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Decode the Coding Interview in Go: Real-World Examples

Feature #11: Generate Movie Viewing Orders

Explore how to generate every possible order of movies for Netflix marathons using the backtracking algorithm. Understand the recursive approach in Go, its time complexity of factorial n, and space complexity related to stack depth. This lesson equips you to handle permutation problems commonly seen in coding interviews.

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Description

We want to offer marathons for our viewers. Each marathon will have a fixed set of movies catering to a specific taste. For a given marathon, different viewing orders of the movies will yield different user satisfaction results. We want to experiment (A/B testing) with different viewing orders of the same marathon.

Your task is to generate all the possible permutations of movies in a given marathon.

Let’s look at an example to better understand this:

Solution

To solve this problem, we will use the backtracking approach.

We will assume a Backtrack function that takes the index of the first movie to consider as an argument Backtrack(first).

  • If the first movie to consider has index n, then that means that the current permutation is done.

  • We will iterate over the marathon from index First to index n - 1.

  • We will place the ith movie first in the permutation, that is, Movies[First], Movies[i] = Movies[i], Movies[First].

  • We will proceed to create all the permutations that start from the ith movie: Backtrack(First + 1).

  • Now we will backtrack, that is, Movies[First], Movies[i] = Movies[i], Movies[First] back.

Let’s look at some illustrations to better understand this:

Note: We are using numbers to represent movies in the following illustration.

Let’s look at the code for this solution:

Go (1.16.5)
package main
import (
   "fmt"
)
func GeneratePermutations(Movies []string) [][]string {
    
    Size := len(Movies)
    var Backtrack func(int)
    var Output [][]string
    Backtrack = func(First int){
        /* If all integers of given array `Movies` are used and 
        and Backtracking is performed add the permutations to Output array. */
        if First == Size {
            temp := make([]string,len(Movies[:]))
            copy(temp,Movies[:])
            Output = append(Output, temp)
        }
        /* Perform Backtracking for the Size of a given array. */
        for i := First; i < Size; i++{
            /* Swap: In the current permutaion place i-th integer first. */
            Movies[First], Movies[i] = Movies[i], Movies[First] 
            /* Complete permuations using the next integers. */
            Backtrack(First+1)
            /* Swap and Backtrack */
            Movies[First], Movies[i] = Movies[i], Movies[First]
        }
    }
    Backtrack(0)
    return Output
}
func main() {
    // Example #1
    Input := []string{"Frozen","Dune","Coco"}
    Output := GeneratePermutations(Input)
    fmt.Println("Output 1:",Output)
    // Example #2
    Input = []string{"Frozen","Dune","Coco","Melificient"}
    Output = GeneratePermutations(Input)
    fmt.Println("Output 2:",Output)
    // Example #3
    Input = []string{"Dune","Coco"}
    Output = GeneratePermutations(Input)
    fmt.Println("Output 3:",Output)
}
Generate movie viewing orders

Complexity measures

Time complexity
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