Find total number of Permutations such that every element becomes an Extrema

Given an array arr[] of size N  non-negative integers, the task is to find the total number of permutations of arr[] such that each element in the array is either strictly greater or smaller than all the elements before it i.e., arr[i] = min(arr[1], arr[2], …, arr[i – 1]) or arr[i] = max(arr[1], arr[2], …, arr[i – 1]).

Examples:

Input: arr[] = {3, 6, 9}, N = 3
Output: 4
Explanation: The array arr[] can be constructed in 4 ways: {3, 6, 9}, {6, 3, 9}, {6, 9, 3}, and {9, 6, 3} where each element is either smaller or greater from the left side.

Input: arr[] = {7, 8}, N = 2
Output: 2
Explanation: The array arr[] can be constructed in 4 ways: {7, 8}, {8, 7} where each element is either smaller or greater from the left side.

Approach: The problem can be solved based on the following idea:

Iterate over each element of all the arrays and check if that element is an extremum, by maintaining two variables that track the min and max encountered so far. Count the number of arrays where the condition is satisfied.

Follow the below steps to implement the idea:

• Find each permutation of the given array.
• For each permutation, Initialise mins and maxs with the greatest and smallest number respectively.
• Iterate over the array and update mins and maxs at each index.
• If the current element is not equal to mins and maxs return False. Otherwise, Return True.
• If True, Count it as a valid permutation.

This approach can be implemented as follows:

C++

// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;

// This function checks if a given
// permutation is valid
bool isValid(vector<int> A)
{
  // Initializing the min and max values
  int mins = *max_element(A.begin(), A.end()) + 1;
  int maxs = -1;

  // Iterating over the array
  for (int i = 0; i < A.size(); i++)
  {

    // Updating the min and max
    mins = min(mins, A[i]);
    maxs = max(maxs, A[i]);

    // If ith element is neither min
    // nor max, then the array is invalid
    if (A[i] != mins && A[i] != maxs)
      return false;
  }
  return true;
}

// This function counts the total number
// of valid permutations
int countTotalValidPermutations(vector<int> S)
{
  set<vector<int> > distinctPermutations;
  sort(S.begin(), S.end());
  do {
    distinctPermutations.insert(S);
  } while (next_permutation(S.begin(), S.end()));



  int count = 0;
  for (vector<int> A : distinctPermutations)
    if (isValid(A))
      count++;

  return count;
}

int main()
{
  vector<int> S1 = { 3, 6, 9 };
  cout << countTotalValidPermutations(S1) << endl;

  vector<int> S2 = { 7, 8 };
  cout << countTotalValidPermutations(S2) << endl;

  vector<int> S3 = { 2, 1, 3, 1 };
  cout << countTotalValidPermutations(S3) << endl;

  return 0;
}

Java

// Java code for the above approach
import java.util.Arrays;
import java.util.Collections;
import java.util.HashSet;
import java.util.Set;
import java.util.Vector;

// This function checks if a given
// permutation is valid

public class Main {
    public static boolean isValid(Vector<Integer> A)
    {
        // Initializing the min and max values
        int mins = Collections.max(A) + 1;
        ;
        int maxs = -1;
        // Iterating over the array
        for (int i = 0; i < A.size(); i++) {
            // Updating the min and max
            mins = Math.min(mins, A.get(i));
            maxs = Math.max(maxs, A.get(i));

            // If ith element is neither min
            // nor max, then the array is invalid
            if (A.get(i) != mins && A.get(i) != maxs)
                return false;
        }
        return true;
    }

    // This function counts the total number
    // of valid permutations

    public static int
    countTotalValidPermutations(Vector<Integer> S)
    {
        Set<Vector<Integer> > distinctPermutations
            = new HashSet<>();
        Collections.sort(S);
        do {
            distinctPermutations.add(new Vector<>(S));
        } while (nextPermutation(S));

        int count = 0;
        for (Vector<Integer> A : distinctPermutations)
            if (isValid(A))
                count++;

        return count;
    }

    public static boolean nextPermutation(Vector<Integer> S)
    {
        int i = S.size() - 2;
        while (i >= 0 && S.get(i) >= S.get(i + 1))
            i--;
        if (i == -1)
            return false;

        int j = S.size() - 1;
        while (S.get(j) <= S.get(i))
            j--;

        int temp = S.get(i);
        S.set(i, S.get(j));
        S.set(j, temp);

        int left = i + 1;
        int right = S.size() - 1;
        while (left < right) {
            temp = S.get(left);
            S.set(left, S.get(right));
            S.set(right, temp);
            left++;
            right--;
        }
        return true;
    }

    public static void main(String[] args)
    {
        Vector<Integer> S1
            = new Vector<>(Arrays.asList(3, 6, 9));
        System.out.println(countTotalValidPermutations(S1));

        Vector<Integer> S2
            = new Vector<>(Arrays.asList(7, 8));
        System.out.println(countTotalValidPermutations(S2));

        Vector<Integer> S3
            = new Vector<>(Arrays.asList(2, 1, 3, 1));
        System.out.println(countTotalValidPermutations(S3));
    }
}

// This code is contributed by rutikbhosale

Python3

# Python code for the above approach
from itertools import permutations

# This function checks if a given
# permutation is valid


def is_valid(A):

