Lexicographically smallest string obtained after concatenating array
Given n strings, concatenate them in an order that produces the lexicographically smallest possible string.
Examples:
Input : a[] = ["c", "cb", "cba"] Output : cbacbc Possible strings are ccbcba, ccbacb, cbccba, cbcbac, cbacbc and cbaccb. Among all these strings, cbacbc is the lexicographically smallest. Input : a[] = ["aa", "ab", "aaa"] Output : aaaaaab
One might think that sorting the given strings in the lexicographical order and then concatenating them produces the correct output. This approach produces the correct output for inputs like [“a”, “ab”, “abc”]. However, applying this method on [“c”, “cb”, “cba”] produces the wrong input and hence this approach is incorrect.
The correct approach is to use a regular sorting algorithm. When two strings a and b are compared to decide if they have to be swapped or not, do not check if a is lexicographically smaller than b or not. Instead check if appending b at the end of a produces a lexicographically smaller string or appending a at the end of b does. This approach works because we want the concatenated string to be lexicographically small, not the individual strings to be in the lexicographical order.
C++
// CPP code to find the lexicographically // smallest string #include <bits/stdc++.h> using namespace std; // Compares two strings by checking if // which of the two concatenations causes // lexicographically smaller string. bool compare(string a, string b) { return (a+b < b+a); } string lexSmallest(string a[], int n) { // Sort strings using above compare() sort(a, a+n, compare); // Concatenating sorted strings string answer = ""; for (int i = 0; i < n; i++) answer += a[i]; return answer; } // Driver code int main() { string a[] = { "c", "cb", "cba" }; int n = sizeof(a)/sizeof(a[0]); cout << lexSmallest(a, n); return 0; } |
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Java
// Java code to find the lexicographically // smallest string class GFG { // function to sort the // array of string static void sort(String a[], int n) { //sort the array for(int i = 0;i < n;i++) { for(int j = i + 1;j < n;j++) { // comparing which of the // two concatenation causes // lexiographically smaller // string if((a[i] + a[j]).compareTo(a[j] + a[i]) > 0) { String s = a[i]; a[i] = a[j]; a[j] = s; } } } } static String lexsmallest(String a[], int n) { // Sort strings sort(a,n); // Concatenating sorted strings String answer = ""; for (int i = 0; i < n; i++) answer += a[i]; return answer; } // Driver code public static void main(String args[]) { String a[] = {"c", "cb", "cba"}; int n = 3; System.out.println("lexiographically smallest string = " + lexsmallest(a, n)); } } // This code is contributed by Arnab Kundu |
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Python 3
# Python 3 code to find the lexicographically
# smallest string
def lexSmallest(a, n):
# Sort strings using above compare()
a.sort(reverse = True)
# Concatenating sorted strings
answer = “”
for i in range( n):
answer += a[i]
return answer
# Driver code
if __name__ == “__main__”:
a = [ “c”, “cb”, “cba” ]
n = len(a)
print(lexSmallest(a, n))
# This code is contributed by ita_c
C#
// C# code to find // the lexicographically // smallest string using System; class GFG { // function to sort the // array of string static void sort(String []a, int n) { //sort the array for(int i = 0;i < n;i++) { for(int j = i + 1;j < n;j++) { // comparing which of the // two concatenation causes // lexiographically smaller // string if((a[i] + a[j]).CompareTo(a[j] + a[i]) > 0) { String s = a[i]; a[i] = a[j]; a[j] = s; } } } } static String lexsmallest(String []a, int n) { // Sort strings sort(a,n); // Concatenating sorted // strings String answer = ""; for (int i = 0; i < n; i++) answer += a[i]; return answer; } // Driver code public static void Main() { String []a = {"c", "cb", "cba"}; int n = 3; Console.Write("lexiographically smallest string = " + lexsmallest(a, n)); } } // This code is contributed by nitin mittal |
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Output:
cbacbc
Time complexity : The above code runs inO(M * N * logN) where N is number of strings and M is maximum length of a string.
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