Maximise the size of consecutive element subsets in an array

Given an integer array and an integer k. The array elements denote positions of points on 1-D number line, find the maximum size of subset of points that can have consecutive values of points which can be formed by placing another k points on the number line. Note that all coordinates should be distinct and elements of array are in increasing order.

Examples:

Input : arr[] = {1, 2, 3, 4, 10, 11, 14, 15},
            k = 4 
Output : 8
For maximum size subset, it is optimal to
choose the points on number line at 
coordinates 12, 13, 16 and 17, so that the
size of the consecutive valued subset will
become 8 which will be maximum .

Input : arr[] = {7, 8, 12, 13, 15, 18}
        k = 5
Output : 10
For maximum size subset, it is optimal to choose
the points on number line at coordinates 9, 10, 
11, 14 and 16, so that the size of the consecutive 
valued subset will become 10 which will be maximum .

Brute force Approach ( Time Complexity – O(N2) ):

A brute force consists of checking all the possible (l, r) pairs for the condition ((arr[r]-arr[l])-(r-l)) <= k. In order to find out if a pair (l, r) is valid, we should check if the number of points that need to be placed between these two initial ones is not greater than K. Since arr[i] is the coordinate of the i-th point in the input array (arr), then we need to check if (arr[r] – arr[l]) – (r – l ) <= k.

This solution has a complexity of O(N2).

C++

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/* C++ program to find the maximum size of subset of 
   points that can have consecutive values using 
   brute force */
#include <bits/stdc++.h>
using namespace std;
  
int maximiseSubset(int arr[], int n, int k)
{
    // Since we can always enforce the solution 
    // to contain all the K added points
    int ans = k; 
  
    for (int l = 0; l < n - 1; l++) 
        for (int r = l; r < n; r++) 
  
            // check if the number of points that 
            // need to be placed between these two 
            // initial ones is not greater than k
            if ((arr[r] - arr[l]) - (r - l) <= k) 
                ans = max(ans, r - l + k + 1);
  
    return (ans);
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 10, 11, 14, 15 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 4;
    printf("%dn", maximiseSubset(arr, n, k));
    return 0;
}

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Java

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/* Java program to find the maximum size of subset of 
points that can have consecutive values using 
brute force */
import java.util.*;
  
class GFG
{
  
    static int maximiseSubset(int[] arr, int n, int k)
    {
        // Since we can always enforce the solution 
        // to contain all the K added points
        int ans = k; 
  
        for (int l = 0; l < n - 1; l++) 
            for (int r = l; r < n; r++) 
  
                // check if the number of points that 
                // need to be placed between these two 
                // initial ones is not greater than k
                if ((arr[r] - arr[l]) - (r - l) <= k) 
                    ans = Math.max(ans, r - l + k + 1);
  
        return (ans);
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int[] arr = { 1, 2, 3, 4, 10, 11, 14, 15 };
        int n = arr.length;
        int k = 4;
        System.out.println(maximiseSubset(arr, n, k));  
    }
}
/* This code is contributed by Mr. Somesh Awasthi */

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Python 3

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# Python 3 program to find the maximum size
# of subset of points that can have consecutive 
# values using brute force 
  
def maximiseSubset(arr , n, k):
  
    # Since we can always enforce the solution 
    # to contain all the K added points
    ans = k
  
    for l in range(n - 1): 
        for r in range(l, n): 
  
            # check if the number of points that 
            # need to be placed between these two 
            # initial ones is not greater than k
            if ((arr[r] - arr[l]) - (r - l) <= k) :
                ans = max(ans, r - l + k + 1)
  
    return (ans)
  
# Driver code
if __name__ == "__main__":
      
    arr = [ 1, 2, 3, 4, 10, 11, 14, 15 ]
    n = len(arr)
    k = 4
    print(maximiseSubset(arr, n, k))
  
# This code is contributed by ita_c

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C#

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/* C# program to find the 
maximum size of subset of 
points that can have 
consecutive values using 
brute force */
using System;
  
class GFG
{
  
    static int maximiseSubset(int[] arr, 
                              int n, int k)
    {
        // Since we can always enforce 
        // the solution to contain all 
        // the K added points
        int ans = k; 
  
        for (int l = 0; l < n - 1; l++) 
            for (int r = l; r < n; r++) 
  
                // check if the number of 
                // points that need to be 
                // placed between these 
                // two initial ones is not
                // greater than k
                if ((arr[r] - arr[l]) - 
                              (r - l) <= k) 
                    ans = Math.Max(ans, r - l +
                                        k + 1);
  
        return (ans);
    }
  
    // Driver code
    public static void Main()
    {
        int[] arr = {1, 2, 3, 4, 
                    10, 11, 14, 15};
        int n = arr.Length;
        int k = 4;
        Console.WriteLine(maximiseSubset(arr, n, k)); 
    }
}
  
// This code is contributed by anuj_67.

