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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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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
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
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
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
python arrays vector indexing
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
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add a comment |
2 Answers
2
active
oldest
votes
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.
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.
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
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.
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.
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.
answered Mar 26 at 17:47
Samuel NdeSamuel Nde
9568 silver badges14 bronze badges
9568 silver badges14 bronze badges
add a comment |
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.
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.
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.
"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.
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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