Count Permutations in a Sequence

Given an array A consisting of N positive integers, find the total number of subsequences of the given array such that the chosen subsequence represents a permutation.

Note:

• Sequence A is a subsequence of B if A can be obtained from B by deleting some(possibly, zero) elements without changing its order. For example, [3,1] is a subsequence of [3,2,1] and [4,3,1], but not a subsequence of [1,3,3,7] and [3,10,4].
• Two permutations are different if they are of different lengths or if there is any element in the two permutations such that its index in the original array is different.
• A permutation of length N is an array of length N in which every element from 1 to N occurs exactly once.

Examples:

Input: N = 5, A[] = {1, 2, 3, 2, 4}
Output: 7
Explanation: We can get 7 permutations: {1}, {1, 2}, {1, 2}, {1, 2, 3}, {1, 2, 3}, {1, 2, 3, 4}, {1, 2, 3, 4}.

Input: N = 4, A[] = {2, 1, 1, 2}
Output: 6
Explanation: We can get 6 permutations: {1}, {1}, {2, 1}, {2, 1}, {1, 2}, {1, 2}

Approach: This can be solved with the following idea:

We can use a map to store the frequency of all the elements in the array A[]. Now, run a loop starting from 1 till we get an element whose frequency is 0. The main idea is to calculate the number of permutations which can be formed using first (i-1) elements, then for each of these permutations we can append every i to form a new permutation.

Below are the steps involved:

• Store the frequency of each element in a map, say freq_arr.
• Maintain a variable prev = 1, to store the number of permutations which can be formed using elements from 1 to (i-1)
• Now, start a loop from 1 till we reach a number whose frequency in the original array is 0.
• Maintain a variable ans = 0, to store the final answer.
• For each iteration i,
  • Check if freq[i] > 0. If not, break
  • Else, Update prev with prev * freq[i] and add prev to ans
• Return ans

Below is the implementation of the code:

// C++ Implementation for the above approach:

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

// Function to count permutation
int solve(int N, vector<int>& A)
{

    // Frequency array to store frequency of
    // all the elements
    map<int, int> freq_arr;
    long long prev = 1;
    for (auto ele : A)
        freq_arr[ele]++;
    long long ans = 0;
    for (int i = 1; freq_arr[i]; i++) {
        prev = prev * freq_arr[i];
        ans += prev;
    }
    return ans;
}

// Driver code
int main()
{

    int N = 4;
    vector<int> A = { 2, 1, 1, 2 };

    // Function call
    cout << solve(N, A);
    return 0;
}

import java.util.HashMap;
import java.util.Map;
import java.util.ArrayList;

public class PermutationCount {
    // Function to count permutations
    static long solve(int N, ArrayList<Integer> A) {
        // Frequency map to store the frequency of all the elements
        Map<Integer, Integer> freqMap = new HashMap<>();
        long prev = 1;
        for (int ele : A) {
            freqMap.put(ele, freqMap.getOrDefault(ele, 0) + 1);
        }
        long ans = 0;
        for (int i = 1; freqMap.containsKey(i); i++) {
            prev *= freqMap.get(i);
            ans += prev;
        }
        return ans;
    }
//Driver code
    public static void main(String[] args) {
        int N = 4;
        ArrayList<Integer> A = new ArrayList<>();
        A.add(2);
        A.add(1);
        A.add(1);
        A.add(2);

        // Function call
        System.out.println(solve(N, A));
    }
}

# Python Implementation for the above approach:

def solve(N, A):
    # Frequency dictionary to store the frequency of all the elements
    freq_dict = {}
    prev = 1

    for ele in A:
        freq_dict[ele] = freq_dict.get(ele, 0) + 1

    ans = 0

    for i in range(1, N + 1):
        if i in freq_dict:
            prev = prev * freq_dict[i]
            ans += prev
        else:
            break

    return ans

# Driver code
if __name__ == "__main__":
    N = 4
    A = [2, 1, 1, 2]

    # Function call
    print(solve(N, A))

using System;
using System.Collections.Generic;

class GFG {
    // Function to count permutation
    static long Solve(int N, List<int> A) {
        // Frequency dictionary to store frequency of all the elements
        Dictionary<int, int> freqDict = new Dictionary<int, int>();
        long prev = 1;

        foreach (var ele in A) {
            if (freqDict.ContainsKey(ele))
                freqDict[ele]++;
            else
                freqDict[ele] = 1;
        }

        long ans = 0;
        for (int i = 1; freqDict.ContainsKey(i); i++) {
            prev = prev * freqDict[i];
            ans += prev;
        }
        return ans;
    }

    // Driver code
    public static void Main(string[] args) {
        int N = 4;
        List<int> A = new List<int> { 2, 1, 1, 2 };

        // Function call
        Console.WriteLine(Solve(N, A));
    }
}

// Function to count permutations
function GFG(N, A) {
    // Frequency map to the store the frequency of all the elements
    const freqMap = new Map();
    let prev = 1;
    for (const ele of A) {
        if (freqMap.has(ele)) {
            freqMap.set(ele, freqMap.get(ele) + 1);
        } else {
            freqMap.set(ele, 1);
        }
    }
    let ans = 0;
    for (let i = 1; freqMap.has(i); i++) {
        prev *= freqMap.get(i);
        ans += prev;
    }
    return ans;
}
// Driver Code
const N = 4;
const A = [2, 1, 1, 2];
// Function call
console.log(GFG(N, A));

6

Time Complexity: O(N), where N is the size of input array A[]
Auxiliary Space: O(N)