Solving Combinatorial Problems Using Go Maps and Slices

Introduction

Hello, coding enthusiast! On our journey to mastering coding and problem-solving, we've arrived at an interesting challenge today. We're going to focus heavily on combinatorial problems . Specifically, we're examining combinatorial problems that involve working with large datasets and multiple pairs of numbers. We'll learn how to solve significant problems efficiently by smartly using data structures like maps and sidestepping expensive operations, such as iterating over large arrays. Are you ready? Let's dive in!

Task Statement

In this unit's task, you'll be given a large slice composed of pairs of distinct, positive integers, including up to 1,000,000 elements. Your challenge is to write a Go function to count the number of indices (i, j) (i != j) where the i-th pair does not share a common element with the j-th pair. A crucial point to remember is that a pair (a, b) is considered identical to (b, a), meaning the order of elements in a pair is irrelevant in this case. It is guaranteed that no two pairs are element-wise equal. For example, given the slice [][]int{{2, 5}, {1, 6}, {3, 2}, {4, 2}, {5, 1}, {6, 3}}, the output should be 8 . The required index pairs are the following: (0, 1), (0, 5), (1, 2), (1, 3), (2, 4), (3, 4), (3, 5), (4, 5) .

Understanding the Solution: The Idea

At the core of our solution, we're going to leverage the power of combinatorics and a smart way of keeping track of occurrences to solve this problem efficiently. The central idea is to calculate the total number of pairs and then subtract from this total the number of pairs that share a common element. This will leave us with the count of pairs that do not share a common element, which is what we're after. Firstly, we calculate the total number of pairs possible in the slice. In a set of n numbers, the number of pairs is given by the formula n * (n - 1) / 2 . This is because each element in the set can pair with every other element, but we divide by 2 because the order of pairs doesn't matter (i.e., pair (a, b) is identical to pair (b, a)). Secondly, we'll count the number of pairs that have at least one common element. To do this, we will use a map to track each number's appearance in the pairs. For each number, we calculate how many pairs it appears in and sum these numbers up.

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package main

import "fmt"

func NonCommonPairs(arr [][]int) int {
    indices := make(map[int][]int)
    totalPairs := len(arr) * (len(arr) - 1) / 2

    // (The rest of the code will be added in subsequent steps)
    return totalPairs // Temporary return to avoid error
}

package main

import "fmt"

func NonCommonPairs(arr [][]int) int {
    indices := make(map[int][]int)
    totalPairs := len(arr) * (len(arr) - 1) / 2

    // (The rest of the code will be added in subsequent steps)
    return totalPairs // Temporary return to avoid error
}

package main

import "fmt"

func NonCommonPairs(arr [][]int) int {
    indices := make(map[int][]int)
    totalPairs := len(arr) * (len(arr) - 1) / 2

    for idx, pair := range arr {
        for _, num := range pair {
            indices[num] = append(indices[num], idx)
        }
    }

    // (The rest of the code will be added in the next step)
    return totalPairs // Temporary return to avoid error
}

package main

import "fmt"

func NonCommonPairs(arr [][]int) int {
    indices := make(map[int][]int)
    totalPairs := len(arr) * (len(arr) - 1) / 2

    for idx, pair := range arr {
        for _, num := range pair {
            indices[num] = append(indices[num], idx)
        }
    }

    // (The rest of the code will be added in the next step)
    return totalPairs // Temporary return to avoid error
}

package main

import "fmt"

func NonCommonPairs(arr [][]int) int {
    indices := make(map[int][]int)
    totalPairs := len(arr) * (len(arr) - 1) / 2

    for idx, pair := range arr {
        for _, num := range pair {
            indices[num] = append(indices[num], idx)
        }
    }

    commonPairs := 0
    for _, occurrences := range indices {
        size := len(occurrences)
        commonPairs += size * (size - 1) / 2
    }

    return totalPairs - commonPairs
}

func main() {
    testArr := [][]int{
        {2, 5},
        {1, 6},
        {3, 2},
        {4, 2},
        {5, 1},
        {6, 3},
    }
    fmt.Println("Count of non-common pairs:", NonCommonPairs(testArr)) // Output: Count of non-common pairs: 8
}

package main

import "fmt"

func NonCommonPairs(arr [][]int) int {
    indices := make(map[int][]int)
    totalPairs := len(arr) * (len(arr) - 1) / 2

    for idx, pair := range arr {
        for _, num := range pair {
            indices[num] = append(indices[num], idx)
        }
    }

    commonPairs := 0
    for _, occurrences := range indices {
        size := len(occurrences)
        commonPairs += size * (size - 1) / 2
    }

    return totalPairs - commonPairs
}

func main() {
    testArr := [][]int{
        {2, 5},
        {1, 6},
        {3, 2},
        {4, 2},
        {5, 1},
        {6, 3},
    }
    fmt.Println("Count of non-common pairs:", NonCommonPairs(testArr)) // Output: Count of non-common pairs: 8
}

package main

import "fmt"

func NonCommonPairs(arr [][]int) int {
    indices := make(map[int][]int)
    totalPairs := len(arr) * (len(arr) - 1) / 2

    // (The rest of the code will be added in subsequent steps)
    return totalPairs // Temporary return to avoid error
}

package main

import "fmt"

func NonCommonPairs(arr [][]int) int {
    indices := make(map[int][]int)
    totalPairs := len(arr) * (len(arr) - 1) / 2

    for idx, pair := range arr {
        for _, num := range pair {
            indices[num] = append(indices[num], idx)
        }
    }

    // (The rest of the code will be added in the next step)
    return totalPairs // Temporary return to avoid error
}

package main

import "fmt"

func NonCommonPairs(arr [][]int) int {
    indices := make(map[int][]int)
    totalPairs := len(arr) * (len(arr) - 1) / 2

    for idx, pair := range arr {
        for _, num := range pair {
            indices[num] = append(indices[num], idx)
        }
    }

    commonPairs := 0
    for _, occurrences := range indices {
        size := len(occurrences)
        commonPairs += size * (size - 1) / 2
    }

    return totalPairs - commonPairs
}

func main() {
    testArr := [][]int{
        {2, 5},
        {1, 6},
        {3, 2},
        {4, 2},
        {5, 1},
        {6, 3},
    }
    fmt.Println("Count of non-common pairs:", NonCommonPairs(testArr)) // Output: Count of non-common pairs: 8
}

Introduction

Task Statement

Understanding the Solution: The Idea

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Initialize Data Structures and Calculate Total Pairs

Populate Indices Map with Pair Data

Calculate and Subtract Common Pairs

Lesson Summary