Short Problem Definition:

You are given a string and your task is to swap cases. In other words, convert all lowercase letters to uppercase letters and vice versa.

Link

sWAP cASE

Complexity:

time complexity is O(N)

space complexity is O(N)

Execution:

I had too much fun with this one, so I refuse to admit that there is a buildin swapcase() function. ASCII can be fun.

Solution:

def swap_case(s):
    result = ""
    for idx in xrange(len(s)):
        ordinal = ord(s[idx])
        if (ordinal >= ord('a') and ordinal <= ord('z')) or \
            (ordinal >= ord('A') and ordinal <= ord('Z')):
            result += chr(ordinal-ord('A')+32)%64+ord('A'))
        else:
            result += s[idx]
    return result

Short Problem Definition:

A prime is a natural number greater than 1 that has no positive divisors other than 1 and itself. Given n integers, determine the primality of each integer and print whether it is Prime or Not prime on a new line.

Note: If possible, try to come up with an O(sqrt(N)) primality algorithm, or see what sort of optimizations you can come up with for an O(N) algorithm.

Link

Time Complexity: Primality

Complexity:

time complexity is O(sqrt(N))

space complexity is O(1)

Execution:

The point of this question is to make sure that you only iterate up to (including) sqrt(N).

Solution:

def primality(n):
    if n == 1:
        return False

    for i in xrange(2, int(math.sqrt(n))+1):
        if n % i == 0:
            return False
    return True

Short Problem Definition:

You are given an unordered array consisting of consecutive integers [1, 2, 3, …, n] without any duplicates. You are allowed to swap any two elements. You need to find the minimum number of swaps required to sort the array in ascending order.

Link

Minimum Swaps 2

Complexity:

time complexity is O(N)

space complexity is O(1)

Execution:

This solution runs in O(N) since it will visit every element at most 2 times. No need for complex cycle algorithms, stacks, etc.

The trick is to put every element in the place it belongs to and swap it with the element at that position. It is possible that the swapped element wont be the correct once, hence the while loop.

Solution:

def minimumSwaps(arr):
    swaps = 0
    n = len(arr)

    for idx in xrange(n):
        while arr[idx]-1 != idx:
            ele = arr[idx]
            arr[ele-1], arr[idx] = arr[idx], arr[ele-1]
            swaps += 1
    return swaps

Short Problem Definition:

Anna and Brian are sharing a meal at a restaurant and they agree to split the bill equally. Brian wants to order something that Anna is allergic to though, and they agree that Anna won’t pay for that item. Brian gets the check and calculates Anna’s portion. You must determine if his calculation is correct.

Link

Bon Appétit

Complexity:

time complexity is O(N)

space complexity is O(N)

Execution:

Anna is being fairly difficult. I hope that their relationship is otherwise healthy. As for the code: just follow the description.

Solution:

def bonAppetit(bill, k, b):
    should_have = (sum(bill) - bill[k])/2
    diff = b - should_have
    if (not diff):
        return "Bon Appetit"
    else:
        return diff

Short Problem Definition:

John works at a clothing store. He has a large pile of socks that he must pair by color for sale. Given an array of integers representing the color of each sock, determine how many pairs of socks with matching colors there are.

Link

Sock Merchant

Complexity:

time complexity is O(N)

space complexity is O(N)

Execution:

Count the occurrence of every element. Use integer division to eliminate unpaired socks.

Solution:

from collections import defaultdict

def sockMerchant(n, ar):
    d = defaultdict(int)

    for k in ar:
        d[k] += 1
        
    cnt = 0
    for ele in d.values():
        cnt += ele // 2
        
    return cnt

Short Problem Definition:

Brie’s Drawing teacher asks her class to open their books to a page number. Brie can either start turning pages from the front of the book or from the back of the book. She always turns pages one at a time. 

Link

Drawing Book

Complexity:

time complexity is O(1)

space complexity is O(1)

Execution:

Think of this as a 0-indexed book. Use integer division.

Solution:

def solve(n, p):
    last_letter = n // 2
    goal_letter = p // 2
    return min(goal_letter, last_letter-goal_letter)

Short Problem Definition:

Gary is an avid hiker. He tracks his hikes meticulously, paying close attention to small details like topography. During his last hike he took exactly n steps. For every step he took, he noted if it was an uphill, U, or a downhill, D step. Gary’s hikes start and end at sea level and each step up or down represents a  unit change in altitude. We define the following terms:

• A mountain is a sequence of consecutive steps above sea level, starting with a step up from sea level and ending with a step down to sea level.
• A valley is a sequence of consecutive steps below sea level, starting with a step down from sea level and ending with a step up to sea level.

Link

Counting Valleys

Complexity:

time complexity is O(N)

space complexity is O(1)

Execution:

I am a hiker myself, so this challenge is kinda funny. The description looks complex, but realistically one only needs to calculate whether the section was above or below sea level.

Solution:

def countingValleys(n, s):
    level = 0
    valleys = 0
    for direction in s:
        if direction == "U":
            level += 1
            if level == 0:
                valleys += 1
        else:
            level -= 1
            
    return valleys

Short Problem Definition:

Monica wants to buy a keyboard and a USB drive from her favorite electronics store. The store has several models of each. Monica wants to spend as much as possible for the 2 items, given her budget.

Given the price lists for the store’s keyboards and USB drives, and Monica’s budget, find and print the amount of money Monica will spend. If she doesn’t have enough money to both a keyboard and a USB drive, print -1 instead. She will buy only the two required items.

Link

Electronics Shop

Complexity:

time complexity is O(N*M)

space complexity is O(1)

Execution:

The solution has a N*M complexity. If the range for B was significantly smaller, one could consider alternatives.

Solution:

def getMoneySpent(keyboards, drives, s):
    spend = -1

    for v1 in keyboards:
        for v2 in drives:
            if v1 + v2 <=s:
                spend = max(spend, v1 + v2)

    return spend

Short Problem Definition:

Two cats and a mouse are at various positions on a line. You will be given their starting positions. Your task is to determine which cat will reach the mouse first, assuming the mouse doesn’t move and the cats travel at equal speed. If the cats arrive at the same time, the mouse will be allowed to move and it will escape while they fight.

Link

Cats and a Mouse

Complexity:

time complexity is O(N)

space complexity is O(1)

Execution:

This one is trivial.

Solution:

def catAndMouse(x, y, z):
    catA = abs(x-z)
    catB = abs(y-z)
    if (catA < catB):
        return "Cat A"
    elif (catB < catA):
        return "Cat B"
    else:
        return "Mouse C"

Short Problem Definition:

Marie invented a Time Machine and wants to test it by time-traveling to visit Russia on the Day of the Programmer (the 256th day of the year) during a year in the inclusive range from 1700 to 2700.

Link

Day of The Programmer

Complexity:

time complexity is O(-1)

space complexity is O(-1)

Execution:

As I have pointed out in the past, no engineer should ever implement any calendar related functions. It should be done natively by the language or by a library.

I am fairly certain that the conventions will change by 2700 🙂 and the calculation will be invalid.

Solution:

#!/bin/python

import math
import os
import random
import re
import sys

# Complete the dayOfProgrammer function below.
def dayOfProgrammer(year):
    raise SystemError("This challenge is stupid")

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    year = int(raw_input().strip())

    result = dayOfProgrammer(year)

    fptr.write(result + '\n')

    fptr.close()