Sometimes, while working with data, we can have a simple problem in which we have ranges in form of tuple, and we need to check if a specific number lies between any of ranges suggested by tuple. This has it’s application in competitive programming. Let’s discuss certain ways in which this task can be performed.
Method #1 : Using loop + enumerate() This task can be performed using combination of above functions. In this, we just need to iterate through each element of list and return the index of tuple pair between which the element exists using enumerate().
Python3
test_list = [( 45 , 90 ), ( 100 , 147 ), ( 150 , 200 )]
print ("The original list is : " + str (test_list))
N = 124
res = None
for idx in (idx for idx, (sec, fir) in enumerate (test_list) if sec < = N < = fir):
res = idx
print ("The index of tuple between which element occurs : " + str (res))
|
Output :
The original list is : [(45, 90), (100, 147), (150, 200)]
The index of tuple between which element occurs : 1
Time Complexity: O(n*n), where n is the length of the list test_list
Auxiliary Space: O(n) additional space of size n is created where n is the number of elements in the res list
Method #2 : Using next() + enumerate() + generator expression This task can also be performed using combination of above functions. In this, we just iterate through using next(). Rest everything is performed similar to the above function.
Python3
test_list = [( 45 , 90 ), ( 100 , 147 ), ( 150 , 200 )]
print ("The original list is : " + str (test_list))
N = 124
res = next ((idx for idx, (sec, fir) in enumerate (test_list) if sec < = N < = fir), None )
print ("The index of tuple between which element occurs : " + str (res))
|
Output :
The original list is : [(45, 90), (100, 147), (150, 200)]
The index of tuple between which element occurs : 1
Filter Function:
Approach:
In this approach, we use the filter function to create a list of tuples where the element is present in the range of each tuple. We then return the index of the first tuple in this list.
Define the function check_range_filter(lst, element) which takes a list of tuples lst and an element element as input.
Use the filter function to create a new list of tuples filtered_lst where the element is present in the range of each tuple in lst.
Check if filtered_lst is empty. If it is, return None.
Otherwise, return the index of the first tuple in lst that matches the first tuple in filtered_lst.
Use the index method of the list lst to find the index of the first tuple in filtered_lst.
Return the index found in the previous step.
Python3
def check_range_filter(lst, element):
filtered_lst = list ( filter ( lambda x: x[ 0 ] < = element < = x[ 1 ], lst))
return lst.index(filtered_lst[ 0 ]) if filtered_lst else None
lst = [( 45 , 90 ), ( 100 , 147 ), ( 150 , 200 )]
element = 110
print (check_range_filter(lst, element))
|
Time Complexity: O(n)
Auxiliary Space: O(n)
METHOD 4:Linear Search
APPROACH:
This approach takes a list of tuples ‘lst’ and an ‘element’ as input. It searches for the tuple which has a range containing the given ‘element’. If such a tuple exists, it returns the index of the tuple. If not, it returns None. The function uses a linear search algorithm to traverse the list of tuples and find the index of the tuple containing the ‘element’.
ALGORITHM:
1.Initialize a variable ‘index’ to None
2.Traverse the list of tuples using a for loop
3.For each tuple, check if the given element lies between the range of the tuple.
4.If the element lies between the range of the tuple, set the ‘index’ variable to the index of the tuple and break out of the loop.
5.If the element is not found in any of the tuples, return None.
Python3
def check_element_range(lst, element):
index = None
for i in range ( len (lst)):
if lst[i][ 0 ] < = element < = lst[i][ 1 ]:
index = i
break
return index
lst = [( 45 , 90 ), ( 100 , 147 ), ( 150 , 200 )]
element = 120
print (check_element_range(lst, element))
|
Time Complexity: O(n) where n is the length of the list of tuples
Auxiliary Space: O(1)
Whether you're preparing for your first job interview or aiming to upskill in this ever-evolving tech landscape,
GeeksforGeeks Courses are your key to success. We provide top-quality content at affordable prices, all geared towards accelerating your growth in a time-bound manner. Join the millions we've already empowered, and we're here to do the same for you. Don't miss out -
check it out now!
Last Updated :
13 Apr, 2023
Like Article
Save Article