Hello, coding enthusiast! On our journey to master programming and problem-solving, we've embraced an exciting challenge today. We’re going to delve into combinatorial problems using Kotlin's distinct features. Our focus is on efficiently solving these problems with large datasets containing multiple pairs of numbers. Throughout this lesson, you'll discover how to implement efficient solutions using Kotlin's powerful data structures like HashMap and avoid time-consuming operations like iterating over large collections. Excited? Let's dive in!

• Input: You are given a list of pairs where each pair consists of distinct, positive integers. The list can contain up to 1,000,000 pairs, and no two pairs are element-wise identical.

• Output: The goal is to find the number of index pairs (i, j) (where i ≠ j) such that the i-th pair does not share a common member with the j-th pair.

• Explanation:

  • A pair (a, b) is considered non-common with another pair (c, d) if they do not share any member, meaning none of a or b is equal to c or d.
  • Pairs (2, 5) and (1, 6) are non-common because they do not share any member.
  • Conversely, pairs (2, 5) and (3, 2) are common because they share the member 2.

• Example:

  • Given the list [(2, 5), (1, 6), (3, 2), (4, 2), (5, 1), (6, 3)], the output should be 8.
  • This is because the non-common index pairs (i, j) are: (0, 1), (0, 5), (1, 2), (1, 3), (2, 4), (3, 4), (3, 5), and (4, 5).

To address this problem efficiently, we take advantage of combinatorial concepts and Kotlin's data structures. Our strategy involves computing the total possible index pairs and then subtracting pairs that share a common element. This will yield the count of pairs without common elements.

First, calculate the total number of index pairs possible with the list. In a set of n elements, the number of combinations is given by the formula n * (n - 1) / 2. This works because each element can pair with every other element once, disregarding order since (a, b) is considered identical to (b, a).

Next, we'll count index pairs that do share at least one common element. We'll use a HashMap in Kotlin to track occurrences of each number in the pairs. For each number, calculate how many pairs it occurs in and sum these occurrences.

To start, we'll need an efficient way to store the occurrence of each number within the pairs. Kotlin's mutableMapOf is perfect for this. It allows us to quickly track occurrences and build our solution from there.

Next, compute the total number of pairs using the n * (n - 1) / 2 formula. This value will be integral to our final calculation.

Let's initialize an empty mutableMapOf and compute the total pairs.

Kotlin
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1fun nonCommonPairs(arr: List<Pair<Int, Int>>): Int {
2    val indices = mutableMapOf<Int, MutableList<Int>>()
3    val totalPairs = arr.size * (arr.size - 1) / 2
4
5    // (The rest of the code will be added in subsequent steps)
6    return totalPairs // Temporary return to avoid compilation error
7}
8
9fun main() {
10    // Testing code will be added later
11}

Moving on, we will populate the mutableMapOf by iterating through each pair. For each pair, examine both elements, adding the current index to the list of indices for this number (starting a new list if necessary).

Here's how we adjust our function to accomplish this:

Kotlin
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1fun nonCommonPairs(arr: List<Pair<Int, Int>>): Int {
2    val indices = mutableMapOf<Int, MutableList<Int>>()
3    val totalPairs = arr.size * (arr.size - 1) / 2
4
5    arr.forEachIndexed { idx, pair ->
6        listOf(pair.first, pair.second).forEach { num ->
7            indices.computeIfAbsent(num) { mutableListOf() }.add(idx)
8        }
9    }
10
11    // (The rest of the code will be added in the next step)
12    return totalPairs // Temporary return to avoid compilation error
13}
14
15fun main() {
16    // Testing code will be added later
17}

With the data set up, our final task is to compute the total number of index pairs that share at least one common element. By examining each number and counting how frequently they occur in different pairs, we use our formula to find the common pairs.

Subtract these pairs from the total pairs to get our answer—the count of non-common pairs:

Kotlin
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1fun nonCommonPairs(arr: List<Pair<Int, Int>>): Int {
2    val indices = mutableMapOf<Int, MutableList<Int>>()
3    val totalPairs = arr.size * (arr.size - 1) / 2
4
5    arr.forEachIndexed { idx, pair ->
6        listOf(pair.first, pair.second).forEach { num ->
7            indices.computeIfAbsent(num) { mutableListOf() }.add(idx)
8        }
9    }
10
11    var commonPairs = 0
12    for (entry in indices.entries) {
13        val size = entry.value.size
14        commonPairs += size * (size - 1) / 2 // Compute the number of common index pairs for this member
15    }
16
17    return totalPairs - commonPairs
18}
19
20fun main() {
21    val arr = listOf(
22        Pair(2, 5),
23        Pair(1, 6),
24        Pair(3, 2),
25        Pair(4, 2),
26        Pair(5, 1),
27        Pair(6, 3)
28    )
29
30    println("Count of non-common pairs: ${nonCommonPairs(arr)}")
31}

Excellent work! Today's challenge was a demanding one, and you've navigated it brilliantly using Kotlin's strengths. You utilized a mutableMapOf to succinctly manage occurrences in a large dataset, and you applied combinatorial logic to isolate and subtract unwanted cases from total possibilities. This approach provided a solution that operates efficiently.

These techniques will serve you well in tackling similar complex problems in the future. As demonstrated, the key to handling large data sets is the application of clever techniques that optimize computation, just like we've achieved today.

It's time to solidify your understanding by working through practice problems related to today's lesson. Dive in to reinforce these concepts. Happy coding!