# Lexicographically largest string for given dictionary order

Given an array arr[] of N strings and a string order which represents the new alphabetical order of the string. The task is to find the lexicographically largest string based on the given order.

Examples:

Input: arr[] = {“abc”, “abd”, “abz”}, order = “abczdefghijklmnopqrstuvwxy”
Output: abd
Explanation:
Compare two words “abc”, “abd”, the first non-matching character is c, d in the order, c comes before d so abd is largest among them.
Similarly, compare abd and abz.

Input: arr[] = {“abc”, “abdz”, “abd”}, order = “abcdefghijklmnopqrstuvwxyz”
Output: abdz
Explanation:
Among all the given strings abdz is the largest.

Naive Approach:

The idea is to check for each string if it is lexicographically largest among the given strings or not. If yes then print that string else check for the next string.

Steps that were to follow the above approach:

•    Define a function named “checkLargest” which takes two string arguments “s1” and “s2”, and an unordered_map “order” as input.
•    In the “checkLargest” function, compare the length of “s1” and “s2” and store it in n1 and n2 respectively.
•    Traverse both strings and find the first mismatching character and check if it is lexicographically largest or not using the unordered_map      “order”.
•    If all characters match and length of s1 is greater than s2, then s1 is lexicographically largest, so return true.
•    If no mismatching character is found, then return the length comparison result.
•    Define another function named “largestString” which takes a vector “arr” and a string “order” as input.
•    Inside the “largestString” function, create an unordered_map “mp” to store the order of characters using “order”.
•   Traverse all strings in the “arr” vector and compare each string with the current lexicographically largest string using the “checkLargest”       function.
•   If the current string is lexicographically larger than the current largest string, update the largest string with the current string.
•   Finally, return the largest string.
•   In the main function, define the input vector “arr” and string “order”.
•   Call the “largestString” function with “arr” and “order” as input and print the returned result.

Below is the code to implement the above steps:

## C++

 // C++ program to find the lexicographically largest// string based on the given order #include using namespace std; // Function to check if a string is lexicographically// largest among the given strings or notbool checkLargest(string s1, string s2,                  unordered_map& order){    int n1 = s1.size();    int n2 = s2.size();     // Traverse both strings and find the first mismatching    // character and check if it is lexicographically    // largest or not    for (int i = 0; i < min(n1, n2); i++) {        if (order[s1[i]] != order[s2[i]])            return order[s1[i]] > order[s2[i]];    }     // If all characters match and length of s1 is greater    // than s2, then s1 is lexicographically largest    if (n1 > n2)        return true;     return false;} // Function to find the lexicographically largest string// based on the given orderstring largestString(vector& arr, string order){    // Create a hash map to store the order of characters    unordered_map mp;    for (int i = 0; i < order.length(); i++)        mp[order[i]] = i;     // Traverse all strings to find the lexicographically    // largest string    string ans = "";    for (string s : arr) {        if (checkLargest(s, ans, mp))            ans = s;    }     return ans;} // Driver codeint main(){    // Given array of strings    vector arr = { "abc", "abd", "abz" };     // Given alphabetical order    string order = "abczdefghijklmnopqrstuvwxy";     // Function call to find the lexicographically    // largest string based on the given order    cout << largestString(arr, order) << endl;     return 0;}

## Java

 // Java program to find the lexicographically largest// string based on the given order import java.util.*; public class GFG {    // Function to check if a string is lexicographically    // largest among the given strings or not    static boolean    checkLargest(String s1, String s2,                 HashMap order)    {        int n1 = s1.length();        int n2 = s2.length();        // Traverse both strings and find the first        // mismatching        // character and check if it is lexicographically        // largest or not        for (int i = 0; i < Math.min(n1, n2); i++) {            if (order.get(s1.charAt(i))                != order.get(s2.charAt(i)))                return order.get(s1.charAt(i))                    > order.get(s2.charAt(i));        }         // If all characters match and length of s1 is        // greater than s2, then s1 is lexicographically        // largest        if (n1 > n2)            return true;         return false;    }     // Function to find the lexicographically largest string    // based on the given order    static String largestString(List arr,                                String order)    {        // Create a hash map to store the order of        // characters        HashMap mp = new HashMap<>();        for (int i = 0; i < order.length(); i++)            mp.put(order.charAt(i), i);         // Traverse all strings to find the        // lexicographically largest string        String ans = "";        for (String s : arr) {            if (checkLargest(s, ans, mp))                ans = s;        }         return ans;    }     // Driver code    public static void main(String[] args)    {        // Given array of strings        List arr            = Arrays.asList("abc", "abd", "abz");         // Given alphabetical order        String order = "abczdefghijklmnopqrstuvwxy";         // Function call to find the lexicographically        // largest string based on the given order        System.out.println(largestString(arr, order));    }}

