Missing Permutations in a list

Given a list of permutations of any word. Find the missing permutation from the list of permutations.

Examples:

Input : Permutation_given[] = {"ABCD", "CABD", "ACDB", 
              "DACB", "BCDA", "ACBD", "ADCB", "CDAB",
              "DABC", "BCAD", "CADB", "CDBA", "CBAD", 
              "ABDC", "ADBC", "BDCA", "DCBA", "BACD",
              "BADC", "BDAC", "CBDA", "DCAB"};
Output : DBAC DBCA

Approach:

We create a set of all given strings.
And one more set of all permutations.
Finally return difference between the two sets.

Implementation:

CPP

#include <bits/stdc++.h>
using namespace std;

void find_missing_strings(string Permutation_given[],
                          size_t Size_Permutation_given)
{
    // vector "permutation" containing all
    // the permutation of input string
    vector<string> permutations;

    // Here we can take any string
    // from the given list and do
    // the necessary permutation
    string input = Permutation_given[0];
    permutations.push_back(input);

    // In the loop we will store
    // all the permutations of the string
    // in the vector "permutation".
    while (true) {

        string p = permutations.back();

        // Getting next permutation of input string
        next_permutation(p.begin(), p.end());
        if (p == permutations.front())
            break;

        permutations.push_back(p);
    }

    // vector containing all the
    // missing strings in permutation
    vector<string> missing;

    // given_permutations contains the
    // permutation of the input string
    set<string> given_permutations(
        Permutation_given,
        Permutation_given + Size_Permutation_given);

    // Through the set difference we will get
    // the missing words in vector missing
    set_difference(permutations.begin(), permutations.end(),
                   given_permutations.begin(),
                   given_permutations.end(),
                   back_inserter(missing));

    // printing all the missing string
    for (auto i = missing.begin(); i != missing.end(); ++i)
        cout << *i << endl;
}

// Driver code
int main()
{
    string Permutation_given[]
        = { "ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD",
            "ADCB", "CDAB", "DABC", "BCAD", "CADB", "CDBA",
            "CBAD", "ABDC", "ADBC", "BDCA", "DCBA", "BACD",
            "BADC", "BDAC", "CBDA", "DCAB" };

    // size of permutation list
    size_t Size_Permutation_given
        = sizeof(Permutation_given)
          / sizeof(*Permutation_given);

    find_missing_strings(Permutation_given,
                         Size_Permutation_given);

    return 0;
}

Java

import java.util.Arrays;
import java.util.HashSet;
import java.util.Set;

public class MissingStrings {
    public static void
        findMissingStrings(String[] permutationGiven)
    {
        // set containing all the permutations of the input
        // string
        Set<String> permutations = new HashSet<String>();
        char[] chars = permutationGiven[0].toCharArray();
        Arrays.sort(chars);
        permutation(permutations, chars, 0,
                    chars.length - 1);

        // set containing the permutation of the input
        // strings
        Set<String> givenPermutations = new HashSet<String>(
            Arrays.asList(permutationGiven));

        // set difference to get the missing strings
        Set<String> missing
            = new HashSet<String>(permutations);
        missing.removeAll(givenPermutations);

        // printing all the missing strings
        for (String s : missing) {
            System.out.println(s);
        }
    }

    // recursive function to find permutations of a string
    public static void permutation(Set<String> permutations,
                                   char[] chars, int l,
                                   int r)
    {
        if (l == r) {
            permutations.add(new String(chars));
        }
        else {
            for (int i = l; i <= r; i++) {
                swap(chars, l, i);
                permutation(permutations, chars, l + 1, r);
                swap(chars, l, i);
            }
        }
    }

    // function to swap two characters of a char array
    public static void swap(char[] chars, int i, int j)
    {
        char temp = chars[i];
        chars[i] = chars[j];
        chars[j] = temp;
    }

    // Driver code
    public static void main(String[] args)
    {
        String[] permutationGiven
            = { "ABCD", "CABD", "ACDB", "DACB", "BCDA",
                "ACBD", "ADCB", "CDAB", "DABC", "BCAD",
                "CADB", "CDBA", "CBAD", "ABDC", "ADBC",
                "BDCA", "DCBA", "BACD", "BADC", "BDAC",
                "CBDA", "DCAB" };

        findMissingStrings(permutationGiven);
    }
}

Python3

import itertools


def find_missing_strings(permutation_given):
    # set containing all the permutations of the input string
    permutations = set([''.join(p)
                        for p in itertools.permutations(permutation_given[0])])

