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Is there any way to find if the index is equal to the element of a vector?


Best way to find if an item is in a JavaScript array?Finding the index of an item given a list containing it in PythonIs there any way to kill a Thread?How to find out if an item is present in a std::vector?How to remove an element from a list by index?How do I erase an element from std::vector<> by index?What is the easiest way to initialize a std::vector with hardcoded elements?Is there a simple way to delete a list element by value?Is there an R function for finding the index of an element in a vector?Swift for loop: for index, element in array?






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0















Using the methodology of Divide and Conquer, I'm trying to do an algorithm that compares if the index of a vector is equal to the element in this position. I've seen that my code is inefficient in large vectors, so I've been thinking to do it by dividing the vector in two halves, but I do not know how to do it...



def indicePosicion (inicio, lista): 
if lista == []:
return -1
elif int(lista[0]) == inicio:
return inicio
else:
return indicePosicion(inicio+1, lista[1:])

numero = int(input())
lista = input().split()
print(indicePosicion(0, lista))


I introduce the number of elements in the vector:
7
Introduce the elements separate by spaces:
-3 -1 2 5 6 7 9
And the output should be
2
where the element is equal to the position










share|improve this question






























    0















    Using the methodology of Divide and Conquer, I'm trying to do an algorithm that compares if the index of a vector is equal to the element in this position. I've seen that my code is inefficient in large vectors, so I've been thinking to do it by dividing the vector in two halves, but I do not know how to do it...



    def indicePosicion (inicio, lista): 
    if lista == []:
    return -1
    elif int(lista[0]) == inicio:
    return inicio
    else:
    return indicePosicion(inicio+1, lista[1:])

    numero = int(input())
    lista = input().split()
    print(indicePosicion(0, lista))


    I introduce the number of elements in the vector:
    7
    Introduce the elements separate by spaces:
    -3 -1 2 5 6 7 9
    And the output should be
    2
    where the element is equal to the position










    share|improve this question


























      0












      0








      0








      Using the methodology of Divide and Conquer, I'm trying to do an algorithm that compares if the index of a vector is equal to the element in this position. I've seen that my code is inefficient in large vectors, so I've been thinking to do it by dividing the vector in two halves, but I do not know how to do it...



      def indicePosicion (inicio, lista): 
      if lista == []:
      return -1
      elif int(lista[0]) == inicio:
      return inicio
      else:
      return indicePosicion(inicio+1, lista[1:])

      numero = int(input())
      lista = input().split()
      print(indicePosicion(0, lista))


      I introduce the number of elements in the vector:
      7
      Introduce the elements separate by spaces:
      -3 -1 2 5 6 7 9
      And the output should be
      2
      where the element is equal to the position










      share|improve this question
















      Using the methodology of Divide and Conquer, I'm trying to do an algorithm that compares if the index of a vector is equal to the element in this position. I've seen that my code is inefficient in large vectors, so I've been thinking to do it by dividing the vector in two halves, but I do not know how to do it...



      def indicePosicion (inicio, lista): 
      if lista == []:
      return -1
      elif int(lista[0]) == inicio:
      return inicio
      else:
      return indicePosicion(inicio+1, lista[1:])

      numero = int(input())
      lista = input().split()
      print(indicePosicion(0, lista))


      I introduce the number of elements in the vector:
      7
      Introduce the elements separate by spaces:
      -3 -1 2 5 6 7 9
      And the output should be
      2
      where the element is equal to the position







      python arrays vector indexing






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Mar 26 at 18:15









      Born Tbe Wasted

      57313 bronze badges




      57313 bronze badges










      asked Mar 26 at 17:10









      Lucas VázquezLucas Vázquez

      1




      1






















          2 Answers
          2






          active

          oldest

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          0














          How about just getting the list of indices where the index is equal to the element?



          a = [1, 2, 3, 3, 4, 4, 6, 6]
          [index for index, element in enumerate(a) if index == element]
          #gives you a list of indices of `a` with a value equal to the index.





          share|improve this answer






























            0















            "I've seen that my code is inefficient in large vector"




            If the input array is not sorted, then the most efficient code will have a time complexity of O(n) since we have to iterate element by element and compare the index to the value. In this case, using divide and conquer is meaningless and is only adding complexity.



            If the input array is sorted, then you can use a modified version of binary search to achieve an O(logn) time complexity.






            share|improve this answer



























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              2 Answers
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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              0














              How about just getting the list of indices where the index is equal to the element?



              a = [1, 2, 3, 3, 4, 4, 6, 6]
              [index for index, element in enumerate(a) if index == element]
              #gives you a list of indices of `a` with a value equal to the index.





              share|improve this answer



























                0














                How about just getting the list of indices where the index is equal to the element?



                a = [1, 2, 3, 3, 4, 4, 6, 6]
                [index for index, element in enumerate(a) if index == element]
                #gives you a list of indices of `a` with a value equal to the index.





                share|improve this answer

























                  0












                  0








                  0







                  How about just getting the list of indices where the index is equal to the element?



                  a = [1, 2, 3, 3, 4, 4, 6, 6]
                  [index for index, element in enumerate(a) if index == element]
                  #gives you a list of indices of `a` with a value equal to the index.





                  share|improve this answer













                  How about just getting the list of indices where the index is equal to the element?



                  a = [1, 2, 3, 3, 4, 4, 6, 6]
                  [index for index, element in enumerate(a) if index == element]
                  #gives you a list of indices of `a` with a value equal to the index.






                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Mar 26 at 17:47









                  Samuel NdeSamuel Nde

                  9568 silver badges14 bronze badges




                  9568 silver badges14 bronze badges























                      0















                      "I've seen that my code is inefficient in large vector"




                      If the input array is not sorted, then the most efficient code will have a time complexity of O(n) since we have to iterate element by element and compare the index to the value. In this case, using divide and conquer is meaningless and is only adding complexity.



                      If the input array is sorted, then you can use a modified version of binary search to achieve an O(logn) time complexity.






                      share|improve this answer





























                        0















                        "I've seen that my code is inefficient in large vector"




                        If the input array is not sorted, then the most efficient code will have a time complexity of O(n) since we have to iterate element by element and compare the index to the value. In this case, using divide and conquer is meaningless and is only adding complexity.



                        If the input array is sorted, then you can use a modified version of binary search to achieve an O(logn) time complexity.






                        share|improve this answer



























                          0












                          0








                          0








                          "I've seen that my code is inefficient in large vector"




                          If the input array is not sorted, then the most efficient code will have a time complexity of O(n) since we have to iterate element by element and compare the index to the value. In this case, using divide and conquer is meaningless and is only adding complexity.



                          If the input array is sorted, then you can use a modified version of binary search to achieve an O(logn) time complexity.






                          share|improve this answer
















                          "I've seen that my code is inefficient in large vector"




                          If the input array is not sorted, then the most efficient code will have a time complexity of O(n) since we have to iterate element by element and compare the index to the value. In this case, using divide and conquer is meaningless and is only adding complexity.



                          If the input array is sorted, then you can use a modified version of binary search to achieve an O(logn) time complexity.







                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited Mar 26 at 17:49

























                          answered Mar 26 at 17:19









                          DeepSpaceDeepSpace

                          44.1k4 gold badges55 silver badges86 bronze badges




                          44.1k4 gold badges55 silver badges86 bronze badges



























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