    # Initializing the min and max values
    mins = max(A) + 1
    maxs = -1

    # Iterating over the array
    for i in range(len(A)):

        # Updating the min and max
        mins = min(mins, A[i])
        maxs = max(maxs, A[i])

        # If ith element is neither min
        # nor max, then the array is invalid
        if A[i] != mins and A[i] != maxs:
            return False
    return True

# This function counts the total number
# of valid permutations


def count_total_valid_permutations(S):

    # Getting the total number of
    # distinct permutations
    distinct_permutations = set(permutations(S))

    # Counting the number of distinct
    # permutations that are valid
    return len([A for A in distinct_permutations if is_valid(A)])


# Driver Code

# Input 1
S1 = [3, 6, 9]

# Function call
print(count_total_valid_permutations(S1))

# Input 2
S2 = [7, 8]

# Function call
print(count_total_valid_permutations(S2))

# Input 3
S3 = [2, 1, 3, 1]

# Function call
print(count_total_valid_permutations(S3))

C#

// C# code for the above approach

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

class GFG{
    // This function checks if a given permutation is valid
    static bool IsValid(List<int> A)
    {
        // Initializing the min and max values
        int mins = A.Max() + 1;
        int maxs = -1;

        // Iterating over the array
        for (int i = 0; i < A.Count; i++)
        {
            // Updating the min and max
            mins = Math.Min(mins, A[i]);
            maxs = Math.Max(maxs, A[i]);

            // If ith element is neither min
            // nor max, then the array is invalid
            if (A[i] != mins && A[i] != maxs)
                return false;
        }
        return true;
    }

    // This function counts the total number
    // of valid permutations
    static int CountTotalValidPermutations(List<int> S)
    {
        HashSet<List<int>> distinctPermutations = new HashSet<List<int>>(new ListComparer<int>());
        S.Sort();
        do
        {
            distinctPermutations.Add(new List<int>(S));
        } while (NextPermutation(S));

        int count = 0;
        foreach (List<int> A in distinctPermutations)
        {
            if (IsValid(A))
                count++;
        }
        return count;
    }

    // Function to generate the next permutation
    static bool NextPermutation(List<int> list)
    {
        int i = list.Count - 2;
        while (i >= 0 && list[i] >= list[i + 1])
            i--;
        if (i < 0)
            return false;
        int j = list.Count - 1;
        while (list[j] <= list[i])
            j--;
        int temp = list[i];
        list[i] = list[j];
        list[j] = temp;
        j = list.Count - 1;
        i++;
        while (i < j)
        {
            temp = list[i];
            list[i] = list[j];
            list[j] = temp;
            i++;
            j--;
        }
        return true;
    }

    // Custom comparer for lists
    class ListComparer<T> : IEqualityComparer<List<T>>
    {
        public bool Equals(List<T> x, List<T> y)
        {
            return x.SequenceEqual(y);
        }

        public int GetHashCode(List<T> obj)
        {
            return obj.Sum(item => item.GetHashCode());
        }
    }

    // Driver code
    static void Main()
    {
        List<int> S1 = new List<int> { 3, 6, 9 };
        Console.WriteLine(CountTotalValidPermutations(S1));

        List<int> S2 = new List<int> { 7, 8 };
        Console.WriteLine(CountTotalValidPermutations(S2));

        List<int> S3 = new List<int> { 2, 1, 3, 1 };
        Console.WriteLine(CountTotalValidPermutations(S3));
    }
}

Javascript

// JavaScript code for the above approach

function isValid(A) {
  // Initializing the min and max values
  let mins = Math.max(...A) + 1;
  let maxs = -1;

  // Iterating over the array
  for (let i = 0; i < A.length; i++) {

    // Updating the min and max
    mins = Math.min(mins, A[i]);
    maxs = Math.max(maxs, A[i]);

    // If ith element is neither min
    // nor max, then the array is invalid
    if (A[i] !== mins && A[i] !== maxs) {
      return false;
    }
  }
  return true;
}

function countTotalValidPermutations(S) {
  let distinctPermutations = new Set();
  S.sort();

  do {
    distinctPermutations.add([...S]);
  } while (nextPermutation(S));

  let count = 0;
  for (let A of distinctPermutations) {
    if (isValid(A)) {
      count++;
    }
  }
  return count;
}

function nextPermutation(arr) {
  let i = arr.length - 2;
  while (i >= 0 && arr[i] >= arr[i + 1]) {
    i--;
  }
  if (i < 0) {
    return false;
  }
  let j = arr.length - 1;
  while (arr[j] <= arr[i]) {
    j--;
  }
  [arr[i], arr[j]] = [arr[j], arr[i]];
  reverse(arr, i + 1);
  return true;
}

function reverse(arr, start) {
  let i = start;
  let j = arr.length - 1;
  while (i < j) {
    [arr[i], arr[j]] = [arr[j], arr[i]];
    i++;
    j--;
  }
}

let S1 = [3, 6, 9];
console.log(countTotalValidPermutations(S1));

let S2 = [7, 8];
console.log(countTotalValidPermutations(S2));

let S3 = [2, 1, 3, 1];
console.log(countTotalValidPermutations(S3));

Output

4
2
7

Time Complexity: O(N! * N)
Auxiliary Space: O(N!)