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: 5
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 = 21Input: arr[] = {4, 1, 6, 7, 8, 2}
Output: 3
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:
- Iterate over all indices and consider only those indices whose values are greater than or equal to the values of the respective values stored in them.
- Keep updating the sum of the differences obtained in the above step.
- For the remaining elements, store their differences with their respective indexes. Sort the differences.
- Include elements into the subset one by one and subtract the difference from the sum. Keep including until an element is encountered whose difference with its index exceeds the remaining sum or all array elements have already been included.
Below is the implementation of the above approach:
// 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;
} |
// 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 |
# 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 ] - i
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 |
// 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 |
<script> // Javascript program for the above approach // Function to find the length // of the longest subset function findSubset(a, n)
{ // Stores the sum of differences
// between elements and
// their respective index
let sum = 0;
// Stores the size of
// the subset
let cnt = 0;
let v = [];
// Iterate over the array
for (let 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(a[i - 1] - i);
}
}
// Sort the differences
// in increasing order
v.sort();
let ptr = 0;
// Include the differences while
// sum remains positive or
while (ptr < v.length &&
sum + v[ptr] <= 0)
{
cnt += 1;
ptr += 1;
sum += v[ptr];
}
// Return the size
return cnt;
} // Driver Code let arr = [ 4, 1, 6, 7, 8, 2 ];
let n = arr.length;
// Function Calling
document.write(findSubset(arr, n));
</script> |
3
Time Complexity: O(N*log(N)), the inbuilt sort function takes N log N time to complete all operations, hence the overall time taken by the algorithm is N log N
Auxiliary Space: O(N), an extra vector is used and in the worst case all elements will be stored inside it the hence algorithm takes up linear space