Given an array, find number of subsets with k elements, where absolute difference between the maximum and mininmum element is at most x

Problem

Given an Array consisting of N integers, we need to find the number of subsets of this Array of size K, where Absolute difference between the Maximum and Minimum element of the subset is at most X.

Explanation

Lets say given Array is [4 5 1 3 2] and K=3 and X=5

number of subsets satisfying given conditions will be 10

One of the subset is [1 2 5] having K elements and absolute difference between minimum and maximum element less then X i.e 5-1<X

Solution

This problem comes under easy-medium level.
Formally a k-combination of a set S is a subset of k distinct elements of S

  1. First sort the Array
    [1 2 3 4 5]
  • Fix the first element.

  • Then traverse the array, counting elements whose difference with fixed element is less than or equal to X.

  • Calculate K-1 combinations of (counted elements - fixed element’s index), if (counted elements - fixed element’s index) is greater then k-1 .
    K-1 because one element is fixed.

  • Fix the next element and repeat from step 3.

  • Repeat above step length(array)-1 times.

Psuedo code

# this function calulates r combinations of n elements

def ncr(n,r):
end

arr=given_array;pointer=0;count=0;answer=0
arr=sort(arr)
N=length(arr)
while(pointer<N):
while(count<N and arr[count]-arr[pointer]<=X):
count++
count--
if (count-pointer>=K):
answer+=ncr(count-pointer,K)
pointer++
print answer

For large arrays using modulo

For arrays small than 20 elemets we can find combinations directly nCr = n! / (r! * (n-r)!)

For arrays large than 20 elements finding combinations goes out of integer limit. So we have to calculate nCr modulo M

If we have to calcuate nCr mod p(where p is a prime), we can calculate factorial mod p and then use modular inverse to find nCr mod p. If we have to find nCr mod m(where m is not prime), we can factorize m into primes and then use Chinese Remainder Theorem(CRT) to find nCr mod m.

For more info on calculating nCr for large numbers

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