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

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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

edited Mar 26 at 18:15

Born Tbe Wasted

5731313 bronze badges

asked Mar 26 at 17:10

Lucas VázquezLucas Vázquez

1

add a comment | 

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

python arrays vector indexing

share|improve this question

edited Mar 26 at 18:15

Born Tbe Wasted

5731313 bronze badges

asked Mar 26 at 17:10

Lucas VázquezLucas Vázquez

1

add a comment | 

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

edited Mar 26 at 18:15

Born Tbe Wasted

5731313 bronze badges

asked Mar 26 at 17:10

Lucas VázquezLucas Vázquez

1

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))

share|improve this question

edited Mar 26 at 18:15

Born Tbe Wasted

5731313 bronze badges

asked Mar 26 at 17:10

Lucas VázquezLucas Vázquez

1

share|improve this question

edited Mar 26 at 18:15

Born Tbe Wasted

5731313 bronze badges

asked Mar 26 at 17:10

Lucas VázquezLucas Vázquez

1

edited Mar 26 at 18:15

Born Tbe Wasted

5731313 bronze badges

edited Mar 26 at 18:15

Born Tbe Wasted

5731313 bronze badges

asked Mar 26 at 17:10

Lucas VázquezLucas Vázquez

1

asked Mar 26 at 17:10

Lucas VázquezLucas Vázquez

1

2 Answers
2

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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

answered Mar 26 at 17:47

Samuel NdeSamuel Nde

95688 silver badges1414 bronze badges

add a comment | 

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

edited Mar 26 at 17:49

answered Mar 26 at 17:19

DeepSpaceDeepSpace

44.1k44 gold badges5555 silver badges8686 bronze badges

add a comment | 

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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

answered Mar 26 at 17:47

Samuel NdeSamuel Nde

95688 silver badges1414 bronze badges

add a comment | 

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

answered Mar 26 at 17:47

Samuel NdeSamuel Nde

95688 silver badges1414 bronze badges

add a comment | 

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

answered Mar 26 at 17:47

Samuel NdeSamuel Nde

95688 silver badges1414 bronze badges

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

answered Mar 26 at 17:47

Samuel NdeSamuel Nde

95688 silver badges1414 bronze badges

answered Mar 26 at 17:47

Samuel NdeSamuel Nde

95688 silver badges1414 bronze badges

answered Mar 26 at 17:47

Samuel NdeSamuel Nde

95688 silver badges1414 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

edited Mar 26 at 17:49

answered Mar 26 at 17:19

DeepSpaceDeepSpace

44.1k44 gold badges5555 silver badges8686 bronze badges

add a comment | 

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

edited Mar 26 at 17:49

answered Mar 26 at 17:19

DeepSpaceDeepSpace

44.1k44 gold badges5555 silver badges8686 bronze badges

add a comment | 

"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

edited Mar 26 at 17:49

answered Mar 26 at 17:19

DeepSpaceDeepSpace

44.1k44 gold badges5555 silver badges8686 bronze badges

share|improve this answer

edited Mar 26 at 17:49

answered Mar 26 at 17:19

DeepSpaceDeepSpace

44.1k44 gold badges5555 silver badges8686 bronze badges

answered Mar 26 at 17:19

DeepSpaceDeepSpace

44.1k44 gold badges5555 silver badges8686 bronze badges

answered Mar 26 at 17:19

DeepSpaceDeepSpace

44.1k44 gold badges5555 silver badges8686 bronze badges