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find complexity of code online

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Drab Dove answered on December 4, 2021 Popularity 6/10 Helpfulness 1/10

answer find complexity of code online

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find complexity of code online

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def count_occurrences_in_text(word, text):
    """
    Return the number of occurrences of the passed word (case insensitive) in text

    complexite : O(n*m)
    """
    # efficaité : donnée et temps de calcul

    # TODO: your code goes here, but it's OK to add new functions or import modules if needed

    # This does not pass the unittests:

    word_lower = word.lower() 
    text_lower = text.lower()

    ponctuations_exclus = string.punctuation.replace("'", "").replace("-", "").replace(".", "")
    traducteur = str.maketrans(ponctuations_exclus, ' '*len(ponctuations_exclus))

    final_text = text_lower.translate(traducteur)
    
    longueur_text = len(final_text)
    longueur_word = len(word_lower)
    
    # Parcourir la phrase avec une fenêtre glissante de la taille de la chaîne
    test = [" ", "\n", "�", "."]
    test_pre = ["'", "_"]
    nombre_occurence = 0
    for i in range(longueur_text - longueur_word+1):
        # Extraire la sous-phrase de la taille de la chaîne à partir de l'index i
        under_word = final_text[i:i+longueur_word]

        suivant = ' ' if i+longueur_word >= longueur_text else final_text[i+longueur_word]

        suiv_suivant = ' ' if i+longueur_word+1 >= longueur_text else final_text[i+longueur_word + 1]

        precedent = ' ' if i-1 < 0 else final_text[i-1]

        pre_precedent = ' ' if i-2 < 0 else final_text[i-2]
        
        # Comparer la sous-phrase avec la chaîne et augmenter le compteur si elles sont égales
        
        if under_word == word_lower and ((suivant in test and precedent in test) or (suivant == "\u201d" and precedent == "\u201c") or (pre_precedent in test_pre and precedent in test_pre) or (suiv_suivant in test_pre and suivant in test_pre)):
            nombre_occurence += 1        

    return nombre_occurence

Popularity 6/10 Helpfulness 1/10 Language whatever

Source: Grepper

Tags: find whatever

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Contributed on Oct 23 2023

Drab Dove

0 Answers Avg Quality 2/10

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sum=0;
for (int j=n ; j >0 ; j=j/2)
{
	for (i= j ; i <=n ; i=i*2)
        {
		sum++;
        }
}

Popularity 9/10 Helpfulness 1/10 Language whatever

Source: www.w3schools.blog

Tags: find

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Contributed on Jan 26 2023

Determined Dogfish

0 Answers Avg Quality 2/10

code complexity calculator online

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MysteryA (int n)

     {     int i,j,k;

           for i=1 to n

    {         for j=1 to n

            {c[i,j]=0;

              }

               for k=1 to n

                 c[i,j]=c[i,j]+a[i,k]*b[k,j];

            } }

Popularity 1/10 Helpfulness 1/10 Language whatever

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Contributed on Dec 04 2021

Funny Ferret

0 Answers Avg Quality 2/10

code complexity calculator online

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for(i=1; i<=n; i++)
{
  for(j=1; j<=i; j++)
  {
    for(k=1;k<=j;k++)
    {}
    k = 2;
    while(k<i)
    {
      k=k*k;
    }
  }
}

Popularity 1/10 Helpfulness 1/10 Language c

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Tags: c calculator code-complexity

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Contributed on Dec 09 2021

Dizzy Dormouse

0 Answers Avg Quality 2/10

code complexity calculator online

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li = [8,3,7,2,3,9,4]
max_element = max(li)
min_element = min(li)

total = max_element - min_element
x = len(li)
for i in li:
    for j in range(x):
        if i - li[j] == total:
            print("buy price is=", li[j],"sell price is= ",i ,"the profit is ", i - li[j])

Popularity 1/10 Helpfulness 1/10 Language python

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Tags: calculator code-complexity python

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Contributed on Apr 28 2022

Pratipal Singh Rajput

0 Answers Avg Quality 2/10

code complexity calculator online

Comment

li = [8,3,7,2,3,9,4]
max_element = max(li)
min_element = min(li)

total = max_element - min_element
x = len(li)
for i in li:
    for j in range(x):
        if i - li[j] == total:
            print("buy price is=", li[j],"sell price is= ",i ,"the profit is ", i - li[j])

Popularity 1/10 Helpfulness 1/10 Language python

Source: Grepper

Tags: calculator code-complexity python

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Contributed on Apr 28 2022

Pratipal Singh Rajput

0 Answers Avg Quality 2/10

find complexity of code online

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