LeetCode in Kotlin

2547. Minimum Cost to Split an Array

Hard

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

Solution

class Solution {
    fun minCost(nums: IntArray, k: Int): Int {
        val n = nums.size
        val dp = IntArray(n)
        dp.fill(-1)
        val len = Array(n) { IntArray(n) }
        for (r in len) r.fill(0)
        for (i in 0 until n) {
            val count = IntArray(n)
            count.fill(0)
            var c = 0
            for (j in i until n) {
                count[nums[j]] += 1
                if (count[nums[j]] == 2) c += 2 else if (count[nums[j]] > 2) c += 1
                len[i][j] = c
            }
        }
        return f(0, nums, k, len, dp)
    }

    private fun f(ind: Int, nums: IntArray, k: Int, len: Array<IntArray>, dp: IntArray): Int {
        if (ind >= nums.size) return 0
        if (dp[ind] != -1) return dp[ind]
        dp[ind] = Int.MAX_VALUE
        for (i in ind until nums.size) {
            val current = len[ind][i] + k
            val next = f(i + 1, nums, k, len, dp)
            dp[ind] = dp[ind].coerceAtMost(current + next)
        }
        return dp[ind]
    }
}

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