Given an array and two integers k and d, find the number of subsets of this array of size k, where difference between the maximum and minimum number of the subset is **atmost** d.

Examples:

Input : a[] = [5, 4, 2, 1, 3], k = 3, d = 5 Output : 10Explanation: {1,2,3}, {1,2,4}, {1,2,5}, {1,3,4}, {1,3,5}, {1,4,5}, {2,3,4}, {2,3,5}, {2,4,5}, {3,4,5}. We can see each subset hasatmostdifference d=5 between the minimum and maximum element of each subset. No of such subsets = 10 Input : a[] = [1, 2, 3, 4, 5, 6], k = 3, d = 5 Output : 20

**Naive approach**: Finding all the subsets of size k and for each subset find the difference between maximum and minimum element. If the difference is less than or equal to d, count them.

**Efficient approach **:

1) **Sorting**: First **sort** the array in increasing order. Now, assume we want to find out for each i**th** element, the number of required subsets in which integer a[i] is present as the minimum element of that subset. The maximum in such a subset will never exceed a[i] + d .

2) **Find maximum index j **: We can apply **binary search** over this array for each i, to find the maximum index j, such that a[j] <= a[i]+d . Now any subset that includes a[i] and any other elements from the range i+1…j will be a required subset as because element a[i] is the minimum of that subset, and the difference between any other element and a[i] is always less than equal to d.

3)** Apply basic combinatorics formula **: Now we want to find the number of required subsets of size k. This will be by using the basic formula of combination when you have to select r items from given n numbers. In the same way we need to choose (k-1) numbers from (j-i) elements already including a[i] which is the minimum number in each subset. The sum of this procedure for each i**th** element will be the final answer.

Here I have used a simple recursive way to find factorial of a number one can use dynamic programming as well to find it.

Illustration :

Input : a = [5, 4, 2, 1, 3],

k = 3, d = 5

Output : 10

Explanation:

Sorted array in ascending order : [1, 2, 3, 4, 5]For a[0] = 1 as minimum element

No. of subset will be 6 which are {1, 2, 3}, {1, 2,

4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {1, 4, 5}.For a[1] = 2 as minimum element

No. of subset will be 3 which are {2, 3, 4}, {2,

3, 5}, {2, 4, 5}For a[2] = 3 as minimum element

No. of subset will be 1 which is {3, 4, 5}

No other subset of size k = 3 will be formed

by taking a[3] = 4 or a[4] = 5 as minimum element

## C++

[sourcecode language=”CPP”]

// C++ code to find no. of subsets with

// maximum difference d between max and

// min of all K-size subsets function to

// calculate factorial of a number

#include <bits/stdc++.h>

using namespace std;

int fact (int n){

if (n==0)

return 1;

else

return n * fact(n-1);

}

// function to count ways to select r

// numbers from n given numbers

int findcombination (int n,int r){

return( fact(n) / (fact(n – r) *

fact(r)));

}

// function to return the total number

// of required subsets :

// n is the number of elements in array

// d is the maximum difference between

// minimum and maximum element in each

// subset of size k

int find(int arr[], int n, int d, int k)

{

sort(arr,arr+n);

int ans = 0, end = n, co = 0,

start = 0;

// loop to traverse from 0-n

for (int i = 0; i < n; i++) {

int val = arr[i] + d;

// binary search to get the position

// which will be stored in start

start = i;

while (start < end – 1){

int mid = (start + end) / 2;

// if mid value greater than

// arr[i]+d do search in

// arr[start:mid]

if (arr[mid] > val)

end = mid;

else

start = mid + 1;

}

if (start != n and arr[start]

<= val)

start += 1;

int c = start-i;

// if the numbers of elements ‘c’

// is greater or equal to the given

// size k, then only subsets of

// required size k can be formed

if (c >= k){

co += findcombination(c – 1, k – 1);}

}

return co;

}

// drive program to test the above

// function

int main()

{

int arr[] = {1, 2, 3, 4, 5, 6},

k = 3, d = 5;

int n = sizeof(arr) / sizeof(arr[0]);

cout << find(arr, n,d,k);

return 0;

}

// This code is contributed by Prerna Saini

[/sourcecode]

## Java

[sourcecode language=”Java”]

// Java code to find no. of subsets

// with maximum difference d between

// max and min of all K-size subsets

import java.util.*;

class GFG {

// function to calculate factorial

// of a number

static int fact (int n){

if (n==0)

return 1;

else

return n * fact(n-1);

}

// function to count ways to select r

// numbers from n given numbers

static int findcombination(int n, int r){

return( fact(n) / (fact(n – r) *

fact(r)));

