08/5/16

# 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.

Non-Divisible Subset

##### 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();
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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08/4/16

# HackerRank ‘Flatland Space Station’ Solution

##### Short Problem Definition:

For each city, determine its distance to the nearest space station and print the maximum of these distances.

Flatland Space Station

##### Complexity:

time complexity is O(N)

space complexity is O(N)

##### Execution:

This is a two pass algorithm. First, measure the distance to the last station on the left. And on the second pass measure the distance to the nearest station on the right. Pick the minimum of both values. Remember that the first and last position are not necessarily stations.

##### Solution:

[rust]
use std::io;
use std::cmp;

fn get_numbers() -> Vec<u32> {
let mut line = String::new();
line.split_whitespace().map(|s| s.parse::<u32>().unwrap()).collect()
}

fn find_distance(c: Vec<u32>, n: usize) -> u32 {
let mut solution = 0;
let mut distances = vec![n as u32; n];

// first pass
let mut last_seen = 0;
let mut seen_one = false;
for i in 0..n {
if c[i] == 1 {
seen_one = true;
last_seen = 0;
} else {
last_seen += 1;
}
if seen_one {
distances[i] = last_seen;
}
}

// second pass
let mut last_seen = 0;
let mut seen_one = false;
for i in (0..n).rev() {
if c[i] == 1 {
seen_one = true;
last_seen = 0;
} else {
last_seen += 1;
}
solution = cmp::max(solution,
match seen_one {
true => cmp::min(last_seen, distances[i]),
false => distances[i],
}
);
}

solution
}

fn main() {
let line = get_numbers();
let n = line[0] as usize;
let c = get_numbers();
let mut v = vec![0; n];
for station in c {
v[station as usize] = 1;
}
println!("{}", find_distance(v, n));
}
[/rust]

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08/3/16

# HackerRank ‘Kangaroo’ Solution

##### Short Problem Definition:

There are two kangaroos on an x-axis ready to jump in the positive direction (i.e, toward positive infinity). The first kangaroo starts at location x1 and moves at a rate of v1 meters per jump. The second kangaroo starts at location x2 and moves at a rate of v2 meters per jump. Given the starting locations and movement rates for each kangaroo, can you determine if they’ll ever land at the same location at the same time?

Kangaroo

##### Complexity:

time complexity is O(1)

space complexity is O(1)

##### Execution:

There is no need to simulate the movement. We can reason that the two kangaroos either meat at the smallest common multiply or never.

##### Solution:

[rust]
use std::io;
fn get_numbers() -> Vec<u32> {
let mut line = String::new();
line.split_whitespace().map(|s| s.parse::<u32>().unwrap()).collect()
}

fn main() {
let numbers = get_numbers();
let x1 = numbers[0];
let v1 = numbers[1];
let x2 = numbers[2];
let v2 = numbers[3];

if x1 == x2 && v1 == v2 {
println!("YES");
}
else if x1 == x2 && v1 > v2 {
println!("NO");
}
else if x1 <= x2 && v1 <= v2 {
println!("NO");
}
else {
if (x2 – x1) % (v1 – v2) == 0 {
println!("YES");
}
else {
println!("NO");
}
}
}
[/rust]

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08/2/16

# HackerRank ‘Save the Prisoner!’ Solution

##### Short Problem Definition:

A jail has N prisoners, and each prisoner has a unique id number,S , ranging from 1 to N. There are M sweets that must be distributed to the prisoners. But wait—there’s a catch—the very last sweet S is poisoned! Can you find and print the ID number of the last prisoner to receive a sweet so he can be warned?

Save the Prisoner!

##### Complexity:

time complexity is O(1)

space complexity is O(1)

##### Execution:

This challenge is painlessly trivial.

##### Solution:

[rust]
use std::io;

fn get_number() -> u32 {
let mut line = String::new();
line.trim().parse::<u32>().unwrap()
}

fn get_numbers() -> Vec<u32> {
let mut line = String::new();
line.split_whitespace().map(|s| s.parse::<u32>().unwrap()).collect()
}

fn solve_prisoner(n: u32, m: u32, s: u32) -> u32 {
((s – 1 + m – 1 ) % n) +1
}

fn main() {
let t = get_number();
for _ in 0..t {
let line = get_numbers();
println!("{}", solve_prisoner(line[0], line[1], line[2]));
}
}
[/rust]

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08/2/16

# HackerRank ‘Jumping on the Clouds’ Solution

##### Short Problem Definition:

Emma is playing a new mobile game involving clouds numbered from 1 to n. There are two types of clouds, ordinary clouds and thunderclouds. The game ends if Emma jumps onto a thundercloud, but if she reaches the last cloud, she wins the game!

