Mastering Array Manipulation and the Two-Pointer Technique in Kotlin

Hello there! Are you ready to solve another engaging problem today? We have a practical task that will enhance your problem-solving skills. It involves critical aspects of programming — dealing with arrays and using techniques such as sorting and the two-pointer method. So, let's jump in!

Alright, our task is as follows. Suppose you have two equally long arrays, $A$ and $B$, with a length varying from $1$ to $1000$, with each element being a unique positive integer ranging from $1$ up to $10^6$. Your challenge is to craft a Kotlin function that identifies the closest number in array $B$ to $2 \cdot B[i]$ for each $i$. Once this number is identified, say, for the specific $i$, it is $B[j]$. We want to create an array from $A[j]$s in the order of increasing $i$.

To illustrate this, let's consider an example. We have:

val A = intArrayOf(10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110)
val B = intArrayOf(4, 12, 3, 9, 6, 1, 5, 8, 37, 25, 100)

After running your function, the resulting array should look like this:

val result = intArrayOf(80, 100, 50, 20, 20, 60, 40, 20, 110, 90, 110)

Let's walk through the first few steps:

The first item in $B$ is $4$ at $index=0$. Double this number is $8$. The closest number to $8$ in array $B$ is $8$, which is at $index=7$. The number at the same index in array $A$ is $80$, so we add $80$ to our new array.

The second item in $B$ is $12$ at $index=1$. Double this number is $24$. The closest number to $24$ in $B$ is $25$, which is at $index=9$. The corresponding index in $A$ has the number $100$. So, we add $100$ to our new array.

The third item in $B$ is $3$ at $index=2$. Double this number is $6$. The closest number to $6$ in $B$ is $6$, which is at $index=4$. The corresponding index in $A$ has the number $50$. So, we add $50$ to our new array.

We continue this process for the rest of the elements in $B$.

Let's embark on our solution-building journey by constructing a sorted array for array B. This array will include pairs of values value and their corresponding indices index from array B. Here, value represents the element in B, while index denotes the index at which value is found in array B.

This sorted array will be similar to an associative array, storing "value-index" pairs. It not only organizes the data for efficient retrieval but also makes it easier for us to traverse the array. Here's the introductory part of our Kotlin function, including the complete sorted array:

fun findAndReplace(A: IntArray, B: IntArray): IntArray {
    val B_sorted = B.mapIndexed { index, value -> index to value }
        .sortedBy { it.second }

In the above code, we generate an array of pairs comprising the values from B and their respective indices using a mapIndexed function. Then, the sortedBy function arranges these pairs in ascending order of their values.