}

// function to return the total number

// of required subsets :

// n is the number of elements in array

// d is the maximum difference between

// minimum and maximum element in each

// subset of size k

static int find(int arr[], int n, int d,

int k)

{

Arrays.sort(arr);

int ans = 0, end = n, co = 0,

start = 0;

// loop to traverse from 0-n

for (int i = 0; i < n; i++) {

int val = arr[i] + d;

// binary search to get the position

// which will be stored in start

start=i;

while (start < end – 1){

int mid = (start + end) / 2;

// if mid value greater than

// arr[i]+d do search in

// arr[start:mid]

if (arr[mid] > val)

end = mid;

else

start = mid+1;

}

if (start !=n && arr[start] <= val)

start += 1;

int c = start-i;

// if the numbers of elements ‘c’ is

// greater or equal to the given size k,

// then only subsets of required size k

// can be formed

if (c >= k){

co += findcombination(c – 1, k – 1);}

}

return co;

}

// drive program to test the above function

public static void main(String[] args)

{

int arr[] = {1, 2, 3, 4, 5, 6}, k = 3,

d = 5;

int n = arr.length;

System.out.println(find(arr, n,d,k));

}

}

// This code is contributed by Prerna Saini

[/sourcecode]

## Python

[sourcecode language=”Python”]

# Python code to find no. of subsets with maximum

# difference d between max and min of all K-size

# subsets function to calculate factorial of a

# number

def fact (n):

if (n==0):

return (1)

else:

return n * fact(n-1)

# function to count ways to select r numbers

# from n given numbers

def findcombination (n,r):

return( fact(n)//(fact(n-r)*fact(r)))

# function to return the total number of required

# subsets :

# n is the number of elements in list l[0..n-1]

# d is the maximum difference between minimum and

# maximum element in each subset of size k

def find (a, n, d, k):

# sort the list first in ascending order

a.sort()

(start, end, co) = (0, n, 0)

for i in range(0, n):

val = a[i]+ d

# binary search to get the position

# which will be stored in start

# such that a[start] <= a[i]+d

start = i

while (start< end-1):

mid = (start+end)//2

# if mid value greater than a[i]+d

# do search in l[start:mid]

if (a[mid] > val):

end = mid

# if mid value less or equal to a[i]+d

# do search in a[mid+1:end]

else:

start = mid+1

if (start!=n and a[start]<=val):

start += 1

# count the numbers of elements that fall

# in range i to start

c = start-i

# if the numbers of elements ‘c’ is greater

# or equal to the given size k, then only

# subsets of required size k can be formed

if (c >= k):

co += findcombination(c-1,k-1)

return co

# Driver code

n = 6 # Number of elements

d = 5 # maximum diff

k = 3 # Size of subsets

print(find([1, 2, 3, 4, 5, 6], n, d, k))

[/sourcecode]

## C#

[sourcecode language=”CSHARP”]

// C# code to find no. of subsets

// with maximum difference d between

// max and min of all K-size subsets

using System;

class GFG {

// function to calculate factorial

// of a number

static int fact (int n)

{

if (n == 0)

return 1;

else

return n * fact(n – 1);

}

// function to count ways to select r

// numbers from n given numbers

static int findcombination(int n, int r)

{

return( fact(n) / (fact(n – r) *

fact(r)));

}

// function to return the total number

// of required subsets :

// n is the number of elements in array

// d is the maximum difference between

// minimum and maximum element in each

// subset of size k

static int find(int []arr, int n, int d,

int k)

{

Array.Sort(arr);

//int ans = 0,

int end = n, co = 0,

start = 0;

// loop to traverse from 0-n

for (int i = 0; i < n; i++)

{

int val = arr[i] + d;

// binary search to get the

// position which will be

// stored in start

start = i;

while (start < end – 1){

int mid = (start + end) / 2;

// if mid value greater than

// arr[i]+d do search in

// arr[start:mid]

if (arr[mid] > val)

end = mid;

else

start = mid+1;

}

if (start !=n && arr[start] <= val)

start += 1;

int c = start-i;

// if the numbers of elements ‘c’ is

// greater or equal to the given size k,

// then only subsets of required size k

// can be formed

if (c >= k)

co += findcombination(c – 1, k – 1);

}

return co;

}

// drive program to test the above function

public static void Main()

{

int []arr = {1, 2, 3, 4, 5, 6};

int k = 3;

int d = 5;

int n = arr.Length;

Console.WriteLine(find(arr, n, d, k));

}

}

// This code is contributed by anuj_67.

[/sourcecode]

Output:

20

This article is contributed by **Sruti Rai **. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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