字典上最小的字符串，它不是给定字符串

给定一个字符串S ，任务是找到按字典顺序最小的字符串，而不是给定字符串S 的子序列。

例子：

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.

方法：解决给定问题的想法是找到给定字符串字符‘a’的频率（比如f ）。如果f的值小于字符串的长度，那么答案将是通过连接‘a’ f+1次而形成的字符串，因为aaa….(f+1 times)将是所需的字典序最小字符串这不是一个子序列。否则，答案将是通过在字符串连接 ‘ a’（f-1 次）和‘b’形成的字符串。

C++

// C++ program for the above approach
#include 
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

输出

aa

时间复杂度： O(N)
辅助空间： O(1)

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