Problem:
Max Score: 350 Points

Problem Statement

There are N buildings along the line. The i-th building from the left is colored in color i, and its height is currently a_i meters.

Chokudai is a mayor of the city, and he loves colorful thigs. And now he wants to see at least K buildings from the left.

You can increase height of buildings, but it costs 1 yens to increase 1 meters. It means you cannot make building that height is not integer.

You cannot decrease height of buildings.

Calculate the minimum cost of satisfying Chokudai's objective.

Note: "Building i can see from the left" means there are no j exists that (height of building j) ≥ (height of building i) and j < i.

Input Format

N K
a_1 a_2 a_3 ... a_N

Output Format

Print the minimum cost in one line. In the end put a line break.

Constraints

1 ≤ K ≤ N ≤ 15

1 ≤ a_i ≤ 10^9

Scoring

Subtask 1 [120 points]

N = K

Subtask 2 [90 points]

N ≤ 5

a_i ≤ 7

Subtask 3 [140 points]

There are no additional constraints.

Sample Input 1

5 5
3949 3774 3598 3469 3424

Sample Output 1

1541

The optimal solution is (height of buildings from the left) = [3949, 3950, 3951, 3952, 3953].

Sample Input 2

5 3
7 4 2 6 4

Sample Output 2

7

The optimal solution is (height of buildings from the left) = [7, 8, 2, 9, 4].

Solution in Kotlin (1.0.0) (judge verdict: Wrong Answer)

import java.util.*

fun main(args: Array<String>) = p14()

fun p14() {
    val (n, k) = readLine()!!.split(' ').map { it.toInt() }
    val aList = readLine()!!.split(' ').map { it.toLong() }

    val maxes = LongArray(n)
    maxes[0] = aList.first()
    for (i in 1 until n) maxes[i] = Math.max(maxes[i - 1], aList[i])

    var answer = Long.MAX_VALUE

    for (i in 0 until (1L shl n)) {
        val bitSet = BitSet.valueOf(longArrayOf(i))
        if (bitSet.cardinality() != k) continue

        val withIndex = aList.withIndex()
        val filtered = withIndex.filter { bitSet[it.index] }
            .map { it.value }

        var sum = 0L
        var currHeight = withIndex.first().let {
            it.value + if (it.index == 0 || it.value > maxes[it.index - 1]) 0 else {
                val add = maxes[it.index - 1] - it.value + 1
                sum += add
                add
            }
        }

        for (f in filtered.drop(1)) {
            val heightDiff = f - currHeight
            if (heightDiff <= 0) {
                sum += -heightDiff + 1
                currHeight++
            } else {
                currHeight = f
            }
        }
        answer = Math.min(answer, sum)
    }

    println(answer)
}