Jumping on the Clouds

##### Complexity:

time complexity is O(N)

space complexity is O(N)

##### Execution:

Theoretically your solution can depend on the fact that the win condition is guaranteed, but I don’t like such solutions. Here I present a semi-DP approach that keeps track of the optimal number of jumps it takes to reach each cloud.

##### Solution:

[rust]
use std::io;
use std::cmp;

fn get_number() -> u32 {
let mut line = String::new();
line.trim().parse::<u32>().unwrap()
}

fn get_numbers() -> Vec<u32> {
let mut line = String::new();
line.split_whitespace().map(|s| s.parse::<u32>().unwrap()).collect()
}

fn calculate_jumping(a: Vec<u32>, n: usize) -> u32{
let mut v = vec![100; n];
v[0] = 0;

for i in 1..n {
//println!("{} {} {:?}", i, a[i], v);
if a[i] == 1 {
continue;
}

if i == 1 {
v[i] = v[i-1] + 1;
} else {
v[i] = cmp::min(v[i-1], v[i-2]) + 1;
}
}

v[n-1]
}

fn main() {
let n = get_number() as usize;
let a = get_numbers();

println!("{}", calculate_jumping(a, n) );
}
[/rust]

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08/1/16

# HackerRank ‘Jumping on the Clouds: Revisited’ Solution

##### Short Problem Definition:

Aerith is playing a cloud game! In this game, there are clouds numbered sequentially from 1 to n. Each cloud is either an ordinary cloud or a thundercloud. Given the values of n and k the configuration of the clouds, can you determine the final value of e after the game ends?

Jumping on the Clouds: Revisited

##### Complexity:

time complexity is O(N)

space complexity is O(1)

##### Execution:

Simulate the game in a loop.

##### Solution:
```#!/bin/python

def solveCloudRevisited(c, n, k):
pos = 0
cnt = 0

while cnt == 0 or pos != 0:
pos += k
pos %= n
if c[pos] == 0:
cnt += 1
else:
cnt += 3

return 100 - cnt

if __name__ == '__main__':
n,k = map(int,raw_input().strip().split(' '))
c = map(int,raw_input().strip().split(' '))
print solveCloudRevisited(c, n, k)
```

[rust]
use std::io;

fn get_numbers() -&gt; Vec&lt;u32&gt; {
let mut line = String::new();
line.split_whitespace().map(|s| s.parse::&lt;u32&gt;().unwrap()).collect()
}

fn calculate_jumping(a: Vec&lt;u32&gt;, n: usize, k: usize) -&gt; u32 {
let mut e = 0;
let mut pos = 0;

while e == 0 || pos != 0 {
e += match a[pos] {
0 =&gt; 1,
1 =&gt; 3,
_ =&gt; panic!("invalid input"),
};

pos += k;
pos %= n;
}

100 – e
}

fn main() {
let line = get_numbers();
let (n, k) = (line[0], line[1]);
let a = get_numbers();
println!("{}", calculate_jumping(a, n as usize, k as usize) );
}
[/rust]

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The provided text passage contains multiple HackerRank problem solutions, but it does not contain a single, unified mathematical question or problem statement. Instead, it consists of several code snippets and problem descriptions for different HackerRank challenges. Here's a summary of the content structure:

- The passage contains solutions to several HackerRank problems, including:
  - "Non-Divisible Subset"
  - "Flatland Space Station"
  - "Kangaroo"
  - "Save the Prisoner!"
  - "Jumping on the Clouds"
  - "Jumping on the Clouds: Revisited"

Each problem is described with a short problem definition, complexity analysis, and a solution in Rust or Python.