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PHP

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<?php
// PHP program to find the maximum size 
// of subset of points that can have  
// consecutive values using brute force 
function maximiseSubset($arr, $n, $k
    // Since we can always enforce 
    // the solution to contain all
    // the K added points 
    $ans = $k
  
    for ($l = 0; $l < $n - 1; $l++) 
        for ($r = $l; $r < $n; $r++) 
  
            // check if the number of points that 
            // need to be placed between these two 
            // initial ones is not greater than k 
            if (($arr[$r] - $arr[$l]) - 
                ($r - $l) <= $k
                $ans = max($ans, $r - $l + $k + 1); 
  
    return ($ans); 
  
// Driver code 
$arr = array(1, 2, 3, 4, 10, 11, 14, 15 ); 
$n = sizeof($arr); 
$k = 4; 
echo (maximiseSubset($arr, $n, $k)); 
  
// This code is contributed 
// by Sach_Code    
?>

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Output:

8

Efficient Approach (Time complexity – O(N))

In order to optimize the brute force, notice that if r increases, then l also increases (or at least stays the same). We can maintain two indexes. Initialise l and r both as 0. Then we start incrementing r. As we do this, at each step we increment l until the condition used in brute force approach becomes true. When r reaches last index, we stop.

C++

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/* C++ program to find the maximum size of subset
   of points that can have consecutive values 
   using efficient approach */
#include <bits/stdc++.h>
using namespace std;
  
int maximiseSubset(int arr[], int n, int k)
{
    // Since we can always enforce the
    // solution to contain all the K added 
    // points
    int ans = k;
  
    int l = 0, r = 0;
    while (r < n) {
  
        // increment l until the number of points 
        // that need to be placed between index l
        // and index r is not greater than k
        while ((arr[r] - arr[l]) - (r - l) > k) 
            l++;
          
        // update the solution as below
        ans = max(ans, r - l + k + 1);
          
        r++;
    }
  
    return (ans);
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 10, 11, 14, 15 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 4;
    printf("%d", maximiseSubset(arr, n, k));
    return 0;
}

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Java

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/* Java program to find the maximum size of subset
of points that can have consecutive values 
using efficient approach */
import java.util.*;
  
class GFG
{
    static int maximiseSubset(int[] arr, int n, int k)
    {
        // Since we can always enforce the
        // solution to contain all the K added 
        // points
        int ans = k;
  
        int l = 0, r = 0;
        while (r < n) {
  
            // increment l until the number of points 
            // that need to be placed between index l
            // and index r is not greater than k
            while ((arr[r] - arr[l]) - (r - l) > k) 
                l++;
          
            // update the solution as below
            ans = Math.max(ans, r - l + k + 1);
          
            r++;
        }
  
        return (ans);
    }
  
    // Driver code  
    public static void main(String[] args)
    {
        int[] arr = { 1, 2, 3, 4, 10, 11, 14, 15 };
        int n = arr.length;
        int k = 4;
        System.out.println(maximiseSubset(arr, n, k));
    }
}
/* This code is contributed by Mr. Somesh Awasthi */

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Python3

# Python 3 program to find the maximum size
# of subset of points that can have consecutive
# values using efficient approach
def maximiseSubset(arr, n, k):

# Since we can always enforce the solution
# to contain all the K added points
ans = k;

l = 0; r = 0;
while (r < n): # increment l until the number of points # that need to be placed between index l # and index r is not greater than k while ((arr[r] - arr[l]) - (r - l) > k):
l = l + 1;

# update the solution as below
ans = max(ans, r – l + k + 1);

r = r + 1;

return (ans);

# Driver code
arr = [ 1, 2, 3, 4, 10, 11, 14, 15 ];
n = len(arr);
k = 4;
print(maximiseSubset(arr, n, k));

# This code is contributed
# by Akanksha Rai

C#

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/* C# program to find the 
maximum size of subset
of points that can have 
consecutive values using
efficient approach */
using System;
  
class GFG
{
    static int maximiseSubset(int[] arr, 
                              int n, int k)
    {
        // Since we can always enforce 
        // the solution to contain all
        // the K added points
        int ans = k;
  
        int l = 0, r = 0;
        while (r < n) 
        {
  
            // increment l until the 
            // number of points that 
            // need to be placed 
            // between index l and 
            // index r is not greater 
            // than k
            while ((arr[r] - arr[l]) - 
                        (r - l) > k) 
                l++;
          
            // update the
            // solution as below
            ans = Math.Max(ans, r - l +
                                k + 1);
          
            r++;
        }
  
        return (ans);
    }
  
    // Driver code 
    public static void Main()
    {
        int[] arr = {1, 2, 3, 4, 
                    10, 11, 14, 15};
        int n = arr.Length;
        int k = 4;
        Console.WriteLine(maximiseSubset(arr, n, k));
    }
}
  
// This code is contributed
// by anuj_67.

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PHP

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<?php
// PHP program to find the maximum size 
// of subset of points that can have
// consecutive values using efficient approach 
function maximiseSubset($arr, $n, $k)
{
    // Since we can always enforce the
    // solution to contain all the K 
    // added points
    $ans = $k;
  
    $l = 0; $r = 0;
    while ($r < $n
    {
  
        // increment l until the number of points 
        // that need to be placed between index l
        // and index r is not greater than k
        while (($arr[$r] - $arr[$l]) - 
               ($r - $l) > $k
            $l++;
      
        // update the solution as below
        $ans = max($ans, $r - $l + $k + 1);
      
        $r++;
    }
  
    return ($ans);
}
  
// Driver code 
$arr = array(1, 2, 3, 4, 10, 11, 14, 15 );
$n = sizeof($arr);
$k = 4;
echo(maximiseSubset($arr, $n, $k));
  
// This code is contributed by Mukul Singh
?>

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Output:

8

Time Complexity: O(N)

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