Find the longest string that can be made up of other strings from the array

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 

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 <bits/stdc++.h>
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<string, bool>& 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<string> listofwords)
{

    // Put all the strings in the map
    map<string, bool> 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<string> listofwords = { "geeks", "for", "geeksfor",
                                   "geeksforgeeks" };
    cout << printlongestword(listofwords);

    return 0;
}

Java

/*package whatever //do not write package name here */

import java.util.*;

class GFG {

  // Function that returns true if string s can be
  // made up of by other two string from the array
  // after concatenating one after another
  static boolean canbuildword(String s, boolean isoriginalword, HashMap<String, Boolean> mp)
  {

    // If current string has been processed before
    if (mp.containsKey(s) && !mp.get(s))
      return false;

    // If current string is found in the map and
    // it is not the string under consideration
    if (mp.containsKey(s) && mp.get(s) && !isoriginalword) {
      return true;
    }

    for (int i = 1; i < s.length(); i++) {

      // Split the string into two
      // contiguous sub-strings
      String left = s.substring(0, i);
      String right = s.substring(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.containsKey(left) && mp.get(left) && canbuildword(right, false, mp)) {
        return true;
      }
    }

    // If everything failed, we return false
    mp.put(s,false);
    return false;
  }

  // Function to return the longest string
  // that can made be made up from the
  // other string of the given array
  static String printlongestword(String[] listofwords)
  {

    // Put all the strings in the map
    HashMap<String, Boolean> mp = new HashMap<>();
    for (String s : listofwords) {
      mp.put(s,true);
    }

    // Sort the string in decreasing
    // order of their lengths
    Arrays.sort(listofwords,(a,b)-> b.length()-a.length());
    // Starting from the longest string
    for (String s : listofwords) {

      // If current string can be made
      // up from other strings
      if (canbuildword(s, true, mp))
        return s;
    }
    return "-1";
  }

  public static void main (String[] args) {
    String []listofwords = { "geeks", "for", "geeksfor", "geeksforgeeks" };
    System.out.println(printlongestword(listofwords));

  }
}

// This code is contributed by aadityaburujwale.

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

C#

using System;
using System.Collections.Generic;

public class GFG {

  // Function that returns true if string s can be
  // made up of by other two string from the array
  // after concatenating one after another
  public static bool
    canbuildword(string s, bool isoriginalword,
                 Dictionary<string, bool> mp)
  {

    // If current string has been processed before
    if (mp.ContainsKey(s) && !mp[s])
      return false;

    // If current string is found in the map and
    // it is not the string under consideration
    if (mp.ContainsKey(s) && mp[s] && !isoriginalword) {
      return true;
    }

    for (int i = 1; i < s.Length; i++) {

      // Split the string into two
      // contiguous sub-strings
      string left = s.Substring(0, i);
      string right = s.Substring(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.ContainsKey(left) && mp[left]
          && canbuildword(right, false, mp)) {
        return true;
      }
    }

    // If everything failed, we return false
    mp[s] = false;
    return false;
  }

  // Function to return the longest string
  // that can made be made up from the
  // other string of the given array
  public static string
    printlongestword(string[] listofwords)
  {

    // Put all the strings in the map
    Dictionary<string, bool> mp
      = new Dictionary<string, bool>();
    foreach (string s in listofwords) {
      mp[s] = true;
    }

    // Sort the string in decreasing
    // order of their lengths
    Array.Sort(listofwords,
               (a, b) => b.Length - a.Length);
    // Starting from the longest string
    foreach (string s in listofwords) {

      // If current string can be made
      // up from other strings
      if (canbuildword(s, true, mp))
        return s;
    }
    return "-1";
  }

  static public void Main()
  {
    string[] listofwords = { "geeks", "for", "geeksfor",
                            "geeksforgeeks" };
    Console.WriteLine(printlongestword(listofwords));
  }
}

// This code is contributed by akashish__

Javascript

<script>

// JavaScript implementation of the approach

// Comparator to sort the string by
// their lengths in decreasing order
function compare(s1, s2)
{
    return s2.length - s1.length;
}

// Function that returns true if string s can be
// made up of by other two string from the array
// after concatenating one after another
function canbuildword(s,isoriginalword,mp)
{

    // If current string has been processed before
    if (mp.has(s) && mp.get(s) == 0)
        return false;

    // If current string is found in the map and
    // it is not the string under consideration
    if (mp.has(s) && mp.get(s) == 1
        && isoriginalword == 0) {
        return true;
    }

    for (let i = 1; i < s.length; i++) {

        // Split the string into two
        // contiguous sub-strings
        let left = s.substring(0, i);
        let right = s.substring(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.has(left) == true && mp.get(left) == 1
            && canbuildword(right, 0, mp)) {
            return true;
        }
    }

    // If everything failed, we return false
    mp.set(s,0);
    return false;
}

// Function to return the longest string
// that can made be made up from the
// other string of the given array
function printlongestword(listofwords)
{

    // Put all the strings in the map
    let mp = new Map();
    for (let s of listofwords) {
        mp.set(s,1);
    }

    // Sort the string in decreasing
    // order of their lengths
    listofwords.sort(compare);

    // Starting from the longest string
    for (let s of listofwords) {

        // If current string can be made
        // up from other strings
        if (canbuildword(s, 1, mp))
            return s;
    }

    return "-1";
}

// Driver code
let listofwords = [ "geeks", "for", "geeksfor", "geeksforgeeks" ];
document.write(printlongestword(listofwords));

// This code is contributed by shinjanpatra

</script>

Output:

geeksforgeeks

Time complexity: O(N^3)
Auxiliary Space: O(N).