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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: 

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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 <bits/stdc++.h>
using namespace std;
 
// Function to calculate the number
// of subsequences that can be formed
int No_Of_subsequences(map<int, int> 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 formed
void maximumsubsequences(int arr[], int n)
{
    // Stores the frequency of arr[]
    map<int, int> 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 Code
int main()
{
    // Given array arr[]
    int arr[] = { 1, 1, 1, 1 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    maximumsubsequences(arr, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function to calculate the number
// of subsequences that can be formed
public static int No_Of_subsequences(HashMap<Integer,
                                             Integer> mp)
{
     
    // Stores the number of subsequences
    int count = 0;
    int left = 0;
 
    // Iterate over the map
    for(Map.Entry<Integer, Integer> 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 formed
public static void maximumsubsequences(int[] arr,
                                       int n)
{
     
    // Stores the frequency of arr[]
    HashMap<Integer, Integer> 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 code
public 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 approach
from collections import defaultdict
 
# Function to calculate
# the number of subsequences
# that can be formed
def 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 formed
def 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 Code
if __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 approach
using System;
using System.Collections.Generic;
class GFG{
     
// Function to calculate the number
// of subsequences that can be formed
public static int No_Of_subsequences(Dictionary<int,
                                                int> mp)
{
  // Stores the number
  // of subsequences
  int count = 0;
  int left = 0;
 
  // Iterate over the map
  foreach(KeyValuePair<int,
                       int> 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 formed
public static void maximumsubsequences(int[] arr,
                                       int n)
{
  // Stores the frequency of []arr
  Dictionary<int,
             int> mp = new Dictionary<int,
                                      int>();
   
  // 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 code
public 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




<script>
 
// Javascript program for the above approach
 
// Function to calculate the number
// of subsequences that can be formed
function No_Of_subsequences(mp)
{
    // Stores the number of subsequences
    var count = 0;
    var left = 0;
 
    // Iterate over the map
    mp.forEach((value, key) => {
         
        value += left;
 
        // Count the number of subsequences
        // that can be formed from key
        count += (value / key);
 
        // Number of occurrences of
        // x.first which are left
        left = value % key;
    });
 
    // Return the number of subsequences
    return count;
}
 
// Function to create the maximum count
// of subsequences that can be formed
function maximumsubsequences(arr, n)
{
 
    // Stores the frequency of arr[]
    var mp = new Map();
 
    // Update the frequency
    for (var i = 0; i < n; i++)
    {
        if(mp.has(arr[i]))
            mp.set(arr[i], mp.get(arr[i])+1)
        else
            mp.set(arr[i], 1);
    }
 
    // Print the number of subsequences
    document.write( No_Of_subsequences(mp));
}
 
// Driver Code
 
// Given array arr[]
var arr = [1, 1, 1, 1];
var N = arr.length;
 
// Function Call
maximumsubsequences(arr, N);
 
// This code is contributed by itsok.
</script>
Output: 
4

 

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




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