Given this structure, the analysis is as follows:
no_question

the passage does not contain a single, unified mathematical question or problem statement. instead, it provides solutions to multiple distinct problems, each with its own context and solution.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
08/5/16

# 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.

Non-Divisible Subset

##### 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();
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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08/4/16

# HackerRank ‘Flatland Space Station’ Solution

##### Short Problem Definition:

For each city, determine its distance to the nearest space station and print the maximum of these distances.

Flatland Space Station

##### Complexity:

time complexity is O(N)

space complexity is O(N)

##### Execution:

This is a two pass algorithm. First, measure the distance to the last station on the left. And on the second pass measure the distance to the nearest station on the right. Pick the minimum of both values. Remember that the first and last position are not necessarily stations.

##### Solution:

[rust]
use std::io;
use std::cmp;

fn get_numbers() -> Vec<u32> {
let mut line = String::new();
line.split_whitespace().map(|s| s.parse::<u32>().unwrap()).collect()
}

fn find_distance(c: Vec<u32>, n: usize) -> u32 {
let mut solution = 0;
let mut distances = vec![n as u32; n];

// first pass
let mut last_seen = 0;
let mut seen_one = false;
for i in 0..n {
if c[i] == 1 {
seen_one = true;
last_seen = 0;
} else {
last_seen += 1;
}
if seen_one {
distances[i] = last_seen;
}
}

// second pass
let mut last_seen = 0;
let mut seen_one = false;
for i in (0..n).rev() {
if c[i] == 1 {
seen_one = true;
last_seen = 0;
} else {
last_seen += 1;
}
solution = cmp::max(solution,
match seen_one {
true => cmp::min(last_seen, distances[i]),
false => distances[i],
}
);
}

solution
}

fn main() {
let line = get_numbers();
let n = line[0] as usize;
let c = get_numbers();
let mut v = vec![0; n];
for station in c {
v[station as usize] = 1;
}
println!("{}", find_distance(v, n));
}
[/rust]

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08/3/16

# HackerRank ‘Kangaroo’ Solution

##### Short Problem Definition:

There are two kangaroos on an x-axis ready to jump in the positive direction (i.e, toward positive infinity). The first kangaroo starts at location x1 and moves at a rate of v1 meters per jump. The second kangaroo starts at location x2 and moves at a rate of v2 meters per jump. Given the starting locations and movement rates for each kangaroo, can you determine if they’ll ever land at the same location at the same time?

Kangaroo

##### Complexity:

time complexity is O(1)

space complexity is O(1)

##### Execution:

There is no need to simulate the movement. We can reason that the two kangaroos either meat at the smallest common multiply or never.

##### Solution:

[rust]
use std::io;
fn get_numbers() -> Vec<u32> {
let mut line = String::new();
line.split_whitespace().map(|s| s.parse::<u32>().unwrap()).collect()
}

fn main() {
let numbers = get_numbers();
let x1 = numbers[0];
let v1 = numbers[1];
let x2 = numbers[2];
let v2 = numbers[3];

if x1 == x2 && v1 == v2 {
println!("YES");
}
else if x1 == x2 && v1 > v2 {
println!("NO");
}
else if x1 <= x2 && v1 <= v2 {
println!("NO");
}
else {
if (x2 – x1) % (v1 – v2) == 0 {
println!("YES");
}
else {
println!("NO");
}
}
}
[/rust]

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08/2/16

# HackerRank ‘Save the Prisoner!’ Solution

##### Short Problem Definition:

A jail has N prisoners, and each prisoner has a unique id number,S , ranging from 1 to N. There are M sweets that must be distributed to the prisoners. But wait—there’s a catch—the very last sweet S is poisoned! Can you find and print the ID number of the last prisoner to receive a sweet so he can be warned?

Save the Prisoner!

##### Complexity:

time complexity is O(1)

space complexity is O(1)

##### Execution:

This challenge is painlessly trivial.

##### Solution:

[rust]
use std::io;

fn get_number() -> u32 {
let mut line = String::new();
line.trim().parse::<u32>().unwrap()
}

fn get_numbers() -> Vec<u32> {
let mut line = String::new();
line.split_whitespace().map(|s| s.parse::<u32>().unwrap()).collect()
}

fn solve_prisoner(n: u32, m: u32, s: u32) -> u32 {
((s – 1 + m – 1 ) % n) +1
}

fn main() {
let t = get_number();
for _ in 0..t {
let line = get_numbers();
println!("{}", solve_prisoner(line[0], line[1], line[2]));
}
}
[/rust]

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08/2/16

# HackerRank ‘Jumping on the Clouds’ Solution

##### Short Problem Definition:

Emma is playing a new mobile game involving clouds numbered from 1 to n. There are two types of clouds, ordinary clouds and thunderclouds. The game ends if Emma jumps onto a thundercloud, but if she reaches the last cloud, she wins the game!

