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Python program to find the character position of Kth word from a list of strings

Last Updated : 28 Apr, 2023
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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 = [“geekforgeeks”, “is”, “best”, “for”, “geeks”], K = 21 
Output : 0
Explanation : 21st index occurs in “geeks” and point to “g” which is 0th element of word.

Input : test_list = [“geekforgeeks”, “is”, “best”, “for”, “geeks”], 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.

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# Python3 code to demonstrate working of
# Word Index for K position in Strings List
# Using enumerate() + list comprehension
 
# initializing list
test_list = ["geekforgeeks", "is", "best", "for", "geeks"]
 
# printing original list
print("The original list is : " + str(test_list))
 
# initializing K
K = 20
 
# enumerate to get indices of all inner and outer list
res = [ele[0] for sub in enumerate(test_list) for ele in enumerate(sub[1])]
 
# getting index of word
res = res[K]
 
# printing result
print("Index of character at Kth position word : " + str(res))
 






                        










Output


The original list is : ['geekforgeeks', 'is', 'best', 'for', 'geeks']
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.




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# Python3 code to demonstrate working of


# Word Index for K position in Strings List


# Using next() + zip() + count()


 


from itertools import count


 


# initializing list


test_list = ["geekforgeeks", "is", "best", "for", "geeks"]


 


# printing original list


print("The original list is : " + str(test_list))


 


# initializing K


K = 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 result


print("Index of character at Kth position word : " + str(res))



















                        







                        










Output


The original list is : ['geekforgeeks', 'is', 'best', 'for', 'geeks']
Index of character at Kth position word : 2




Time Complexity: O(n2)
Auxiliary Space: O(n)


Method 3:  Using nested loop




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# Python3 code to demonstrate working of


# Word Index for K position in Strings List


# Using nested loop


 


# initializing list


test_list = ["geekforgeeks", "is", "best", "for", "geeks"]


 


# printing original list


print("The original list is : " + str(test_list))


 


# initializing K


K = 20


 


# initializing index counter


idx = 0


 


# iterating over each word in the list


for 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 list


else:


    print("K is beyond the end of the list")



















                        







                        










Output


The original list is : ['geekforgeeks', 'is', 'best', 'for', 'geeks']
Index of character at Kth position word : 2






                                    


                                                                        

                                        

                                            

                                                

                                                    
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                                                    manjeet_04                                                

                                            

                                            

            
                                        

                                        

                                            
    

        
        
    
                                            

                                                
                                                
                                            

                                            

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