Sometimes, while working with records, we might have a problem in which we need to find all the indices of elements for a particular value at a particular Kth position of tuple. This seems to be a peculiar problem but while working with many keys in records, we encounter this problem. Let’s discuss certain ways in which this problem can be solved.
Method #1 : Using loop This is the brute force method by which this problem can be solved. In this, we keep a counter for indices and append to list if we find the specific record at Kth position in tuple.
Python3
# Python3 code to demonstrate working of# Indices of Kth element value# Using loop# initialize listtest_list = [(3, 1, 5), (1, 3, 6), (2, 5, 7), (5, 2, 8), (6, 3, 0)]# printing original listprint("The original list is : " + str(test_list))# initialize eleele = 3# initialize KK = 1# Indices of Kth element value# Using loop# using y for K = 1res = []count = 0for x, y, z in test_list: if y == ele: res.append(count) count = count + 1# printing resultprint("The indices of element at Kth position : " + str(res)) |
The original list is : [(3, 1, 5), (1, 3, 6), (2, 5, 7), (5, 2, 8), (6, 3, 0)] The indices of element at Kth position : [1, 4]
Time Complexity: O(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 enumerate() + list comprehension
The combination of the above functions can be used to solve this problem. In this, we enumerate for the indices using enumerate(), rest is performed as in the above method but in a compact way.
Python3
# Python3 code to demonstrate working of# Indices of Kth element value# Using enumerate() + list comprehension# initialize listtest_list = [(3, 1, 5), (1, 3, 6), (2, 5, 7), (5, 2, 8), (6, 3, 0)]# printing original listprint("The original list is : " + str(test_list))# initialize eleele = 3# initialize KK = 1# Indices of Kth element value# Using enumerate() + list comprehensionres = [a for a, b in enumerate(test_list) if b[K] == ele]# printing resultprint("The indices of element at Kth position : " + str(res)) |
The original list is : [(3, 1, 5), (1, 3, 6), (2, 5, 7), (5, 2, 8), (6, 3, 0)] The indices of element at Kth position : [1, 4]
Method 3: Using sorted() function
Approach:
This function takes in an input list and an integer k, and returns a list of indices where the kth smallest element appears in the original input list.
Python3
def indices_of_kth_element_sorted(input_list, k): sorted_list = sorted(input_list) indices = [i for i, x in enumerate(input_list) if sorted_list[k-1] == x] return indicesinput_list = [(3, 1, 5), (1, 3, 6), (2, 5, 7), (5, 2, 8), (6, 3, 0)]k = 2indices = indices_of_kth_element_sorted(input_list, k)print(f"The original list is : {input_list}")print(f"The indices of element at Kth position : {indices}") |
The original list is : [(3, 1, 5), (1, 3, 6), (2, 5, 7), (5, 2, 8), (6, 3, 0)] The indices of element at Kth position : [2]
Time complexity: Sorting the list takes O(n log n) time in the average and worst cases, where n is the length of the input list. The list comprehension to find the indices takes O(n) time in the worst case. Therefore, the overall time complexity is O(n log n).
Auxiliary Space: Sorting the list takes O(n) extra space, where n is the length of the input list. The list comprehension to find the indices takes O(n) extra space in the worst case. Therefore, the overall space complexity is O(n).