## Python3

 from typing import Listfrom collections import defaultdict  def checkLargest(s1: str, s2: str, order: dict) -> bool:    n1, n2 = len(s1), len(s2)    for i in range(min(n1, n2)):        if order[s1[i]] != order[s2[i]]:            return order[s1[i]] > order[s2[i]]    return n1 > n2  def largestString(arr: List[str], order: str) -> str:    # Create a dictionary to store the order of characters    mp = defaultdict(int)    for i, char in enumerate(order):        mp[char] = i     # Traverse all strings to find the lexicographically largest string    ans = ''    for s in arr:        if checkLargest(s, ans, mp):            ans = s    return ans  # Driver codeif __name__ == '__main__':    # Given array of strings    arr = ['abc', 'abd', 'abz']     # Given alphabetical order    order = 'abczdefghijklmnopqrstuvwxy'     # Function call to find the lexicographically largest string    print(largestString(arr, order))

## C#

 using System;using System.Collections.Generic; class Program {  // Function to check if a string is lexicographically  // largest among the given strings or not  static bool CheckLargest(string s1, string s2,                           Dictionary order)  {    int n1 = s1.Length;    int n2 = s2.Length;     // Traverse both strings and find the first    // mismatching character and check if it is    // lexicographically largest or not    for (int i = 0; i < Math.Min(n1, n2); i++) {      if (order[s1[i]] != order[s2[i]])        return order[s1[i]] > order[s2[i]];    }     // If all characters match and length of s1 is    // greater than s2, then s1 is lexicographically    // largest    if (n1 > n2)      return true;     return false;  }   // Function to find the lexicographically largest string  // based on the given order  static string LargestString(List arr,                              string order)  {    // Create a dictionary to store the order of    // characters    Dictionary mp      = new Dictionary();    for (int i = 0; i < order.Length; i++)      mp[order[i]] = i;     // Traverse all strings to find the    // lexicographically largest string    string ans = "";    foreach(string s in arr)    {      if (CheckLargest(s, ans, mp))        ans = s;    }     return ans;  }   static void Main(string[] args)  {    // Given array of strings    List arr      = new List{ "abc", "abd", "abz" };     // Given alphabetical order    string order = "abczdefghijklmnopqrstuvwxy";     // Function call to find the lexicographically    // largest string based on the given order    Console.WriteLine(LargestString(arr, order));  }}

## Javascript

 function checkLargest(s1, s2, order) {  let n1 = s1.length;  let n2 = s2.length;   // Traverse both strings and find the first mismatching  // character and check if it is lexicographically  // largest or not  for (let i = 0; i < Math.min(n1, n2); i++) {    if (order[s1[i]] !== order[s2[i]]) {      return order[s1[i]] > order[s2[i]];    }  }   // If all characters match and length of s1 is greater  // than s2, then s1 is lexicographically largest  if (n1 > n2) {    return true;  }   return false;} function largestString(arr, order) {  // Create a hash map to store the order of characters  let mp = {};  for (let i = 0; i < order.length; i++) {    mp[order[i]] = i;  }   // Traverse all strings to find the lexicographically  // largest string  let ans = "";  for (let s of arr) {    if (checkLargest(s, ans, mp)) {      ans = s;    }  }   return ans;} // Driver codelet arr = ["abc", "abd", "abz"];let order = "abczdefghijklmnopqrstuvwxy"; // Function call to find the lexicographically// largest string based on the given orderconsole.log(largestString(arr, order));

Output
abd

Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: The idea is to implement a comparator function according to the given string order find the string which is lexicographically largest. Below are the steps:

1. Create a map to store the index of the character in the given order of string.
2. Consider first string of the array as the lexicographically largest string as ans.
3. Now traverse the given string in the range [1, N] and compare each string with string ans using the indexes stored in the map.
4. Keep updating the largest lexicographically string in the above step and print the string.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach#include   using namespace std;     int compare(string word1, string word2,            int order[]);             // Find the lexicographically// largest stringstring largestString(string a[], int n,                     string order){         // Create a map of characters    int map[26];     // Value of each character is    // string is given priority    // according to their occurrence    // in the string    for(int i = 0; i < order.length(); i++)        map[order[i] - 'a'] = i;     // Take first String as maximum    string ans = a[0];     for(int i = 1; i < n; i++)    {                 // Compare two strings each time        if (compare(ans, a[i], map) < 0)                 // Update answer            ans = a[i];    }    return ans;} // Implement compare function// to get the dictionary orderint compare(string word1, string word2,             int order[]){    int i = 0, j = 0, charcompareval = 0;     while (i < word1.length() &&            j < word2.length())     {                 // Compare each char        // according to the order        charcompareval = order[word1[i] - 'a'] -                         order[word2[i] - 'a'];             // Find the first non matching        // character in the string        if (charcompareval != 0)            return charcompareval;                     i++;        j++;    }     // If one word is prefix of    // other return shortest word    if (charcompareval == 0)        return (word1.length() -                 word2.length());    else        return charcompareval;} // Driver Codeint main(){    int n = 3;     // Given array of strings arr    string arr[] = { "abc", "abd", "abz" };     // Given order of string    string order = "abczdefghijklmnopqrstuvwxy";     // Function call    string ans = largestString(arr, n, order);     cout << ans;         return 0;}  // This code is contributed by rutvik_56

