Create a string with unique characters from the given N substrings

Last Updated : 24 Feb, 2023

Given an array arr[] containing N substrings consisting of lowercase English letters, the task is to return the minimum length string that contains all given parts as a substring. All characters in this new answer string should be distinct. If there are multiple strings with the following property present, you have to print the string with minimum length.

Examples:

Input: N = 3, arr[] = {"bcd", "ab", "cdef"}
Output: "abcdef"
Explanation:

abcdef contains "bcd" from [1-3] index (0-based indexing)
abcdef contains "ab" from [0-2]
abcdef contains "cdef" from [2-5]

It can be proven that "abcdef" is the smallest string consisting of the given 3 substring fragments from input.

Input: N = 3, arr[]= {"dfghj", "ghjkl", "asdfg"];
Output: "asdfghjkl"
Explanation:

asdfghjkl contains "dfghj" from [2-6] index (0-based indexing)
asdfghjkl contains "ghjkl" from [4-8]
asdfghjkl contains "asdfg" from [0-4]

It can be proven that "asdfghjkl " is the smallest string consisting of the given 3 substring fragments from input.

Approach: Implement the below idea to solve the problem:

While traversing a string, if a character in left and right exists we will be storing this, information in the left[] and right[] array. If an element of a substring does not have a left element,
this means that it's the first element. We will be marking these elements as -1(indicating the start of a substring).
Finally, print the final answer. We will be starting with the character having a - 1 value in the left[] array because it's the start of the string and will gonna print its right element from the right[] array, now this right element is the new element and now we will be finding it's the right element from right[] array the same way until we reach -2 (as right) for an element, which indicates the end of the string.

Follow the steps to solve the above idea:

Create two arrays left[] and right[] having size = 26 (a to z)
Initialize each value for left[] and right[] = -2 (indicating the following characters has no left and right elements yet)
Iterate over each substring given in the input array
- if it's the first character mark it's left = -1 (indicating the start of a substring)
- else if, it's left == -2 (does not exist yet) update it with the left character if it exists in the current substring
- update the right[] value by accessing s[i+1]^th character of the current substring if it exists.
To print the final result find out the index containing -2 in the left[] array
- print this element and update current = right[current]
- repeat until right != -2

Below is the code to implement the approach:

C++

// C++ code for the above approach:  #include <bits/stdc++.h> using namespace std;  // Function to create string containing // all substrings of array arr[] void formedString(int n, vector<string>& arr) {      // Defining left and right array to     // store the left and right element     // for each character (a to z)     vector<int> left(26), right(26);      // Initializing left and right values     // for each character = -2 (indicating     // the following characters has no left     // and right elements yet)     for (int i = 0; i < 26; i++)         left[i] = right[i] = -2;      for (string s : arr) {          // Iterating over each substring s         // given in the array         for (int j = 0; j < s.size(); j++) {              // Finding the character index             // to store             int t = s[j] - 'a';              // If it's the first character             // of substring (means left             // does not exist)             if (j > 0) {                 int t1 = s[j - 1] - 'a';                 left[t] = t1;             }             else if (left[t] == -2)                 left[t] = -1;              // If right element exists for             // jth character find it and             // store in right[t]             if (j < s.size() - 1) {                 int t1 = s[j + 1] - 'a';                 right[t] = t1;             }         }     }      // Code to output the final string     int j;      // Looping over all characters from     // (a to z) to find out the first     // index (having left = -1)     for (int i = 0; i < 26; i++)          if (left[i] == -1) {              // If 1st index is found print             // it and update j = right[j]             // and repeat until -2 is             // encountered -2 indicates             // the end             int j = i;              while (j != -2) {                 // Converting the integer                 // value into character                 cout << (char)(j + 97);                 j = right[j];             }         } }  // Driver code int main() {      int n = 3;     vector<string> arr = { "dfghj", "ghjkl", "asdfg" };      // Function call     formedString(n, arr); }

Java

// Java code for the above approach:  import java.io.*; import java.util.*;  class GFG {      // Function to create string containing     // all substrings of array arr[]     static void formedString(int n, String arr[])     {          // Defining left and right array to         // store the left and right element         // for each character (a to z)         int[] left = new int[26];         int[] right = new int[26];         // Initializing left and right values         // for each character = -2 (indicating         // the following characters has no left         // and right elements yet)         for (int i = 0; i < 26; i++)             left[i] = right[i] = -2;          for (String s : arr) {              // Iterating over each substring s             // given in the array             for (int j = 0; j < s.length(); j++) {                  // Finding the character index                 // to store                 int t = s.charAt(j) - 'a';                 // If it's the first character                 // of substring (means left                 // does not exist)                 if (j > 0) {                     int t1 = s.charAt(j - 1) - 'a';                     left[t] = t1;                 }                 else if (left[t] == -2)                     left[t] = -1;                  // If right element exists for                 // jth character find it and                 // store in right[t]                 if (j < s.length() - 1) {                     int t1 = s.charAt(j + 1) - 'a';                     right[t] = t1;                 }             }         }          // Code to output the final string         int j;         // Looping over all characters from         // (a to z) to find out the first         // index (having left = -1)         for (int i = 0; i < 26; i++)              if (left[i] == -1) {                  // If 1st index is found print                 // it and update j = right[j]                 // and repeat until -2 is                 // encountered -2 indicates                 // the end                 j = i;                  while (j != -2) {                     // Converting the integer                     // value into character                     System.out.print((char)(j + 97));                     j = right[j];                 }             }     }      // Driver code     public static void main(String[] args)     {         int n = 3;          String[] arr = { "dfghj", "ghjkl", "asdfg" };         // Function Call         formedString(n, arr);     } }

