09/13/18

# HackerRank ‘CamelCase’ Solution

##### Short Problem Definition:

Alice wrote a sequence of words in CamelCase as a string of letters, , having the following properties:

• It is a concatenation of one or more words consisting of English letters.
• All letters in the first word are lowercase.
• For each of the subsequent words, the first letter is uppercase and rest of the letters are lowercase.

Given s , print the number of words in s on a new line.

For example, s = OneTwoThree . There are 3 words in the string.

CamelCase

##### Complexity:

time complexity is O(N)

space complexity is O(1)

##### Execution:

Since the input always has at least 1 character, we can assume that there will always be at least one word. Each upper case later identifies the next word. So the result is number of capital letters + 1.

##### Solution:
s = raw_input().strip()

cnt = 1

for c in s:
if c.isupper():
cnt += 1

print cnt


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09/12/18

# HackerRank ‘Super Reduced String’ Solution

##### Short Problem Definition:

Steve has a string of lowercase characters in range ascii[‘a’..’z’]. He wants to reduce the string to its shortest length by doing a series of operations. In each operation he selects a pair of adjacent lowercase letters that match, and he deletes them. For instance, the string aab could be shortened to b in one operation.

Steve’s task is to delete as many characters as possible using this method and print the resulting string. If the final string is empty, print Empty String

Super Reduced String

##### Complexity:

time complexity is O(N)

space complexity is O(N)

##### Execution:

The solution creates a stack of values. If the top value on the stack is equivalent to the next value, simply remove both. This solution assumes that there are only ever 2 values next to each other. If any adjacent values were to be removed, I would require more loops.

##### Solution:
i = raw_input()

s = []

for c in i:
if not s:
s.append(c)
else:
if s[-1] == c:
s.pop()
else:
s.append(c)

if not s:
print "Empty String"
else:
print ''.join(s)


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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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07/29/16

# HackerRank ‘Circular Array Rotation’ Solution

##### Short Problem Definition:

John Watson performs an operation called a right circular rotation on an array of integers.

Circular Array Rotation

##### Complexity:

time complexity is O(Q)

space complexity is O(1)

##### Execution:

Calculate the offset for every query. Watch out for index overflows and negative modulo.

##### 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 main() {
let line = get_numbers();
let (n,k,q) = (line[0], line[1], line[2]);
let a = get_numbers();
for _ in 0..q {
let m = get_number();
let idx = ((m-(k%n)+n)%n) as usize;
println!("{}", a[idx]);
}
}


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07/29/16

# HackerRank ‘Compare Triplets’ Solution

##### Short Problem Definition:

Alice and Bob each created one problem for HackerRank. A reviewer rates the two challenges, awarding points on a scale from 1 to 100  for three categories: problem clarity, originality, and difficulty.

Compare Triplets

##### Complexity:

time complexity is O(1)

space complexity is O(1)

##### Execution:

This is a warmup. Follow specification.

##### Solution:
use std::io;
use std::cmp::Ordering;

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

fn main() {
let a = get_numbers();
let b = get_numbers();

let mut alice = 0;
let mut bob = 0;

for idx in 0..3 {
match a[idx].cmp(&b[idx]) {
Ordering::Less      => bob += 1,
Ordering::Greater   => alice += 1,
Ordering::Equal     => {},
}
}

println!("{} {}", alice, bob);
}


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