# HackerRank ‘Non-Divisible Subset’ Solution

##### Short Problem Definition:

Given a set *S* of *n* distinct integers, print the size of a maximal subset *S’* of *S* where the sum of any 2 numbers in *S’* are not evenly divisible by* k*.

##### Link

##### Complexity:

time complexity is O(N)

space complexity is O(N)

##### Execution:

This is by all means not an easy task and is also reflected by the high failure ratio of the participants. For a sum of two numbers to be evenly divisible by k the following condition has to hold. If the remainder of *N1%k == r* then *N2%k = k-r* for* N1+N2 % k == 0*. Let us calculate the set of all numbers with a remainder of *r* and *k-r *and pick the larger set. If the remainder is half of k such as 2 % 4 = 2 or exactly k such as 4 % 4 = 0, just one number from each of these sets can be contained in* S’.*

##### Solution:

def solveSubset(S, k, n): r = [0] * k for value in S: r[value%k] += 1 result = 0 for a in xrange(1, (k+1)//2): result += max(r[a], r[k-a]) if k % 2 == 0 and r[k//2]: result += 1 if r[0]: result += 1 return result n, k = map(int, raw_input().split()) S = map(int, raw_input().split()) print solveSubset(S, k, n)

[rust]

use std::io;

fn get_numbers() -> Vec<u32> {

let mut line = String::new();

io::stdin().read_line(&mut line).ok().expect("Failed to read line");

line.split_whitespace().map(|s| s.parse::<u32>().unwrap()).collect()

}

fn calculate_nondivisible(a: Vec<u32>, n: usize, k: usize) -> u32 {

let mut result = 0;

let mut r = vec![0; k];

for val in a {

r[(val as usize)%k] += 1;

}

for idx in 1..(k+1)/2 {

result += std::cmp::max(r[idx as usize], r[(k-idx) as usize]);

}

if k % 2 == 0 && r[k/2] != 0 {

result += 1;

}

if r[0] != 0 {

result += 1;

}

result

}

fn main() {

let line = get_numbers();

let a = get_numbers();

println!("{}", calculate_nondivisible(a, line[0] as usize, line[1] as usize) );

}

[/rust]

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