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#94 The Hashtag Generator

The marketing team are spending way too much time typing in hashtags.

Let's help them with out own Hashtag Generator!

Here's the deal:

- If the final result is longer than 140 chars it must return false.

- If the input is a empty string it must return false.

- It must start with a hashtag (#).

- All words must have their first letter capitalized.

Example Input to Output:

" Hello there thanks for trying my Kata" => "#HelloThereThanksForTryingMyKata"

" Hello World " => "#HelloWorld"

function capitalize(str) {
    return str.charAt(0).toUpperCase() + str.slice(1);
}

function generateHashtag (str) {
  if(!str)
    return false
  const result = "#"+str.split(" ").map(word => capitalize(word).trim()).join("");
  if(result.length > 140)
    return false;
  return result;
}

#93 Number Format

Format any integer provided into a string with "," (commas) in the correct places.

Example:

numberFormat(100000); // return '100,000'

numberFormat(5678545); // return '5,678,545'

numberFormat(-420902); // return '-420,902'

function numberFormat(number) {
  //return number.toString().replace(/\B(?=(\d{3})+(?!\d))/g, ",");
  return number.toLocaleString();
}

#92 Bit calculator 

In this kata your task is to create bit calculator. Function arguments are two bit representation of numbers ("101","1","10"...), and you must return their sum in decimal representation.

Test.expect(calculate("10","10") == 4);

Test.expect(calculate("10","0") == 2);

Test.expect(calculate("101","10") == 7);

parseInt and some Math functions are disabled.

Those Math functions are enabled: pow,round,random

function convert(num) {
  var idx =0;
  var result =0;
  var temp = num.toString().split("");
  for(var i = temp.length-1; i>=0; i--){
    if(temp[idx++] === '1')
      result += Math.pow(2,i);
  }

  return result;
}

function calculate(num1,num2){
  return convert(num1)+convert(num2);
}

#91 Find X 

We have a function that takes in an integer n, and returns a number x.

Lets call this function findX(n)/find_x(n) (depending on your language):

function findX(n) {

  let x = 0;

  for (let i = 0; i < n; i++) {

    for (let j = 0; j < 2*n; j++)

      x += i + j;

  }

  return x;

}

The functions loops throught the number n and at every iteration, performs a nested loop on 2*n, at each iteration of this nested loop it increments x with the (nested loop index + parents loop index).

This works well when the numbers are reasonably small.

findX(2) //=> 16

findX(3) //=> 63

findX(5) //=> 325

But may be slow for numbers > 103

So your task is to optimize the function findX/find_x, so it works well for large numbers.

Input Range

1 <= n <= 106 (105 in JS)

Note: This problem is more about logical reasoning than it is about finding a mathematicial formula, infact there are no complex math formula involved

const findX = n => n * n * (3 * n - 2);

 #90 Cure Cancer

Now you are a doctor.

You are working with a patient's body which has many cells.

The patient's body is a matrix where every row represents a cell.

Each cell contains just uppercase and lowercase letters,

and every cell in the body should be the same.

Oh no! It seems that one of the cells have mutated!

It is your job to locate the mutation so that the chemo specialists can fix it!

return the location [i,j] within the matrix...

before it's too late! :(

example:

cellscellscellscodecodecells

cellscellscellscodecodecells

cellscellscellscodecodecells

cellscellscellscodecodecells

cellscellscellscodecodecells

cellscellscellscodecodecells

cellscellscellscodecodecells

cellscellscellscodecodecells

cellscellscellscodecodecells

cellscellscellscodecadecells <- here it is! [9, 20]

cellscellscellscodecodecells

cellscellscellscodecodecells

cellscellscellscodecodecells

cellscellscellscodecodecells

no bodies will have less than 3 cells.

if the diagnose was a false alarm, return an empty array.

import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;

class JomoPipi {
    public static int[] mutationLocation(char[][] body) {
    	Map<string, integer=""> map = new HashMap<>();
    	for (int i = 0; i < 3; i++) {
    		map.put(String.valueOf(body[i]), 1 + map.getOrDefault(String.valueOf(body[i]),0));
		}
    	
    	char [] dna = map.entrySet().stream()
    	.sorted(Map.Entry.<string, integer="">comparingByValue().reversed())
    	.findFirst().get().getKey().toCharArray();

    	for (int x = 0; x < body.length; x++) {
    		if(!Arrays.equals(dna, body[x]))
    			for (int y = 0; y < body[x].length; y++) {
    				if(dna[y] != body[x][y]) {
    					return new int [] {x,y};
    				}
				}
			
		}
    	return new int [0];
    }
}

#89 Mexican Wave

Introduction

 The wave (known as the Mexican wave in the English-speaking world outside North America) is an example of metachronal rhythm achieved in a packed stadium when successive groups of spectators briefly stand, yell, and raise their arms. Immediately upon stretching to full height, the spectator returns to the usual seated position. The result is a wave of standing spectators that travels through the crowd, even though individual spectators never move away from their seats. In many large arenas the crowd is seated in a contiguous circuit all the way around the sport field, and so the wave is able to travel continuously around the arena; in discontiguous seating arrangements, the wave can instead reflect back and forth through the crowd. When the gap in seating is narrow, the wave can sometimes pass through it. Usually only one wave crest will be present at any given time in an arena, although simultaneous, counter-rotating waves have been produced. (Source Wikipedia)

Task

In this simple Kata your task is to create a function that turns a string into a Mexican Wave. You will be passed a string and you must return that string in an array where an uppercase letter is a person standing up.

