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)


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) );
}


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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:
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));
}


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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:
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");
}
}
}


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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:
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]));
}
}


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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:
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) );
}


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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)


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) );
}


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07/3/16

# HackerRank ‘Divisible Sum Pairs’ Solution

##### Short Problem Definition:

You are given an array of n integers and a positive integer, k. Find and print the number of (i,j) pairs where i < j and ai + aj is evenly divisible by k.

Divisible Sum Pairs

##### Complexity:

time complexity is O(N^2)

space complexity is O(1)

##### Execution:

Brute force search.

##### Solution:
n,k = raw_input().strip().split(' ')
n,k = [int(n),int(k)]
a = map(int,raw_input().strip().split(' '))
count=0
for i in xrange(len(a)):
for j in xrange(i+1,len(a)):
if (a[i]+a[j]) % k == 0:
count+=1

print count


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06/22/16

# HackerRank ‘Lisas Workbook’ Solution

##### Short Problem Definition:

Lisa just got a new math workbook. A workbook contains exercise problems, grouped into chapters.

• …..

Lisa believes a problem to be special if its index (within a chapter) is the same as the page number where it’s located. Given the details for Lisa’s workbook, can you count its number of special problems?

Lisa’s Workbook

##### Complexity:

time complexity is O(N)

space complexity is O(1)

##### Execution:

This brute force solution iterates over all pages in the final book keeping track of the page(offset) and chapter it is on. There can only be one special problem per page and therefore I check if there is any problem that would match the criteria.

##### Solution:
def findPages(N, K, P):
cnt = 0
offset = 1
for chapter in P:
pages = (chapter + K -1)/K
for idx in xrange(pages):
if offset >= (idx * K)+1 and offset <= min((idx+1)*K, chapter):
cnt += 1
offset += 1

return cnt

if __name__ == '__main__':
N, K = map(int, raw_input().split())
P = map(int, raw_input().split())
print findPages(N, K, P)


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06/21/16

# HackerRank ‘Sherlock and The Beast’ Solution

##### Short Problem Definition:

Sherlock Holmes suspects his archenemy, Professor Moriarty, is once again plotting something diabolical. Sherlock’s companion, Dr. Watson, suggests Moriarty may be responsible for MI6’s recent issues with their supercomputer, The Beast.

Shortly after resolving to investigate, Sherlock receives a note from Moriarty boasting about infecting The Beastwith a virus; however, he also gives him a clue—a number, . Sherlock determines the key to removing the virus is to find the largest Decent Number having digits.

Sherlock and The Beast

##### Complexity:

time complexity is O(1)

space complexity is O(1)

##### Execution:

I am presenting two solutions. The naive brute force solution executes in O(N) and passes all the tests. Yet further optimisations and a runtime complexity of O(1) are possible.

When observing the possible output space we realise that there can only be 0, 5 or 10 threes in the output. If there would be 15 threes, it is better to use fives instead. The number of trailing threes can therefore be defined by K = 5*((2*N)%3). Let us plug some numbers into the equation:

• 1 -> 5*(2%3) = 10 -> INVALID
• 2 -> 5*(4%3) = 5 -> INVALID
• 3 -> 5*(3%3) = 0 -> 555
• 4 -> 5*(8%3) = 10 -> INVALID
• 5 -> 5*(10%3) = 5 -> 33-33-3
• 8 -> 5*(16%3) = 5 -> 555-33-33-3
• 10 -> 5*(20%3) = 10 -> 33-33-33-33-33
• 15 -> 5*(30%3) = 0 -> 555-555-555-555-555
##### Solution:
def sherlockBeast(N):
K = 5*((2*N)%3)
if K > N:
return -1
else:
return '5' * (N-K) + '3'*K

if __name__ == '__main__':
t = input()
for _ in xrange(t):
n = input()
print sherlockBeast(n)

def sherlockBeastNaive(N):
if (N < 3): return "-1" three_cnt = N//3 five_cnt = 0 while three_cnt >=0:
rem = N - three_cnt*3
if rem % 5 == 0:
five_cnt = rem/5
break
three_cnt -= 1

if three_cnt <= 0 and five_cnt == 0:
return "-1"

return "555"*three_cnt + "33333"*five_cnt

if __name__ == '__main__':
t = input()
for _ in xrange(t):
n = input()
print sherlockBeastNaive(n)


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05/13/16

# HackerRank ‘Bigger is Greater’ Solution

##### Short Problem Definition:

Given a word w, rearrange the letters of w to construct another word in such a way that is lexicographically greater than w. In case of multiple possible answers, find the lexicographically smallest one among them.

Bigger is Greater

##### Complexity:

time complexity is O(N)

space complexity is O(N)

##### Execution:

This task challenges us to find the next permutation of any given array. There are many implementations available online and it is worthwhile comparing them . I would recommend reading the article by Nayuki or re-implementing the std::next_permutation.

##### Solution:
def next_permutation(arr):
# Find non-increasing suffix
i = len(arr) - 1
while i > 0 and arr[i - 1] >= arr[i]:
i -= 1
if i <= 0:
return False

# Find successor to pivot
j = len(arr) - 1
while arr[j] <= arr[i - 1]:
j -= 1
arr[i - 1], arr[j] = arr[j], arr[i - 1]

# Reverse suffix
arr[i : ] = arr[len(arr) - 1 : i - 1 : -1]
return True

def main():
t = input()
for _ in xrange(t):
s = list(raw_input())
if next_permutation(s):
print "".join(s)
else: