Given a list of strings. The task is to find the index of the character position for the word, which lies at the Kth index in the list of strings.
Examples:
Input : test_list = [“geekforLazyroar”, “is”, “best”, “for”, “Lazyroar”], K = 21
Output : 0
Explanation : 21st index occurs in “Lazyroar” and point to “g” which is 0th element of word.Input : test_list = [“geekforLazyroar”, “is”, “best”, “for”, “Lazyroar”], K = 15
Output : 1
Explanation : 15th index occurs in “best” and point to “e” which is 1st element of word.
Method #1 : Using enumerate() + list comprehension
In this, we use nested enumerate() to check indices for words, and strings in the list, list comprehension is used to encapsulate logic in 1 liner.
Python3
# Python3 code to demonstrate working of# Word Index for K position in Strings List# Using enumerate() + list comprehension# initializing listtest_list = ["geekforLazyroar", "is", "best", "for", "Lazyroar"]# printing original listprint("The original list is : " + str(test_list))# initializing KK = 20# enumerate to get indices of all inner and outer listres = [ele[0] for sub in enumerate(test_list) for ele in enumerate(sub[1])]# getting index of wordres = res[K]# printing resultprint("Index of character at Kth position word : " + str(res)) |
The original list is : ['geekforLazyroar', 'is', 'best', 'for', 'Lazyroar'] Index of character at Kth position word : 2
Time Complexity: O(n2)
Auxiliary Space: O(n)
Method #2 : Using next() + zip() + count()
In this, we pair up the number of words with their counts using zip(), and accumulate till we don’t reach Kth Index.
Python3
# Python3 code to demonstrate working of# Word Index for K position in Strings List# Using next() + zip() + count()from itertools import count# initializing listtest_list = ["geekforLazyroar", "is", "best", "for", "Lazyroar"]# printing original listprint("The original list is : " + str(test_list))# initializing KK = 20# count() for getting count# pairing using zip()cnt = count()res = next(j for sub in test_list for j, idx in zip( range(len(sub)), cnt) if idx == K)# printing resultprint("Index of character at Kth position word : " + str(res)) |
The original list is : ['geekforLazyroar', 'is', 'best', 'for', 'Lazyroar'] Index of character at Kth position word : 2
Time Complexity: O(n2)
Auxiliary Space: O(n)
Method 3: Using nested loop
Python3
# Python3 code to demonstrate working of# Word Index for K position in Strings List# Using nested loop# initializing listtest_list = ["geekforLazyroar", "is", "best", "for", "Lazyroar"]# printing original listprint("The original list is : " + str(test_list))# initializing KK = 20# initializing index counteridx = 0# iterating over each word in the listfor word in test_list: # if the kth position is in the current word if idx + len(word) > K: # printing result print("Index of character at Kth position word : " + str(K - idx)) break # if the kth position is not in the current word else: idx += len(word)# if the kth position is beyond the end of the listelse: print("K is beyond the end of the list") |
The original list is : ['geekforLazyroar', 'is', 'best', 'for', 'Lazyroar'] Index of character at Kth position word : 2
