k size subsets with maximum difference d between max and min

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 : 10
Explanation:
{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 has atmost 
difference 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 ith 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 ith 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

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