Rope Cutting Problem in Kotlin

The rope cutting problem is a classic problem in mathematics and computer science that involves cutting a rope into pieces of a certain length. The problem is usually stated as follows:

Given a rope of length L, and a set of integers representing the desired lengths of the pieces, what is the minimum number of cuts required to obtain the desired pieces?

For example, consider a rope of length 10 and the desired piece lengths of 2, 3, and 5. The minimum number of cuts required to obtain the desired pieces would be 3: we could first cut the rope into two pieces of length 5, then cut one of the pieces into two pieces of length 2 and 3, and finally cut the other piece into a piece of length 5 and a piece of length 2.

The rope cutting problem can be solved using dynamic programming. We can define a function cutRope(n: Int, lengths: IntArray): Int that takes the length of the rope n and an array of desired piece lengths lengths, and returns the minimum number of cuts required to obtain the desired pieces.

We can use a dynamic programming approach to solve this problem by building a table cuts of size (lengths.size + 1) x (n + 1), where cuts[i][j] represents the minimum number of cuts required to obtain the desired pieces of lengths lengths[0]...lengths[i-1] from a rope of length j.

We can fill in the table using the following recurrence relation:

cuts[i][j] = min(cuts[i-1][j], cuts[i][j-lengths[i-1]] + 1)

The base case is cuts[0][j] = j, which represents the minimum number of cuts required to obtain a rope of length j from pieces of length 1.

Here is the Kotlin implementation of the cutRope function:

fun cutRope(n: Int, lengths: IntArray): Int {
    val m = lengths.size
    val cuts = Array(m+1) { IntArray(n+1) }

    for (i in 0..m) {
        for (j in 0..n) {
            if (i == 0) {
                cuts[i][j] = j
            } else if (j < lengths[i-1]) {
                cuts[i][j] = cuts[i-1][j]
            } else {
                cuts[i][j] = minOf(cuts[i-1][j], cuts[i][j-lengths[i-1]] + 1)
            }
        }
    }
    return cuts[m][n]
}
fun main() {
    val n = 10
    val lengths = intArrayOf(2, 3, 5)
    val minCuts = cutRope(n, lengths)
    println("The minimum number of cuts required is $minCuts.")
}

The cutRope function takes two arguments: n is the length of the rope, and lengths is an array of the desired piece lengths. The function builds the cuts table using the recurrence relation described above, and returns the value of cuts[m][n], where m is the size of the lengths array.

This program calculates the minimum number of cuts required to obtain the desired pieces from a rope of length 10 and desired piece lengths of 2, 3, and 5. The result is printed as “The minimum number of cuts required is 2.”