Python – Check if Kth index elements are unique

Last Updated : 26 Apr, 2023

Given a String list, check if all Kth index elements are unique.

Input : test_list = [“gfg”, “best”, “for”, “geeks”], K = 1
Output : False
Explanation : e occurs as 1st index in both best and geeks.
Input : test_list = [“gfg”, “best”, “geeks”], K = 2
Output : True
Explanation : g, s, e, all are unique.

Method #1 : Using loop

This is brute way to solve this problem. In this, we iterate for each string and flag off when any element is repeated, and return false.

Python3

   # Python3 code to demonstrate working of
# Check if Kth index elements are unique
# Using loop
 
# initializing list
test_list = ["gfg", "best", "for", "geeks"]
 
# printing original list
print("The original list is : " + str(test_list))
 
# initializing K
K = 2
 
res = []
flag = True
for ele in test_list:
 
    # checking if element is repeated
    if ele[K] in res:
        flag = False
        break
    else:
        res.append(ele[K])
 
# printing result
print("Is Kth index all unique : " + str(flag))
 
  

Output

The original list is : ['gfg', 'best', 'for', 'geeks'] Is Kth index all unique : True

Method #2 : Using Counter() + all()

In this, we count frequency of each char. at Kth index, and all() is used to check if Counter is less than 2 for all.

Python3

   # Python3 code to demonstrate working of
# Check if Kth index elements are unique
# Using Counter() + all()
from collections import Counter
 
# initializing list
test_list = ["gfg", "best", "for", "geeks"]
 
# printing original list
print("The original list is : " + str(test_list))
 
# initializing K
K = 2
 
# getting count of each Kth index item
count = Counter(sub[K] for sub in test_list)
 
# extracting result
res = all(val < 2 for val in count.values())
 
# printing result
print("Is Kth index all unique : " + str(res))
 
  

Output

The original list is : ['gfg', 'best', 'for', 'geeks'] Is Kth index all unique : True

The Time and Space Complexity for all the methods are the same:

Time Complexity: O(n)
Auxiliary Space: O(n)

Method #3 : Using list(),set() methods

Python3

   # Python3 code to demonstrate working of
# Check if Kth index elements are unique
 
# initializing list
test_list = ["gfg", "best", "for", "geeks"]
 
# printing original list
print("The original list is : " + str(test_list))
 
# initializing K
K = 2
res = False
if(test_list[K] != list(set(test_list[K]))):
    res = True
 
# printing result
print("Is Kth index all unique : " + str(res))
 
  

Output

The original list is : ['gfg', 'best', 'for', 'geeks'] Is Kth index all unique : True

Method #4: Using list comprehension and set()

Algorithm:

Initialize an empty set and an empty list to keep track of unique elements.
Traverse through the given list and extract the Kth index element of each string.
Check if the element is already present in the set.
If it is already present, return False as it means there is a duplicate.
If not, add the element to the set and the list.
After traversing the whole list, compare the length of the set and the list.
If they are equal, it means all elements were unique. Return True.
If not, it means there was at least one duplicate. Return False.

Python3

   #Python3 code to demonstrate working of
#Check if Kth index elements are unique
#Using set
#initializing list
test_list = ["gfg", "best", "for", "geeks"]
 
#printing original list
print("The original list is : " + str(test_list))
 
#initializing K
K = 2
 
#Using set to check unique Kth index elements
unique = len(set([ele[K] for ele in test_list])) == len([ele[K] for ele in test_list])
 
#printing result
print("Is Kth index all unique : " + str(unique))
#This code is contributed by Vinay Pinjala.
 
  

Output

The original list is : ['gfg', 'best', 'for', 'geeks'] Is Kth index all unique : True

Time Complexity: O(n), where n is the number of strings in the list. We need to traverse through the entire list only once.

Space Complexity: O(n), where n is the number of strings in the list. In the worst case, all Kth index elements are unique, so the set and the list will contain n elements.

Method 5 : using dictionary.

steps to implement this method:

Initialize an empty dictionary to keep track of the frequency of each element at the Kth index.
Loop through the elements in the input list and check if the element at the Kth index is already in the dictionary.
If the element is already in the dictionary, increment its frequency by 1.
If the element is not in the dictionary, add it with frequency 1.
Finally, check if any element in the dictionary has a frequency greater than 1. If so, return False, else return True.

Python3

   test_list = ["gfg", "best", "for", "geeks"]
 
print("The original list is : " + str(test_list))
 
K = 2
 
freq_dict = {}
 
for ele in test_list:
    if ele[K] in freq_dict:
        freq_dict[ele[K]] += 1
    else:
        freq_dict[ele[K]] = 1
 
unique = all(value == 1 for value in freq_dict.values())
 
print("Is Kth index all unique : " + str(unique))

Output

The original list is : ['gfg', 'best', 'for', 'geeks'] Is Kth index all unique : True

The time complexity of this approach is O(N), where N is the length of the input list.

The auxiliary space required by this approach is O(K), where K is the number of unique elements at the Kth index.

Python | Check if all elements in a list are identical

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