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Find the longest string that can be made up of other strings from the array
• Difficulty Level : Hard
• Last Updated : 16 Jul, 2019

Given an array of strings arr[], the task is to find the largest string in the array which is made up of the other strings from the array after concatenating one after another. If no such string exists then print -1.

Examples:

Input: arr[] = {“geeks”, “for”, “geeksfor”, “geeksforgeeks”}
Output: geeksforgeeks
“geeksforgeeks” is made up of (“geeks” + “for” + “geeks”).
Even though “geeksfor” is also made up of other strings
but it is not the largest string.

Input: arr[] = {“Hey”, “you”, “stop”, “right”, “there”}
Output : -1

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

1. Sort all the strings based on their lengths in decreasing order.
2. Now, starting from the longest string. Check for all possible prefix of the string whether it is present in the given array and for the remaining part of the string, recursively check whether it can be made up from other strings from the array.
3. Map can be used to check whether a string exists in the array or not. The first string which satisfies the above conditions is the answer.
4. If no such string exists then print -1.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;`` ` `// Comparator to sort the string by``// their lengths in decreasing order``bool` `compare(string s1, string s2)``{``    ``return` `s1.size() > s2.size();``}`` ` `// Function that returns true if string s can be``// made up of by other two string from the array``// after concatenating one after another``bool` `canbuildword(string& s, ``bool` `isoriginalword,``                  ``map& mp)``{`` ` `    ``// If current string has been processed before``    ``if` `(mp.find(s) != mp.end() && mp[s] == 0)``        ``return` `false``;`` ` `    ``// If current string is found in the map and``    ``// it is not the string under consideration``    ``if` `(mp.find(s) != mp.end() && mp[s] == 1``        ``&& isoriginalword == 0) {``        ``return` `true``;``    ``}`` ` `    ``for` `(``int` `i = 1; i < s.length(); i++) {`` ` `        ``// Split the string into two``        ``// contiguous sub-strings``        ``string left = s.substr(0, i);``        ``string right = s.substr(i);`` ` `        ``// If left sub-string is found in the map and``        ``// the right sub-string can be made from``        ``// the strings from the given array``        ``if` `(mp.find(left) != mp.end() && mp[left] == 1``            ``&& canbuildword(right, 0, mp)) {``            ``return` `true``;``        ``}``    ``}`` ` `    ``// If everything failed, we return false``    ``mp[s] = 0;``    ``return` `false``;``}`` ` `// Function to return the longest string``// that can made be made up from the``// other string of the given array``string printlongestword(vector listofwords)``{`` ` `    ``// Put all the strings in the map``    ``map mp;``    ``for` `(string s : listofwords) {``        ``mp[s] = 1;``    ``}`` ` `    ``// Sort the string in decreasing``    ``// order of their lengths``    ``sort(listofwords.begin(), listofwords.end(), compare);`` ` `    ``// Starting from the longest string``    ``for` `(string s : listofwords) {`` ` `        ``// If current string can be made``        ``// up from other strings``        ``if` `(canbuildword(s, 1, mp))``            ``return` `s;``    ``}`` ` `    ``return` `"-1"``;``}`` ` `// Driver code``int` `main()``{``    ``vector listofwords = { ``"geeks"``, ``"for"``, ``"geeksfor"``,``                                   ``"geeksforgeeks"` `};``    ``cout << printlongestword(listofwords);`` ` `    ``return` `0;``}`

## Python3

 `# Python implementation of the approach`` ` `# Function that returns true if string s can be``# made up of by other two string from the array``# after concatenating one after another``def` `canbuildword(s, isoriginalword, mp):`` ` `    ``# If current string has been processed before``    ``if` `s ``in` `mp ``and` `mp[s] ``=``=` `0``:``        ``return` `False`` ` `    ``# If current string is found in the map and``    ``# it is not the string under consideration``    ``if` `s ``in` `mp ``and` `mp[s] ``=``=` `1` `and` `isoriginalword ``=``=` `0``:``        ``return` `True`` ` `    ``for` `i ``in` `range``(``1``, ``len``(s)):`` ` `        ``# Split the string into two``        ``# contiguous sub-strings``        ``left ``=` `s[:i]``        ``right ``=` `s[i:]`` ` `        ``# If left sub-string is found in the map and``        ``# the right sub-string can be made from``        ``# the strings from the given array``        ``if` `left ``in` `mp ``and` `mp[left] ``=``=` `1` `and` `canbuildword(right, ``0``, mp):``            ``return` `True`` ` `    ``# If everything failed, we return false``    ``mp[s] ``=` `0``    ``return` `False`` ` `# Function to return the longest string``# that can made be made up from the``# other string of the given array``def` `printlongestword(listofwords):`` ` `    ``# Put all the strings in the map``    ``mp ``=` `dict``()``    ``for` `i ``in` `listofwords:``        ``mp[i] ``=` `1`` ` `    ``# Sort the string in decreasing``    ``# order of their lengths``    ``listofwords.sort(key``=``lambda` `x: ``len``(x), reverse``=``True``)`` ` `    ``# Starting from the longest string``    ``for` `i ``in` `listofwords:`` ` `        ``# If current string can be made``        ``# up from other strings``        ``if` `canbuildword(i, ``1``, mp):``            ``return` `i`` ` `    ``return` `"-1"`` ` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ``listofwords ``=` `[``"geeks"``, ``"for"``, ``"geeksfor"``,``                ``"geeksforgeeks"``]`` ` `    ``print``(printlongestword(listofwords))`` ` `# This code is contributed by``# sanjeev2552`
Output:
```geeksforgeeks
```

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