Three algorithms exercises

Exercise 1

# Algo_FindLongestRepeatedSequence.py

# Tiene un arreglo (llamado myArray) con 10 elementos (enteros en el rango de 1 a 9). 
# Escriba un programa en Java que imprima el número que tiene más ocurrencias seguidas en el arreglo y también imprimir la cantidad de veces que aparece en la secuencia.

# El código que llena el arreglo ya está escrito, pero puede editarlo para probar con otros valores. 
# Con el botón de refrescar puede recuperar el valor original que será utilizado para evaluar la pregunta como correcta o incorrecta durante la ejecución.

# Su programa debe analizar el arreglo de izquierda a derecha para que en caso de que dos números cumplan la condición, 
# el que aparece por primera vez de izquierda a derecha será el que se imprima. 
# La salida de los datos para el arreglo en el ejemplo (1,2,2,5,4,6,7,8,8,8) sería la siguiente:
# Longest: 3
# Number: 8

# En el ejemplo, la secuencia más larga la tiene el número 8 con una secuencia de tres ochos seguidos. 
# Tenga en cuenta que el código que escriba debe imprimir los resultados exactamente como se muestra con el fin de que la pregunta sea considerada válida.

import random

LENGHT = 10


def fillArray():
    a = []
    for i in range(LENGHT):
        a.append(random.randint(1, 9))
    # return a
    return [1,2,2,5,4,6,7,8,8,8]

def findLongestRepeatedSequence(a):
    print("myArray", a)
    longest = {}
    current = {}

    for elm in a:
        if not current or not longest:
            current['Number'] = elm
            current['Longest'] = 1
            longest['Number'] = current['Number']
            longest['Longest'] = current['Longest']
            continue

        if elm == current['Number']:
            current['Longest'] = current['Longest'] + 1
            if current['Longest'] > longest['Longest']:
                longest['Number'] = current['Number']
                longest['Longest'] = current['Longest']
        else:
            current['Number'] = elm
            current['Longest'] = 1

    return longest


if __name__ == '__main__':
    myArray = fillArray()
    print("Result", findLongestRepeatedSequence(myArray))

Exercise 2

# https://community.turing.com/
# Jump Start event, Thursday 2023-03-30
# Problem 1

# Cleanup Strings
# A string s can have consecutive repeated letters. 
# Your job is to remove the least number of letters to make the number of consecutive repeated letters less than four. 
# Return the final string.
# The answer will always be unique.

# Example:
# Input: s = "tuuuuuriiiiiing"
# Output: s = "tuuuriiing"

# Constraints:
# 1 <= s.lenght <= 10^5
# s consists only of lowercase letters


def solution(s: str) -> str:
    # write your solution here
    print("Input", s)
    clone = []
    current = {}

    for elm in a:
        if not current:
            current['element'] = elm
            current['sequence'] = 1
            clone.append(elm)
            continue

        if elm == current['element']:
            current['sequence'] = current['sequence'] + 1
            if current['sequence'] > 3:
                continue
        else:
            current['element'] = elm
            current['sequence'] = 1
        clone.append(elm)

    return ''.join(str(x) for x in clone)

# R E A D M E
# DO NOT CHANGE the code below, we use it to grade your submission. If changed your submission will be failed automatically.
if __name__ == '__main__':
    # a = input()
    # a = 'abbcccddddeeeeeffffffggggggghhhhhhhh'
    # a = '122333444455555666666777777788888888999999999'
    a = 'tuuuuuriiiiiing'
    output = solution(a)
    print("Output", output)

Exercise 3

# https://community.turing.com/
# Jump Start event, Thursday 2023-03-30
# Problem 2

# Suppose you are given two strings s and t String t is generated by random shuffling string s and then adding one more letter at a random position. 
# Return the letter that was added to t.

# Example 1:
# Input: s = "abcd", t = "ceadb"
# Output: "e"

# Explanation:
# 'cadb' was produced by random shuffling "abcd" and "e" was added in as the second letter.

# Example 2:
# Input: s = "abbdd", t = "dabadb"
# Output: "a"

# Example 3:
# Input: s = "", t = "y"
# Output: "y"

# Example 4:
# Input: s = "aaaa", t = "aabaa"
# Output: "b"

# Example 5:
# Input: s = "ac", t = "aac"
# Output: "a"

import random


def frequencyTable(lst):
    # print("lst", lst)
    freq = {}
    for k in lst:
        if k in freq:
            freq[k] = freq[k] + 1
        else:
            freq[k] = 1
    # print("freq", freq)
    return freq

def compareTables(dict1, dict2):
    print("dict1", dict1)
    print("dict2", dict2)
    sorted1 = sorted(dict1)
    sorted2 = sorted(dict2)
    print("sorted1", sorted1)
    print("sorted2", sorted2)
    result = sorted2[-1]

    print("result", result)
    for i in range(len(sorted1)):
        if sorted1[i] != sorted2[i]:
            result = sorted2[i]
        else:
            if dict1[sorted1[i]] != dict2[sorted2[i]]:
                result = sorted2[i]
                break

    return result

def findTheDifference(s: str, t: str) -> str:
    return compareTables(frequencyTable(s), frequencyTable(t))


if __name__ == '__main__':
    # s = input().strip()
    # concatenated = (s + chr(random.randint(97, 122)))
    # t = ''.join(random.sample(concatenated, len(concatenated)))

    s = "abcd"
    t = "ceadb"
    
    # s = "abbdd"
    # t = "dabadb"

    # s = ""
    # t = "y"

    # s = "aaaa"
    # t = "aabaa"

    # s = "ac"
    # t = "aac"
    print("s", s)
    print("t", t)

    print(findTheDifference(s, t))

References

• https://community.turing.com/