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++ program to find the lexicographically largest // string based on the given order #include <bits/stdc++.h> using namespace std;
// Function to check if a string is lexicographically // largest among the given strings or not bool checkLargest(string s1, string s2,
unordered_map< char , int >& 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 order string largestString(vector<string>& arr, string order) { // Create a hash map to store the order of characters
unordered_map< char , int > 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 code int main()
{ // Given array of strings
vector<string> 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 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<Character, Integer> 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<String> arr,
String order)
{
// Create a hash map to store the order of
// characters
HashMap<Character, Integer> 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<String> 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));
}
} |
from typing import List
from 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 code if __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))
|
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< char , int > 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< string > arr,
string order)
{
// Create a dictionary to store the order of
// characters
Dictionary< char , int > mp
= new Dictionary< char , int >();
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< string > arr
= new List< string >{ "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));
}
} |
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 code let arr = [ "abc" , "abd" , "abz" ];
let order = "abczdefghijklmnopqrstuvwxy" ;
// Function call to find the lexicographically // largest string based on the given order console.log(largestString(arr, order)); |
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:
- Create a map to store the index of the character in the given order of string.
- Consider first string of the array as the lexicographically largest string as ans.
- Now traverse the given string in the range [1, N] and compare each string with string ans using the indexes stored in the map.
- Keep updating the largest lexicographically string in the above step and print the string.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
int compare(string word1, string word2,
int order[]);
// Find the lexicographically // largest string string 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 order 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 Code int 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 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 program for the above approach # Find the lexicographically # largest string def 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 order def 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 Code if __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# program for the above approach using System;
class GFG{
// 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[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[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 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);
Console.WriteLine(ans);
} } // This code is contributed by Amit Katiyar |
<script> // Javascriptam for the above approach // Find the lexicographically // largest string function largestString(a, n, order)
{ // Create a map of characters
let map = new Array(26);
// Value of each character is
// string is given priority
// according to their occurrence
// in the string
for (let i = 0; i < order.length; i++)
map[order[i].charCodeAt(0) -
'a' .charCodeAt(0)] = i;
// Take first String as maximum
let ans = a[0];
for (let 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 function compare(word1, word2, order)
{ let i = 0, j = 0, charcompareval = 0;
while (i < word1.length &&
j < word2.length)
{
// Compare each char
// according to the order
charcompareval = order[word1[i].charCodeAt(0) -
'a' .charCodeAt(0)] -
order[word2[i].charCodeAt(0) -
'a' .charCodeAt(0)];
// 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 let n = 3; let arr = [ "abc" , "abd" , "abz" ];
let order = "abczdefghijklmnopqrstuvwxy" ;
let ans = largestString(arr, n, order); document.write(ans); // This code is contributed by avanitrachhadiya2155 </script> |
abd
Time Complexity: O(N *max_word_length)
Auxiliary Space: O(1)