Largest subset having with sum less than equal to sum of respective indices

Given an array arr[], the task is to find the length of the largest subset with the sum of elements less than or equal to the sum of its indexes(1-based indexing).

Examples:

Input: arr[] = {1, 7, 3, 5, 9, 6, 6} 
Output:
Explanation: 
Largest Subset is {1, 3, 5, 6, 6} 
Sum of indexes = 1 + 3 + 4 + 6 + 7 = 21 
Sum of elements = 1 + 3 + 5 + 6 + 6 = 21

Input: arr[] = {4, 1, 6, 7, 8, 2} 
Output:

Naive Approach: 
The simplest approach to solve the problem is to generate all possible subsets and calculate the length of the subsets which have the sum of elements less than or equal to the sum of its respective indices. 



Time Complexity: O(N*2N
Auxiliary Space: O(N)

Efficient Approach: 
Follow the steps below to solve the problem: 

Below is the implementation of the above approach:

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// C++ Program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the length
// of the longest subset
int findSubset(int* a, int n)
{
    // Stores the sum of differences
    // between elements and
    // their respective index
    int sum = 0;
 
    // Stores the size of
    // the subset
    int cnt = 0;
 
    vector<int> v;
 
    // Iterate over the array
    for (int i = 1; i <= n; i++) {
 
        // If an element which is
        // smaller than or equal
        // to its index is encountered
        if (a[i - 1] - i <= 0) {
 
            // Increase count and sum
            sum += a[i - 1] - i;
            cnt += 1;
        }
 
        // Store the difference with
        // index of the remaining
        // elements
        else {
            v.push_back(a[i - 1] - i);
        }
    }
 
    // Sort the differences
    // in increasing order
    sort(v.begin(), v.end());
 
    int ptr = 0;
 
    // Include the differences while
    // sum remains positive or
    while (ptr < v.size()
           && sum + v[ptr] <= 0) {
        cnt += 1;
        ptr += 1;
        sum += v[ptr];
    }
 
    // Return the size
    return cnt;
}
 
// Driver Code
int main()
{
    int arr[] = { 4, 1, 6, 7,
                  8, 2 };
 
    int n = sizeof(arr)
            / sizeof(arr[0]);
 
    // Function Calling
    cout << findSubset(arr, n)
         << endl;
}
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// Java program to implement
// the above approach
import java.util.*;
class GFG{
     
// Function to find the length
// of the longest subset
public static int findSubset(int[] a, int n)
{
     
    // Stores the sum of differences
    // between elements and
    // their respective index
    int sum = 0;
 
    // Stores the size of
    // the subset
    int cnt = 0;
 
    Vector<Integer> v = new Vector<>();
 
    // Iterate over the array
    for(int i = 1; i <= n; i++)
    {
         
        // If an element which is
        // smaller than or equal
        // to its index is encountered
        if (a[i - 1] - i <= 0)
        {
             
            // Increase count and sum
            sum += a[i - 1] - i;
            cnt += 1;
        }
 
        // Store the difference with
        // index of the remaining
        // elements
        else
        {
            v.add(a[i - 1] - i);
        }
    }
 
    // Sort the differences
    // in increasing order
    Collections.sort(v);
 
    int ptr = 0;
 
    // Include the differences while
    // sum remains positive or
    while (ptr < v.size() &&
           sum + v.get(ptr) <= 0)
    {
        cnt += 1;
        ptr += 1;
        sum += v.get(ptr);
    }
 
    // Return the size
    return cnt;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 4, 1, 6, 7, 8, 2 };
    int n = arr.length;
 
    // Function Calling
    System.out.println(findSubset(arr, n));
}
}
 
// This code is contributed by divyeshrabadiya07
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# Python3 program to implement 
# the above approach
 
# Function to find the length 
# of the longest subset
def findSubset(a, n):
     
    # Stores the sum of differences 
    # between elements and 
    # their respective index 
    sum = 0
   
    # Stores the size of 
    # the subset 
    cnt = 0
   
    v = [] 
   
    # Iterate over the array 
    for i in range(1, n + 1):
         
        # If an element which is 
        # smaller than or equal 
        # to its index is encountered 
        if (a[i - 1] - i <= 0):
             
            # Increase count and sum 
            sum += a[i - 1] -
            cnt += 1
     
        # Store the difference with 
        # index of the remaining 
        # elements 
        else:
            v.append(a[i - 1] - i)
             
    # Sort the differences 
    # in increasing order 
    v.sort()
   
    ptr = 0
   
    # Include the differences while 
    # sum remains positive or 
    while (ptr < len(v) and
           sum + v[ptr] <= 0):
        cnt += 1
        ptr += 1 
        sum += v[ptr] 
     
    # Return the size 
    return cnt
 
# Driver code
if __name__=="__main__":
     
    arr = [ 4, 1, 6, 7, 8, 2
    n = len(arr)
   
    # Function calling 
    print(findSubset(arr, n))
 
# This code is contributed by rutvik_56
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// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to find the length
// of the longest subset
public static int findSubset(int[] a, int n)
{
     
    // Stores the sum of differences
    // between elements and
    // their respective index
    int sum = 0;
 
    // Stores the size of
    // the subset
    int cnt = 0;
 
    List<int> v = new List<int>();
 
    // Iterate over the array
    for(int i = 1; i <= n; i++)
    {
         
        // If an element which is
        // smaller than or equal
        // to its index is encountered
        if (a[i - 1] - i <= 0)
        {
             
            // Increase count and sum
            sum += a[i - 1] - i;
            cnt += 1;
        }
 
        // Store the difference with
        // index of the remaining
        // elements
        else
        {
            v.Add(a[i - 1] - i);
        }
    }
 
    // Sort the differences
    // in increasing order
    v.Sort();
 
    int ptr = 0;
 
    // Include the differences while
    // sum remains positive or
    while (ptr < v.Count &&
           sum + v[ptr] <= 0)
    {
        cnt += 1;
        ptr += 1;
        sum += v[ptr];
    }
 
    // Return the size
    return cnt;
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 4, 1, 6, 7, 8, 2 };
    int n = arr.Length;
 
    // Function calling
    Console.WriteLine(findSubset(arr, n));
}
}
 
// This code is contributed by amal kumar choubey
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Output: 
3

 

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

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