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 2 years ago
Given a list of words, L, that are all the same length, and a string, S, find the starting position of the substring of S that is a concatenation of each word in L exactly once and without any intervening characters. This substring will occur exactly once in S.
The first line of input specifies L, it will contain between 1 and 100 spaceseparated words, each between 1 and 10 characters long.
The second line of input specifies S, the line will contain a single word up to 1 million characters long.
The characters in both L and S should be treated as casesensitive.
 2 years ago
Given a list of words, L, that are all the same length, and a string, S, find the starting position of the substring of S that is a concatenation of each word in L exactly once and without any intervening characters. This substring will occur exactly once in S. The first line of input specifies L, it will contain between 1 and 100 spaceseparated words, each between 1 and 10 characters long. The second line of input specifies S, the line will contain a single word up to 1 million characters long. The characters in both L and S should be treated as casesensitive.

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g152xx
 2 years ago
Best ResponseYou've already chosen the best response.0Example: L: "fooo", "barr", "wing", "ding", "wing" S: "lingmindraboofooowingdingbarrwingmonkeypoundcake" fooowingdingbarrwing Answer: 13 L: "mon", "key" S: "monkey monkey Answer: 0 L: "a", "b", "c", "d", "e" S: "abcdfecdba" ecdba Answer: 5

A.Avinash_Goutham
 2 years ago
Best ResponseYou've already chosen the best response.0i dint get u..... why is that answer for the second case is zero? like monkey s got mon and key r8?

rsmith6559
 2 years ago
Best ResponseYou've already chosen the best response.1One thought, parse the substings in L into a set or the keys of a map, iterate through S, checking for S.substring( iterator, iterator+len( setEntry ) ) to be in the set/map. If found, hand a copy of the set/map, len( setEntry ) and S.substring(iterator) to a checker function that deletes each word as it's found. Success would be an empty set/map.
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