Count Subsequences with ordered integers in Array

Given an array nums[] of N positive integers, the task is to find the number of subsequences that can be created from the array where each subsequence contains all integers from 1 to its size in any order. If two subsequences have different chosen indices, then they are considered different.

Examples:

Input: N = 5, nums = {1, 2, 3, 2, 4}
Output: 7
Explanation: There are 7 subsequences present in the given array which has the size of M and contains integers from 1 to M. The subsequences are: [1], [1, 2], [1, 2], [1, 2, 3], [1, 3, 2], [1, 2, 3, 4], and [1, 3, 2, 4]

Input: N = 7, nums = {1, 3, 5, 8, 9, 8, 2}
Output: 3
Explanation: The three subsequences present in the given array are: [1], [1, 2], and [1, 3, 2].

Approach: This can be solved with the following idea:

First of all, we will make a frequency map using the hash map. Suppose we have given the array, nums = [1, 2, 3, 2, 4] then the frequency map will be like: 
[[1 -> 1], [2 -> 2], [3 -> 1], [4 -> 1]].

• Now we will start iterating from i = 1, we will continue the iteration till the frequency map contains a key equal to i.
• As for i = 2, the above case has a frequency value of 2, now we will check how many arrays we can make using i = 1, and we know it is 1, therefore the number of arrays we can make of size 2 such that it contains elements 1 and 2 will be: (number of arrays we can make of size 1) * (frequency value of 2). For the above case, it will be: 1 * 2 = 2, so till i = 2 we can make two arrays of size 2 and 1 array of size 1, the total value(answer variable) will be: 2 + 1 = 3.
• For i = 3, the above case has a frequency value of 1, and the number of arrays we can create of size 2 is 2, so the number of arrays of size 3 we can create will be: (2 * 1) = 2, answer variable will get updated as: 3 + 2 = 5.
• For i = 4, the above case has a frequency value of 1, and the number of arrays we can create of size 2 is 2, so the number of arrays of size 4 we can create will be: (2 * 1) = 2, answer variable will get updated as: 5 + 2 = 7.

To implement the above intuition follow the below-given steps:

• Firstly, make a hash map freq, which stores the frequency value of every element.
• Initialize our answer variable with a frequency value of 1 and also initialize a last variable which contains the value of a number of arrays we can make of size i – 1.
• Start iterating with i = 2, until it doesn’t contain in the frequency map.
• For every i, find a number of arrays you can make and update it in the answer variable also don’t forget to take modulo with 1000000007.
• Return the answer.

Below is the implementation of the above approach:

C++

// C++ algorithm for the above approach

#include <iostream>
#include <unordered_map>
using namespace std;

// Function which will return number
// of arrays we can make
int makeArrays(int nums[], int N)
{
    // Making frequency map of the given array nums
    unordered_map<int, int> freq;

    for (int i = 0; i < N; i++) {
        freq[nums[i]] = freq[nums[i]] + 1;
    }

    long long mod = 1000000007;

    // Initializing ans variable
    long long ans = (freq.count(1)) ? freq[1] : 0;

    // Initializing last variable
    long long last = ans;

    int i = 2;

    // Iterating after i = 1
    while (freq.count(i)) {

        last = (last * freq[i]) % mod;
        ans = (ans + last) % mod;
        i++;
    }

    // Returning answer
    return static_cast<int>(ans);
}

// Driver Code
int main()
{

    int nums[] = { 1, 3, 5, 8, 9, 8, 2 };
    int N = sizeof(nums) / sizeof(nums[0]);
    cout << makeArrays(nums, N) << endl;
    return 0;
}

Java

// Java algorithm for above approach
import java.util.*;

class GFG {

    // Driver Code
    public static void main(String[] args)
    {
        int[] nums = { 1, 3, 5, 8, 9, 8, 2 };
        int N = nums.length;
        System.out.println(makeArrays(nums, N));
    }

    // Function which will return number
    // of arrays we can make
    public static int makeArrays(int[] nums, int N)
    {

        // Making frequency map of
        // the given array nums
        Map<Integer, Integer> freq = new HashMap<>();

        for (int i = 0; i < N; i++) {
            freq.put(nums[i],
                     freq.getOrDefault(nums[i], 0) + 1);
        }

        long mod = 1000000007;

        // Initializing ans variable
        long ans = (freq.containsKey(1)) ? freq.get(1) : 0;

        // Initializing last variable
        long last = ans;

        int i = 2;

        // Iterating after i = 1
        while (freq.containsKey(i)) {

            last = (last * freq.get(i)) % mod;
            ans = (ans + last) % mod;
            i++;
        }

        // Returning answer
        return (int)(ans);
    }
}

Python3

def make_arrays(nums):
    freq = {}

    # Count the frequency of each number
    for num in nums:
        freq[num] = freq.get(num, 0) + 1

    mod = 1000000007
    ans = freq.get(1, 0)
    last = ans
    i = 2

    # Iterate over the numbers and calculate the answer
    while i in freq:
        last = (last * freq[i]) % mod
        ans = (ans + last) % mod
        i += 1

    # Return the answer
    return ans

# Example usage
nums = [1, 3, 5, 8, 9, 8, 2]
print(make_arrays(nums)) # Output: 7

C#

// C# program for above approach
using System;
using System.Collections.Generic;

class GFG {

    // Function which will return number
    // of arrays we can make
    static int makeArrays(int[] nums, int N)
    {

        // Making frequency map of
        // the given array nums
        Dictionary<int, int> freq
            = new Dictionary<int, int>();

        int i;
        for (i = 0; i < N; i++) {
            if (freq.ContainsKey(nums[i]))
                freq[nums[i]]++;
            else
                freq.Add(nums[i], 1);
        }

        long mod = 1000000007;

        // Initializing ans variable
        long ans = (freq.ContainsKey(1)) ? freq[1] : 0;

        // Initializing last variable
        long last = ans;

        i = 2;

        // Iterating after i = 1
        while (freq.ContainsKey(i)) {

            last = (last * freq[i]) % mod;
            ans = (ans + last) % mod;
            i++;
        }

        // Returning answer
        return (int)(ans);
    }

    // Driver Code
    public static void Main()
    {
        int[] nums = { 1, 3, 5, 8, 9, 8, 2 };
        int N = nums.Length;
        Console.Write(makeArrays(nums, N));
    }
}

// This code is contributed by Tapesh(tapeshdua420)

Javascript

// Function which will return number of arrays we can make
function makeArrays(nums) {
  // Making frequency map of the given array nums
  const freq = new Map();

  for (let i = 0; i < nums.length; i++) {
    freq.set(nums[i], (freq.get(nums[i]) || 0) + 1);
  }

  const mod = 1000000007;

  // Initializing ans variable
  let ans = freq.has(1) ? freq.get(1) : 0;

  // Initializing last variable
  let last = ans;

  let i = 2;

  // Iterating after i = 1
  while (freq.has(i)) {
    last = (last * freq.get(i)) % mod;
    ans = (ans + last) % mod;
    i++;
  }

  // Returning answer
  return ans;
}

// Driver Code
const nums = [1, 3, 5, 8, 9, 8, 2];
console.log(makeArrays(nums));

Output

3

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