## Java

 // Java program for the above approach import java.util.*;public class Main {     // Find the lexicographically    // largest string    public static String    largestString(String[] a, int n,                String order)    {        // Create a map of characters        int map[] = new int[26];         // Value of each character is        // string is given priority        // according to their occurrence        // in the string        for (int i = 0; i < order.length(); i++)            map[order.charAt(i) - 'a'] = i;         // Take first String as maximum        String ans = a[0];         for (int i = 1; i < n; i++) {             // Compare two strings each time            if (compare(ans, a[i], map) < 0)                 // Update answer                ans = a[i];        }        return ans;    }     // Implement compare function    // to get the dictionary order    public static int    compare(String word1, String word2,            int[] order)    {        int i = 0, j = 0, charcompareval = 0;         while (i < word1.length()            && j < word2.length()) {             // Compare each char            // according to the order            charcompareval                = order[word1.charAt(i) - 'a']                - order[word2.charAt(i) - 'a'];             // Find the first non matching            // character in the string            if (charcompareval != 0)                 return charcompareval;            i++;            j++;        }         // If one word is prefix of        // other return shortest word        if (charcompareval == 0)            return (word1.length()                    - word2.length());        else            return charcompareval;    }     // Driver Code    public static void main(String args[])    {        int n = 3;         // Given array of strings arr        String arr[] = { "abc", "abd", "abz" };         // Given order of string        String order            = "abczdefghijklmnopqrstuvwxy";         // Function call        String ans            = largestString(arr, n, order);         System.out.println(ans);    }}

## Python3

 # Python3 program for the above approach # Find the lexicographically# largest stringdef largestString(a, n, order):     # Create a map of characters    map = [0] * 26     # Value of each character is    # string is given priority    # according to their occurrence    # in the string    for i in range(len(order)):            map[ord(order[i]) - ord('a')] = i     # Take first String as maximum    ans = a[0]     for i in range(1, n):                 # Compare two strings each time        if (compare(ans, a[i], map) < 0):             # Update answer            ans = a[i]             return ans # Implement compare function# to get the dictionary orderdef compare(word1, word2, order):     i = 0    j = 0    charcompareval = 0;     while (i < len(word1) and        j < len(word2)):         # Compare each char        # according to the order        charcompareval = (order[ord(word1[i]) - ord('a')] -                        order[ord(word2[i]) - ord('a')])         # Find the first non matching        # character in the string        if (charcompareval != 0):            return charcompareval                     i += 1        j += 1     # If one word is prefix of    # other return shortest word    if (charcompareval == 0):        return (len(word1) - len(word2))    else:         return charcompareval     # Driver Codeif __name__ == "__main__":     n = 3     # Given array of strings arr    arr = [ "abc", "abd", "abz" ]     # Given order of string    order = "abczdefghijklmnopqrstuvwxy"     # Function call    ans = largestString(arr, n, order)         print(ans) # This code is contributed by chitranayal

## C#

 // C# program for the above approachusing System;class GFG{ // Find the lexicographically// largest stringpublic static String largestString(String[] a, int n,                                    String order){    // Create a map of characters    int []map = new int[26];     // Value of each character is    // string is given priority    // according to their occurrence    // in the string    for (int i = 0; i < order.Length; i++)    map[order[i] - 'a'] = i;     // Take first String as maximum    String ans = a[0];     for (int i = 1; i < n; i++)    {     // Compare two strings each time    if (compare(ans, a[i], map) < 0)         // Update answer        ans = a[i];    }    return ans;} // Implement compare function// to get the dictionary orderpublic static int compare(String word1,                             String word2,                             int[] order){    int i = 0, j = 0, charcompareval = 0;     while (i < word1.Length &&         j < word2.Length)     {     // Compare each char    // according to the order    charcompareval = order[word1[i] - 'a'] -                    order[word2[i] - 'a'];     // Find the first non matching    // character in the string    if (charcompareval != 0)         return charcompareval;    i++;    j++;    }     // If one word is prefix of    // other return shortest word    if (charcompareval == 0)    return (word1.Length -             word2.Length);    else    return charcompareval;} // Driver Codepublic static void Main(String []args){    int n = 3;     // Given array of strings arr    String []arr = { "abc", "abd", "abz" };     // Given order of string    String order = "abczdefghijklmnopqrstuvwxy";     // Function call    String ans = largestString(arr, n, order);     Console.WriteLine(ans);}} // This code is contributed by Amit Katiyar

## Javascript



Output:
abd

Time Complexity: O(N *max_word_length)
Auxiliary Space: O(1)

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