Efficient Combinatorial Problem Solving with Kotlin

Introduction

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!

Task Statement

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 \neq 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) .

Understanding the Solution: The Idea

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.

Join the 1M+ learners on CodeSignal

Be a part of our community of 1M+ users who develop and demonstrate their skills on CodeSignal

Solution Building: Step 3

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

fun nonCommonPairs(arr: List<Pair<Int, Int>>): Int {
    val indices = mutableMapOf<Int, MutableList<Int>>()
    val totalPairs = arr.size * (arr.size - 1) / 2

    arr.forEachIndexed { idx, pair ->
        listOf(pair.first, pair.second).forEach { num ->
            indices.computeIfAbsent(num) { mutableListOf() }.add(idx)
        }
    }

    var commonPairs = 0
    for (entry in indices.entries) {
        val size = entry.value.size
        commonPairs += size * (size - 1) / 2 // Compute the number of common index pairs for this member
    }

    return totalPairs - commonPairs
}

fun main() {
    val arr = listOf(
        Pair(2, 5),
        Pair(1, 6),
        Pair(3, 2),
        Pair(4, 2),
        Pair(5, 1),
        Pair(6, 3)
    )

    println("Count of non-common pairs: ${nonCommonPairs(arr)}")
}

fun nonCommonPairs(arr: List<Pair<Int, Int>>): Int {
    val indices = mutableMapOf<Int, MutableList<Int>>()
    val totalPairs = arr.size * (arr.size - 1) / 2

    arr.forEachIndexed { idx, pair ->
        listOf(pair.first, pair.second).forEach { num ->
            indices.computeIfAbsent(num) { mutableListOf() }.add(idx)
        }
    }

    var commonPairs = 0
    for (entry in indices.entries) {
        val size = entry.value.size
        commonPairs += size * (size - 1) / 2 // Compute the number of common index pairs for this member
    }

    return totalPairs - commonPairs
}

fun main() {
    val arr = listOf(
        Pair(2, 5),
        Pair(1, 6),
        Pair(3, 2),
        Pair(4, 2),
        Pair(5, 1),
        Pair(6, 3)
    )

    println("Count of non-common pairs: ${nonCommonPairs(arr)}")
}

Introduction

Task Statement

Understanding the Solution: The Idea

Join the 1M+ learners on CodeSignal

Solution Building: Step 1

Solution Building: Step 2

Solution Building: Step 3

Lesson Summary