    # set containing the permutation of the input strings
    given_permutations = set(permutation_given)

    # set difference to get the missing strings
    missing = permutations - given_permutations

    # printing all the missing strings
    for s in missing:
        print(s)


# Driver code
if __name__ == '__main__':
    permutation_given = [
        "ABCD", "CABD", "ACDB", "DACB",
        "BCDA", "ACBD", "ADCB", "CDAB",
        "DABC", "BCAD", "CADB", "CDBA",
        "CBAD", "ABDC", "ADBC", "BDCA",
        "DCBA", "BACD", "BADC", "BDAC",
        "CBDA", "DCAB"
    ]

    find_missing_strings(permutation_given)

using System;
using System.Collections.Generic;
using System.Linq;

class MissingStrings {
    public static void
        FindMissingStrings(string[] permutationGiven)
    {
        // set containing all the permutations of the input
        // string
        HashSet<string> permutations
            = new HashSet<string>();
        char[] chars = permutationGiven[0].ToCharArray();
        Array.Sort(chars);
        Permutation(permutations, chars, 0,
                    chars.Length - 1);

        // set containing the permutation of the input
        // strings
        HashSet<string> givenPermutations
            = new HashSet<string>(permutationGiven);

        // set difference to get the missing strings
        HashSet<string> missing
            = new HashSet<string>(permutations);
        missing.ExceptWith(givenPermutations);

        // printing all the missing strings
        foreach(string s in missing)
        {
            Console.WriteLine(s);
        }
    }

    // recursive function to find permutations of a string
    public static void
    Permutation(HashSet<string> permutations, char[] chars,
                int l, int r)
    {
        if (l == r) {
            permutations.Add(new string(chars));
        }
        else {
            for (int i = l; i <= r; i++) {
                Swap(chars, l, i);
                Permutation(permutations, chars, l + 1, r);
                Swap(chars, l, i);
            }
        }
    }

    // function to swap two characters of a char array
    public static void Swap(char[] chars, int i, int j)
    {
        char temp = chars[i];
        chars[i] = chars[j];
        chars[j] = temp;
    }

    // Driver code
    public static void Main(string[] args)
    {
        string[] permutationGiven
            = { "ABCD", "CABD", "ACDB", "DACB", "BCDA",
                "ACBD", "ADCB", "CDAB", "DABC", "BCAD",
                "CADB", "CDBA", "CBAD", "ABDC", "ADBC",
                "BDCA", "DCBA", "BACD", "BADC", "BDAC",
                "CBDA", "DCAB" };

        FindMissingStrings(permutationGiven);
    }
}

JavaScript

function findMissingStrings(PermutationGiven) {
    // Array "permutations" containing all the permutations of input string
    let permutations = [];

    // Here we can take any string from the given list and do the necessary permutation
    let input = PermutationGiven[0];
    permutations.push(input);

    // In the loop, we will store all the permutations of the string in the array "permutations"
    while (true) {
        let p = permutations[permutations.length - 1];

        // Getting the next permutation of the input string
        p = p.split('');
        let nextPerm = getNextPermutation(p);
        if (nextPerm.join('') === permutations[0])
            break;

        permutations.push(nextPerm.join(''));
    }

    // Array containing all the missing strings in permutations
    let missing = [];

    // Convert PermutationGiven to a Set
    let givenPermutations = new Set(PermutationGiven);

    // Find the missing words by performing set difference
    for (let i = 0; i < permutations.length; i++) {
        if (!givenPermutations.has(permutations[i])) {
            missing.push(permutations[i]);
        }
    }

    // Printing all the missing strings
    for (let i = 0; i < missing.length; i++) {
        console.log(missing[i]);
    }
}

// Function to get the next permutation of a string
function getNextPermutation(arr) {
    let i = arr.length - 2;
    while (i >= 0 && arr[i] >= arr[i + 1]) {
        i--;
    }

    if (i === -1) {
        return arr.reverse();
    }

    let j = arr.length - 1;
    while (arr[j] <= arr[i]) {
        j--;
    }

    swap(arr, i, j);

    let left = i + 1;
    let right = arr.length - 1;
    while (left < right) {
        swap(arr, left, right);
        left++;
        right--;
    }

    return arr;
}

// Function to swap two elements in an array
function swap(arr, i, j) {
    let temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
}

// Driver code
const PermutationGiven = [
    "ABCD", "CABD", "ACDB", "DACB", "BCDA", "ACBD",
    "ADCB", "CDAB", "DABC", "BCAD", "CADB", "CDBA",
    "CBAD", "ABDC", "ADBC", "BDCA", "DCBA", "BACD",
    "BADC", "BDAC", "CBDA", "DCAB"
];

findMissingStrings(PermutationGiven);

Output

DBAC
DBCA

Time Complexity: O(n!) where n is the length of the first string in the input
Auxiliary Space: O(n!), because we need to store all possible permutations of the input string in memory

Missing Permutations in a list

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