Rules

1.  The input string will always be lower case but maybe empty.

2.  If the character in the string is whitespace then pass over it as if it was an empty seat.

Example

wave("hello") => ["Hello", "hEllo", "heLlo", "helLo", "hellO"]

Good luck and enjoy!

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class MexicanWave {

	public static String capitalize(String str, int i) {
		String [] array = str.split("");
		array[i] = array[i].toUpperCase();
		return String.join("", array);
	}
	

    public static String[] wave(String str) {
  	   int size = str.length();
  	   List<String> result = new ArrayList<>();
  	   for (int i = 0; i < size; i++) {
  	   	   String temp = capitalize(str,i);
  		   if(!temp.equals(str))
  		   	   result.add(temp);
	   }
       return result.stream().toArray(String[]::new);
    }
    
}

#88 Which are in?

Given two arrays of strings a1 and a2 return a sorted array r in lexicographical order of the strings of a1 which are substrings of strings of a2.

#Example 1: a1 = ["arp", "live", "strong"]

a2 = ["lively", "alive", "harp", "sharp", "armstrong"]

returns ["arp", "live", "strong"]

#Example 2: a1 = ["tarp", "mice", "bull"]

a2 = ["lively", "alive", "harp", "sharp", "armstrong"]

returns []

Notes:

Arrays are written in "general" notation. See "Your Test Cases" for examples in your language.

In Shell bash a1 and a2 are strings. The return is a string where words are separated by commas.

Beware: r must be without duplicates.

Don't mutate the inputs.

import java.util.Arrays;

public class WhichAreIn { 
	
	public static boolean contains(String str, String [] target) {
		return Arrays.stream(target).parallel().anyMatch(x->x.contains(str));
	}
	
	public static String[] inArray(String[] array1, String[] array2) {
		return Arrays.stream(array1).filter(x -> contains(x,array2)).sorted().toArray(value -> new String[value]);
	}
  
}

#87 Extract the IDs from the data set 

Complete the method so that it returns an array of all ID's passed in. The data structure will be similar to the following:

var data = {
  id: 1,
  items: [
    {id: 2},
    {id: 3, items: [
      {id: 4},
      {id: 5}
    ]}
  ]
}

extractIds(data) // should return [1,2,3,4,5]

The method should be able to handle the case of empty data being passed in.

Note: The only arrays that need to be traversed are those assigned to the "items" property.

function extractIds(data){
  var result = [];
  for(var key in data){
    if(typeof data[key] === "object")
      result = result.concat(extractIds(data[key]));
    else 
      if(key ==="id")
          result.push(data[key]);
  }
  return result;
}

#86 Remove Duplicates

Remove Duplicates

You are to write a function called unique that takes an array of integers and returns the array with duplicates removed. It must return the values in the same order as first seen in the given array. Thus no sorting should be done, if 52 appears before 10 in the given array then it should also be that 52 appears before 10 in the returned array.

Assumptions

All values given are integers (they can be positive or negative).

You are given an array but it may be empty.

They array may have duplicates or it may not.

You cannot use the uniq method on Arrays (don't even try it), or the nub function from Data.List.

Example

UniqueArray.unique([1, 5, 2, 0, 2, -3, 1, 10]) 

// -> [1, 5, 2, 0, -3, 10]

import java.util.*;

public class UniqueArray {
  public static int[] unique(int[] integers) {
      return Arrays.stream(integers).distinct().toArray();
  }
}

 #85 Circularly Sorted Array

Write a method, isCircleSorted(int[] A) (Java, JavaScript), or Array#circularly_sorted? (Ruby) that determines if A is circularly sorted. An Array is circularly sorted if the elements are sorted in ascending order, but displaced, or rotated, by any number of steps.

For Example:

// True:

isCircleSorted([2,3,4,5,0,1]);

isCircleSorted([4,5,6,9,1]);

isCircleSorted([10,11,6,7,9]);

isCircleSorted([1,2,3,4,5]);

isCircleSorted([5,7,43,987,-9,0]);

// False:

isCircleSorted([4,1,2,5]);

isCircleSorted([8,7,6,5,4,3]);

isCircleSorted([6,7,4,8]);

isCircleSorted([7,6,5,4,3,2,1]);

public class CircleSorted {

    public int getSmallestNumberIndex(int [] a){
  	   int min = a[0];
	     int idx=0;

	     for(int i = 0; i < a.length; i++) {
	       if(min > a[i]) {
	         min = a[i];
               idx=i;
	       }
	     }
	    return idx;
    
    }

    public boolean isCircleSorted(int[] a) {
       int idx = getSmallestNumberIndex(a);
       for(int i = 0; i < a.length-1; i++) {
         if(a[((idx+i)%a.length)]>a[((idx+i+1)%a.length)])
           return false;
	     }
       return true;
    }
}