Jumping on the Clouds

##### Complexity:

time complexity is O(N)

space complexity is O(N)

##### Execution:

Theoretically your solution can depend on the fact that the win condition is guaranteed, but I don’t like such solutions. Here I present a semi-DP approach that keeps track of the optimal number of jumps it takes to reach each cloud.

##### Solution:

[rust]
use std::io;
use std::cmp;

fn get_number() -> u32 {
let mut line = String::new();
line.trim().parse::<u32>().unwrap()
}

fn get_numbers() -> Vec<u32> {
let mut line = String::new();
line.split_whitespace().map(|s| s.parse::<u32>().unwrap()).collect()
}

fn calculate_jumping(a: Vec<u32>, n: usize) -> u32{
let mut v = vec![100; n];
v[0] = 0;

for i in 1..n {
//println!("{} {} {:?}", i, a[i], v);
if a[i] == 1 {
continue;
}

if i == 1 {
v[i] = v[i-1] + 1;
} else {
v[i] = cmp::min(v[i-1], v[i-2]) + 1;
}
}

v[n-1]
}

fn main() {
let n = get_number() as usize;
let a = get_numbers();

println!("{}", calculate_jumping(a, n) );
}
[/rust]

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08/1/16

# HackerRank ‘Jumping on the Clouds: Revisited’ Solution

##### Short Problem Definition:

Aerith is playing a cloud game! In this game, there are clouds numbered sequentially from 1 to n. Each cloud is either an ordinary cloud or a thundercloud. Given the values of n and k the configuration of the clouds, can you determine the final value of e after the game ends?

Jumping on the Clouds: Revisited

##### Complexity:

time complexity is O(N)

space complexity is O(1)

##### Execution:

Simulate the game in a loop.

##### Solution:
```#!/bin/python

def solveCloudRevisited(c, n, k):
pos = 0
cnt = 0

while cnt == 0 or pos != 0:
pos += k
pos %= n
if c[pos] == 0:
cnt += 1
else:
cnt += 3

return 100 - cnt

if __name__ == '__main__':
n,k = map(int,raw_input().strip().split(' '))
c = map(int,raw_input().strip().split(' '))
print solveCloudRevisited(c, n, k)
```

[rust]
use std::io;

fn get_numbers() -&gt; Vec&lt;u32&gt; {
let mut line = String::new();
line.split_whitespace().map(|s| s.parse::&lt;u32&gt;().unwrap()).collect()
}

fn calculate_jumping(a: Vec&lt;u32&gt;, n: usize, k: usize) -&gt; u32 {
let mut e = 0;
let mut pos = 0;

while e == 0 || pos != 0 {
e += match a[pos] {
0 =&gt; 1,
1 =&gt; 3,
_ =&gt; panic!("invalid input"),
};

pos += k;
pos %= n;
}

100 – e
}

fn main() {
let line = get_numbers();
let (n, k) = (line[0], line[1]);
let a = get_numbers();
println!("{}", calculate_jumping(a, n as usize, k as usize) );
}
[/rust]

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user
The provided text passage contains multiple HackerRank problem solutions, but it does not contain a single, unified mathematical question or problem statement. Instead, it consists of several code snippets and problem descriptions for different HackerRank challenges. Here's a summary of the content structure:

- The passage contains solutions to several HackerRank problems, including:
  - "Non-Divisible Subset"
  - "Flatland Space Station"
  - "Kangaroo"
  - "Save the Prisoner!"
  - "Jumping on the Clouds"
  - "Jumping on the Clouds: Revisited"

Each problem is described with a short problem definition, complexity analysis, and a solution in Rust or Python.

Given this structure, the analysis is as follows:

TYPE: NO_QUESTION

The passage does not contain a single, unified mathematical question or problem statement. Instead, it provides solutions to multiple distinct problems, each with its own context and solution.