Maximize subsequences having array elements not exceeding length of the subsequence

• Difficulty Level : Basic
• Last Updated : 19 May, 2021

Given an array arr[] consisting of N positive integers, the task is to maximize the number of subsequences that can be obtained from an array such that every element arr[i] that is part of any subsequence does not exceed the length of that subsequence.

Examples:

Input: arr[] = {1, 1, 1, 1}
Output:
Explanation:
Form 4 subsequences of 1 which satisfy the given condition {1}, {1}, {1}, {1}.

Input: arr[] = {2, 2, 3, 1, 2, 1}
Output:
Explanation:
The possible group are {1}, {1}, {2, 2}
So, the output is 3.

Approach: The idea is to use Greedy Technique to solve this problem. Follow the steps below to solve the problem:

1. Initialize a map to store the frequency of every array element.
2. Initialize a variable, say count to 0, to store the total number of subsequences obtained.
3. Keep track of the number of elements that are left to be added.
4. Now, iterate over the map and count the number of elements that can be included in a particular group.
5. Keep on adding the elements on the valid subsequences.
6. After completing the above steps, print the count of subsequences.

Below is the implementation of the above approach:

C++

 // C++ program for the above approach#include using namespace std; // Function to calculate the number// of subsequences that can be formedint No_Of_subsequences(map mp){    // Stores the number of subsequences    int count = 0;    int left = 0;     // Iterate over the map    for (auto x : mp) {         x.second += left;         // Count the number of subsequences        // that can be formed from x.first        count += (x.second / x.first);         // Number of occurrences of        // x.first which are left        left = x.second % x.first;    }     // Return the number of subsequences    return count;} // Function to create the maximum count// of subsequences that can be formedvoid maximumsubsequences(int arr[], int n){    // Stores the frequency of arr[]    map mp;     // Update the frequency    for (int i = 0; i < n; i++)        mp[arr[i]]++;     // Print the number of subsequences    cout << No_Of_subsequences(mp);} // Driver Codeint main(){    // Given array arr[]    int arr[] = { 1, 1, 1, 1 };     int N = sizeof(arr) / sizeof(arr);     // Function Call    maximumsubsequences(arr, N);     return 0;}

Java

 // Java program for the above approachimport java.util.*; class GFG{     // Function to calculate the number// of subsequences that can be formedpublic static int No_Of_subsequences(HashMap mp){         // Stores the number of subsequences    int count = 0;    int left = 0;     // Iterate over the map    for(Map.Entry x : mp.entrySet())    {        mp.replace(x.getKey(), x.getValue() + left);                 // Count the number of subsequences        // that can be formed from x.first        count += (x.getValue() / x.getKey());         // Number of occurrences of        // x.first which are left        left = x.getValue() % x.getKey();    }     // Return the number of subsequences    return count;} // Function to create the maximum count// of subsequences that can be formedpublic static void maximumsubsequences(int[] arr,                                       int n){         // Stores the frequency of arr[]    HashMap mp = new HashMap<>();     // Update the frequency    for(int i = 0; i < n; i++)    {        if (mp.containsKey(arr[i]))        {            mp.replace(arr[i], mp.get(arr[i]) + 1);        }        else        {            mp.put(arr[i], 1);        }    }     // Print the number of subsequences    System.out.println(No_Of_subsequences(mp));} // Driver codepublic static void main(String[] args){         // Given array arr[]    int arr[] = { 1, 1, 1, 1 };     int N = arr.length;     // Function call    maximumsubsequences(arr, N);}} // This code is contributed divyeshrabadiya07

Python3

 # Python3 program for# the above approachfrom collections import defaultdict # Function to calculate# the number of subsequences# that can be formeddef No_Of_subsequences(mp):     # Stores the number    # of subsequences    count = 0    left = 0     # Iterate over the map    for x in mp:         mp[x] += left         # Count the number of        # subsequences that can        # be formed from x.first        count += (mp[x] // x)         # Number of occurrences of        # x.first which are left        left = mp[x] % x        # Return the number    # of subsequences    return count # Function to create the# maximum count of subsequences# that can be formeddef maximumsubsequences(arr, n):     # Stores the frequency of arr[]    mp = defaultdict (int)     # Update the frequency    for i in range (n):        mp[arr[i]] += 1     # Print the number of subsequences    print(No_Of_subsequences(mp)) # Driver Codeif __name__ == "__main__":       # Given array arr[]    arr = [1, 1, 1, 1]     N = len(arr)     # Function Call    maximumsubsequences(arr, N) # This code is contributed by Chitranayal

C#

 // C# program for// the above approachusing System;using System.Collections.Generic;class GFG{     // Function to calculate the number// of subsequences that can be formedpublic static int No_Of_subsequences(Dictionary mp){  // Stores the number  // of subsequences  int count = 0;  int left = 0;   // Iterate over the map  foreach(KeyValuePair x in mp)  {    if(!mp.ContainsKey(x.Key))      mp.Add(x.Key, x.Value + left);     // Count the number of subsequences    // that can be formed from x.first    count += (x.Value / x.Key);     // Number of occurrences of    // x.first which are left    left = x.Value % x.Key;  }   // Return the number of subsequences  return count;} // Function to create the maximum count// of subsequences that can be formedpublic static void maximumsubsequences(int[] arr,                                       int n){  // Stores the frequency of []arr  Dictionary mp = new Dictionary();     // Update the frequency  for(int i = 0; i < n; i++)  {    if (mp.ContainsKey(arr[i]))    {      mp[arr[i]] =  mp[arr[i]] + 1;    }    else    {      mp.Add(arr[i], 1);    }  }   // Print the number of subsequences  Console.WriteLine(No_Of_subsequences(mp));} // Driver codepublic static void Main(String[] args){  // Given array []arr  int []arr = {1, 1, 1, 1};   int N = arr.Length;   // Function call  maximumsubsequences(arr, N);}} // This code is contributed by Rajput-Ji

Javascript


Output:
4

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

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