int ans = 1 << 30;
//         for (int j = i; j < n; ++j) {
//             int x = ++cnt[nums[j]];
//             if (x == 1) {
//                 ++one;
//             } else if (x == 2) {
//                 --one;
//             }
//             ans = Math.min(ans, k + j - i + 1 - one + dfs(j + 1));
//         }
//         return f[i] = ans;
//     }
// }

// Accepted solution for LeetCode #2547: Minimum Cost to Split an Array
function minCost(nums: number[], k: number): number {
    const n = nums.length;
    const f = new Array(n).fill(0);
    const dfs = (i: number) => {
        if (i >= n) {
            return 0;
        }
        if (f[i]) {
            return f[i];
        }
        const cnt = new Array(n).fill(0);
        let one = 0;
        let ans = 1 << 30;
        for (let j = i; j < n; ++j) {
            const x = ++cnt[nums[j]];
            if (x == 1) {
                ++one;
            } else if (x == 2) {
                --one;
            }
            ans = Math.min(ans, k + j - i + 1 - one + dfs(j + 1));
        }
        f[i] = ans;
        return f[i];
    };
    return dfs(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^2)

Space

O(n)

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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.