Count of Subsequences with distinct elements

Given an array arr[] (1<=a[i]<=1e9) containing N (1<=N<=1e5) elements, the task is to find the total number of subsequences such that all elements in that subsequences are distinct. Since the answer can be very large print and modulo 1e9+7.

Examples:

Input: arr[] = [1, 1, 2, 2]
Output: 8
Explanation: The subsequences are [1], [1], [2], [2], [1, 2], [1, 2], [1, 2], [1, 2]

Input: arr[] = [1, 3, 2, 2]
Output: 11
Explanation: The subsequences are [1], [3], [2], [2], [1, 3], [1, 2], [1, 2], [3, 2], [3, 2], [1, 3, 2], [1, 3, 2]

Approach: To solve the problem follow the below idea:

This problem can be solved using unordered_map.

Follow the steps to solve the problem:

First, let's store the frequency of all elements in a map
Suppose an element has a frequency of x then we have to either choose one element from this or exclude this element.
Remember we need all distinct elements so we can't choose more than 1 same element.
So for an element with frequency x, we have (x+1) options (1 for excluding the elements)
Now for counting the number of subsequences, we can multiply the options that we have.
At last, we need to subtract 1 because one empty subsequence is generated so to exclude that we need to subtract 1.

Below is the code for the above approach:

C++

// C++ code for the above approach:
#include <bits/stdc++.h>
using namespace std;
int mod = 1e9 + 7;

// Function to calculate subsequences with
// distinct elements
void distSubs(vector<int> v, int n)
{
    long long ans = 1;

    // Creating a map
    unordered_map<int, int> m;

    // Calculating frequency of each element
    for (int i = 0; i < n; i++) {
        m[v[i]]++;
    }

    // Calculating ans
    for (auto it : m) {

        ans = (ans % mod * (it.second + 1) % mod) % mod;
    }

    ans = (ans - 1 + mod) % mod;

    cout << "Total subsequences with all elements distinct "
            "are : "
         << ans << endl;
}

// Drivers code
int main()
{

    vector<int> v = { 1, 1, 1, 3, 3, 3 };
    int n = v.size();

    // Function Call
    distSubs(v, n);
    return 0;
}

Java

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

public class DistinctSubsequences {

    private static final int MOD = 1000000007;

    // Function to calculate subsequences with distinct elements
    private static void distSubs(int[] arr, int n) {
        long ans = 1;

        // Creating a map to store element frequencies
        Map<Integer, Integer> frequencyMap = new HashMap<>();

        // Calculating frequency of each element
        for (int i = 0; i < n; i++) {
            frequencyMap.put(arr[i], frequencyMap.getOrDefault(arr[i], 0) + 1);
        }

        // Calculating ans
        for (Map.Entry<Integer, Integer> entry : frequencyMap.entrySet()) {
            int elementFrequency = entry.getValue();
            ans = (ans % MOD * (elementFrequency + 1) % MOD) % MOD;
        }

        ans = (ans - 1 + MOD) % MOD;

        System.out.println("Total subsequences with all elements distinct are: " + ans);
    }

    public static void main(String[] args) {
        int[] arr = {1, 1, 1, 3, 3, 3};
        int n = arr.length;

        // Function Call
        distSubs(arr, n);
    }
}

Python3

def distSubs(v):
    mod = 10**9 + 7
    ans = 1

    # Creating a dictionary
    m = {}

    # Calculating frequency of each element
    for i in range(len(v)):
        m[v[i]] = m.get(v[i], 0) + 1

    # Calculating ans
    for key, value in m.items():
        ans = (ans % mod * (value + 1) % mod) % mod

    ans = (ans - 1 + mod) % mod

    print("Total subsequences with all elements distinct are:", ans)

# Drivers code
if __name__ == "__main__":
    v = [1, 1, 1, 3, 3, 3]

    # Function Call
    distSubs(v)

// C# implementation of the above approach
using System;
using System.Collections.Generic;

public class DistinctSubsequences
{
    private const int MOD = 1000000007;

    // Function to calculate subsequences with distinct elements
    private static void DistSubs(int[] arr, int n)
    {
        long ans = 1;

        // Creating a dictionary to store element frequencies
        Dictionary<int, int> frequencyMap = new Dictionary<int, int>();

        // Calculating frequency of each element
        for (int i = 0; i < n; i++)
        {
            if (frequencyMap.ContainsKey(arr[i]))
            {
                frequencyMap[arr[i]]++;
            }
            else
            {
                frequencyMap[arr[i]] = 1;
            }
        }

        // Calculating ans
        foreach (KeyValuePair<int, int> entry in frequencyMap)
        {
            int elementFrequency = entry.Value;
            ans = (ans % MOD * (elementFrequency + 1) % MOD) % MOD;
        }

        ans = (ans - 1 + MOD) % MOD;

        Console.WriteLine("Total subsequences with all elements distinct are: " + ans);
    }

    public static void Main(string[] args)
    {
        int[] arr = { 1, 1, 1, 3, 3, 3 };
        int n = arr.Length;

        // Function Call
        DistSubs(arr, n);
    }
}

// This code is contributed by Tapesh(tapeshdua420)

JavaScript

function distSubs(v) {
    const mod = 1e9 + 7;
    let ans = 1;
    // Creating a map
    const map = new Map();
    // Calculating frequency of each element
    for (let i = 0; i < v.length; i++) {
        if (map.has(v[i])) {
            map.set(v[i], map.get(v[i]) + 1);
        } else {
            map.set(v[i], 1);
        }
    }
    // Calculating ans
    for (const [key, value] of map) {
        ans = (ans % mod * (value + 1) % mod) % mod;
    }
    ans = (ans - 1 + mod) % mod;
    console.log("Total subsequences with all elements distinct are:", ans);
}
// Driver code
const v = [1, 1, 1, 3, 3, 3];
// Function call
distSubsv);

Output

Total subsequences with all elements distinct are : 15

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

Count of Subsequences with distinct elements

Explore