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 <bits/stdc++.h>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<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 orderstring 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 codeint 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
// Java program to find the lexicographically largest// string based on the given orderimport 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)); }} |
Python3
from typing import Listfrom collections import defaultdictdef 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 > n2def 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<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)); }} |
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:
- 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++
// 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 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 approachimport 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
<script>// Javascriptam for the above approach// Find the lexicographically// largest stringfunction 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 orderfunction 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 Codelet 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> |
Output:
abd
Time Complexity: O(N *max_word_length)
Auxiliary Space: O(1)
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