Shaddyjr

// source: https://www.hackerrank.com/challenges/almost-sorted/problem
// video: https://youtu.be/deEW0pxWwsA

function almostSorted(arr) {
    const isSortedAroundIndex = index => {
        /*
        Return true if value at and around index are sorted.
        Handles edge "bounces".
        */
        const val = arr[index]

Shaddyjr

// source: https://www.hackerrank.com/challenges/larrys-array/problem
// video: https://youtu.be/XWXVRdpYE0I

function larrysArray(A) {
    // Since each value in the given array is unique we can map backwards
    // the index of each value (allows us to quickly "find" where a desired val is)
    const value_index_map = {}
    for(let i = 0; i < A.length; i++){ // O(n)
        const val = A[i]
        value_index_map[val] = i

Shaddyjr

// source: https://www.hackerrank.com/challenges/minimum-distances/problem
// video: https://youtu.be/gKp8Aq9mm9E

function minimumDistances(a) {
    // stores each unique value as a key with a value of last index
    const val_last_i = {}
    
    let min_distance = Infinity;
    // loop through array and populate val_last_i
    for(let i = 0; i < a.length; i++){ // O(n)

Shaddyjr

# source: https://www.hackerrank.com/challenges/absolute-permutation/problem
# video: https://youtu.be/Z0bpJiVz0no

def absolutePermutation(n, k):
    if k == 0:  # no need to swap anything
        return list(range(1, n + 1))
    cache = set(range(1, n + 1)) # O(n)
    output = []
    for i in range(1, n + 1): # O(n)
        lower = i - k

Shaddyjr

# source: https://www.hackerrank.com/challenges/kaprekar-numbers/problem
# video: https://youtu.be/UoHhr2dwP6g

def is_mod_kap(num):
    '''Returns True if num is a modified kaprekar number'''
    sq_num_str = str(num ** 2)
    mid = len(sq_num_str) // 2  # floor division
    left = sq_num_str[:mid] or 0
    right = sq_num_str[mid:] or 0

Shaddyjr

# source: https://www.hackerrank.com/challenges/sparse-arrays/problem
# Glitched Failure video: https://youtu.be/n9w7Nuy-FmA

function matchingStrings(strings, queries) {
    const memory = {};
    
    // O(n), where n = number of string
    for(const string of strings) memory[string] = (memory[string] || 0) + 1;

    // O(m), where m = number of queries (the lookup is constant time, O(1))

Shaddyjr

# source: https://www.hackerrank.com/challenges/richie-rich/problem
# video: https://youtu.be/fcf4gy5V3L4
def count_mismatches(string_1, string_2):
    """
    Returns count of mismatches between two given strings
    """
    assert len(string_1) == len(string_2)
    count = 0
    for char_1, char_2 in zip(string_1, string_2):
        count += char_1 != char_2

Shaddyjr

# source: https://www.hackerrank.com/challenges/lisa-workbook/problem
# video: https://youtu.be/G2_o-FqdCTs

def workbook(n, k, arr):
    count = 0

    current_page = 1 # NOTE: we're indexing starting at 1
    for question_count in arr: # O(n)
        pages_added, remaining_questions = divmod(question_count, k)
        for new_chapter_page in range(1, pages_added + 1): # O(pmodk) => O(100mod1) => O(100) => O(1)

Shaddyjr

// source: https://www.hackerrank.com/challenges/coin-change
// video: https://www.youtube.com/watch?v=b34GYbwwrJ8
function getWays(n, c) {
    const store = new Array(n + 1).fill(0);
    store[0] = 1;

    for(const coin of c){
        for(let total = coin; total < store.length; total++){
            const remainder = total - coin;
            store[total] += store[remainder];

Shaddyjr

# source: https://www.hackerrank.com/challenges/abbr/problem
# video: https://youtu.be/4WzCFTmjKu4

def abbreviation(a, b):
    n = len(a)
    m = len(b)

    # init dp_list => O(n * m)
    dp_list = [[False] * (m + 1) for _ in range(n + 1)] # + 1 b/c including empty string
    dp_list[0][0] = True # abbreviation("", "") => True (base case)