Python3

# Python3 code for the above approach  # Function to create string containing # all substrings of array arr[] def formedstring(n, arr):      # Defining left and right array to     # store the left and right element     # for each character (a to z)     left = [-2 for i in range(26)]     right = [-2 for i in range(26)]      for s in arr:          # Iterating over each substring s         #  given in the array         for j in range(len(s)):              # Find the character index to store             t = ord(s[j]) - ord('a')              # If it is the first character             # of substring (means left does not exist)             if j > 0:                 t1 = ord(s[j - 1]) - ord('a')                 left[t] = t1                              elif left[t] == -2:                 left[t] = -1              # If right element exists for jth character              # find it and store in right[t]             if j < len(s) - 1:                 t1 = ord(s[j + 1]) - ord('a')                 right[t] = t1      # Code to output the final string     j = None      # Looping over all characters from     # (a to z) to find out the first     # index (having left = -1)     for i in range(26):          if left[i] == -1:              # If 1st index is found print             # it and update j = right[j]             # and repeat until -2 is             # encountered -2 indicates the end             j = i              while j != -2:                 # Converting the integer                 # value into character                 print(chr(j + 97), end="")                 j = right[j]  # Driver code if __name__ == "__main__":     n = 3     arr = ["dfghj", "ghjkl", "asdfg"]      # Function call     formedstring(n, arr)  #This code is contributed by nikhilsainiofficial546

// C# code for the above approach: using System;  public class GFG{      // Function to create string containing     // all substrings of array arr[]     static void formedString(int n, string[] arr)     {          // Defining left and right array to         // store the left and right element         // for each character (a to z)         int[] left = new int[26];         int[] right = new int[26];         // Initializing left and right values         // for each character = -2 (indicating         // the following characters has no left         // and right elements yet)         for (int i = 0; i < 26; i++)             left[i] = right[i] = -2;          foreach (string s in arr) {              // Iterating over each substring s             // given in the array             for (int k = 0; k < s.Length;k++) {                  // Finding the character index                 // to store                 int t = s[k] - 'a';                 // If it's the first character                 // of substring (means left                 // does not exist)                 if (k > 0) {                     int t1 = s[k - 1] - 'a';                     left[t] = t1;                 }                 else if (left[t] == -2)                     left[t] = -1;                  // If right element exists for                 // jth character find it and                 // store in right[t]                 if (k < s.Length - 1) {                     int t1 = s[k + 1] - 'a';                     right[t] = t1;                 }             }         }          // Code to output the final string         int j;         // Looping over all characters from         // (a to z) to find out the first         // index (having left = -1)         for (int i = 0; i < 26; i++)              if (left[i] == -1) {                  // If 1st index is found print                 // it and update j = right[j]                 // and repeat until -2 is                 // encountered -2 indicates                 // the end                 j = i;                  while (j != -2) {                     // Converting the integer                     // value into character                     Console.Write((char)(j + 97));                     j = right[j];                 }             }     }      // Driver code     static public void Main (){          int n = 3;          String[] arr = { "dfghj", "ghjkl", "asdfg" };         // Function Call         formedString(n, arr);     } }

JavaScript

function formedString(n, arr) {   // Defining left and right array to   // store the left and right element   // for each character (a to z)   let left = Array(26).fill(-2);   let right = Array(26).fill(-2);    arr.forEach((s) => {     // Iterating over each substring s     // given in the array     for (let j = 0; j < s.length; j++) {       // Find the character index to store       let t = s.charCodeAt(j) - 'a'.charCodeAt(0);         // If it is the first character       // of substring (means left does not exist)       if (j > 0) {         let t1 = s.charCodeAt(j - 1) - 'a'.charCodeAt(0);         left[t] = t1;       } else if (left[t] == -2) {         left[t] = -1;       }        // If right element exists for jth character       // find it and store in right[t]       if (j < s.length - 1) {         let t1 = s.charCodeAt(j + 1) - 'a'.charCodeAt(0);         right[t] = t1;       }     }   });    // Code to output the final string   let j = null;    // Looping over all characters from   // (a to z) to find out the first   // index (having left = -1)   for (let i = 0; i < 26; i++) {     if (left[i] == -1) {     // If 1st index is found print     // it and update j = right[j]     // and repeat until -2 is     // encountered -2 indicates the end     j = i;     while (j != -2) {       // Converting the integer       // value into character       process.stdout.write(String.fromCharCode(j + 97));       j = right[j];       }     }   } }  // Driver code if (require.main === module) { let n = 3; let arr = ["dfghj", "ghjkl", "asdfg"];  // Function call formedString(n, arr); }

Output

asdfghjkl

Time Complexity: O(N*C), Where C is equal to the maximum size of a substring.
Auxiliary Space: O(1) (two extra arrays left[] and right[] of size 26 are used which is constant for all input sizes).

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Create a string with unique characters from the given N substrings

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