2547. Minimum Cost to Split an Array

Description

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

Solutions

Solution 1: Memoization Search

We design a function $dfs(i)$, which represents the minimum cost of splitting from index $i$. So the answer is $dfs(0)$.

The calculation process of the function $dfs(i)$ is as follows:

If $i \ge n$, it means that the splitting has reached the end of the array, and $0$ is returned at this time. Otherwise, we enumerate the end $j$ of the subarray. During the process, we use an array or hash table cnt to count the number of times each number appears in the subarray, and use a variable one to count the number of numbers in the subarray that appear once. So the importance of the subarray is $k + j - i + 1 - one$, and the cost of splitting is $k + j - i + 1 - one + dfs(j + 1)$. We enumerate all $j$ and take the minimum value as the return value of $dfs(i)$. During the process, we can use memoization search, that is, use an array $f$ to memorize the return value of the function $dfs(i)$ to avoid repeated calculations.

The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $nums$.

• 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;
}
}

• class Solution {
public:
int minCost(vector<int>& nums, int k) {
int n = nums.size();
int f[n];
memset(f, 0, sizeof f);
function<int(int)> dfs = [&](int i) {
if (i >= n) {
return 0;
}
if (f[i]) {
return f[i];
}
int cnt[n];
memset(cnt, 0, sizeof cnt);
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 = min(ans, k + j - i + 1 - one + dfs(j + 1));
}
return f[i] = ans;
};
return dfs(0);
}
};

• 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)


• 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)
}

• 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);
}