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Lexicographically smallest string which is not a subsequence of given string

  • Last Updated : 05 Jul, 2021

Given a string S, the task is to find the string which is lexicographically smallest and not a subsequence of the given string S.

 Examples:

Input: S = “abcdefghijklmnopqrstuvwxyz”
Output: 
aa
Explanation:
String “aa” is the lexicographically smallest string which is not present in the given string as a subsequence.

Input: S = “aaaa”
Output: 
aaab
Explanation:
String “aaa” is the lexicographically smallest string which is not present in the given string as a subsequence.

Approach: The idea is to solve the given problem is to find the frequency of the character ‘a’ in the given string (say f). If the value of f is less than the length of the string, then the answer would be the string formed by concatenating ‘a’ f+1 number of times as aaa….(f+1 times) will be the required lexicographically smallest string which is not a subsequence. Otherwise, the answer would be the string formed by concatenating ‘a’ (f-1 times) and ‘b’ at the end of the string.



C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find lexicographically smallest string that
// is not a subsequence of S
string smallestNonSubsequence(string S, int N)
{
    // variable to store frequency of 'a'
    int freq = 0;
    // calculate frequency of 'a'
    for (int i = 0; i < N; i++)
        if (S[i] == 'a')
            freq++;
    // Initialize string consisting of freq number of 'a'
    string ans(freq, 'a');
    // change the last digit to 'b'
    if (freq == N)
        ans[freq - 1] = 'b';
    // add another 'a' to the string
    else
        ans += 'a';
    // return tha answer string
    return ans;
}
// Driver code
int main()
{
    // Input
    string S = "abcdefghijklmnopqrstuvwxyz";
    int N = S.length();
 
    // Function call
    cout << smallestNonSubsequence(S, N) << endl;
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
public class GFG
{
 
// Function to find lexicographically smallest string that
// is not a subsequence of S
static String smallestNonSubsequence(String S, int N)
{
   
    // variable to store frequency of 'a'
    int freq = 0;
   
    // calculate frequency of 'a'
    for (int i = 0; i < N; i++)
        if (S.charAt(i) == 'a')
            freq++;
   
    // Initialize string consisting of freq number of 'a'
 String ans="";
    for(int i = 0; i < f req; i++){
        ans += 'a';
    }
   
    // change the last digit to 'b'
    if (freq == N)
       ans = ans.replace(ans.charAt(freq - 1), 'b');
   
    // add another 'a' to the string
    else
        ans += 'a';
   
    // return tha answer string
    return ans;
}
 
// Driver Code
public static void main(String args[])
{
    String S = "abcdefghijklmnopqrstuvwxyz";
    int N = S.length();
 
    // Function call
    System.out.println(smallestNonSubsequence(S, N));
    }
}
 
// This code is contributed by SoumikMondal

Python3




# Python3 program for the above approach
 
# Function to find lexicographically smallest
# string that is not a subsequence of S
def smallestNonSubsequence(S, N):
     
    # Variable to store frequency of 'a'
    freq = 0
     
    # Calculate frequency of 'a'
    for i in range(N):
        if (S[i] == 'a'):
            freq += 1
             
    # Initialize string consisting
    # of freq number of 'a'
    ans = ""
    for i in range(freq):
        ans += 'a'
 
    # Change the last digit to 'b'
    if (freq == N):
        ans = ans.replace(ans[freq - 1], 'b')
         
    # Add another 'a' to the string
    else:
        ans += 'a'
         
    # Return tha answer string
    return ans
 
# Driver code
if __name__ == '__main__':
     
    # Input
    S = "abcdefghijklmnopqrstuvwxyz"
    N = len(S)
 
    # Function call
    print(smallestNonSubsequence(S, N))
     
# This code is contributed by SURENDRA_GANGWAR
Output
aa

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

 

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