Smallest Greater (than S) String of length K whose letters are subset of S

Given a string S of length N consisting of lowercase English letters and an integer K. Find the lexicographically smallest string T of length K, such that its set of letters is a subset of the set of letters of S and T is lexicographically greater than S. Note: The set of letters is a set, not a multiset. For example, the set of letters of abadaba is {a, b, d}.

Examples:

Input : S = “abc”, K = 3
Output : T = “aca”
Explanation: The list of strings T of length 3, such that the set of letters of T is a subset of letters of S is as follows: “aaa”, “aab”, “aac”, “aba”, “abb”, “abc”, “aca”, “acb”, …. Among them, those which are lexicographically greater than “abc”:
“aca”, “acb”, …. Out of those the lexicographically smallest is “aca”.

Input : S = “abc”, K = 2
Output : T = “ac”

A simple solution is to one by on try all strings of length k in increasing order. For every string, check if it is greater than S, if yes, then return it.

Below is an efficient method to solve this problem.

Let’s consider two cases:
1. If N < K: For this case, we can simply copy the string S to string T for upto N characters and for the remaining (K – N) characters we can append the minimum character (least ASCII value) in string S to string T (K – N) times since we have to find the lexicographically smallest string which is just greater than string S.

2. If N ≥ K: For this case, we need to copy the string S to string T for upto K characters and then for string T by iterating in reverse direction we have to replace all those characters by some character until the string T becomes the lexicographically smallest string greater than string S. For this we can do the following:

  1. Store the characters of string S in a STL set (ofcourse, in sorted order)
  2. Copy the string S to string T upto K characters.
  3. Iterating in reverse direction, find a character (let it be found at position ‘p’) for which there exists some character having greater ASCII value in the set.
  4. Replace this character with that found in the set.
  5. Replace all characters with minimum character after (p + 1)th index upto Kth index.

Illustration: Let S = “bcegikmyyy”, N = 10, K = 9. Set = { b, c, e, g, i, k, m, y }. Copy string S to T upto K characters T = “bcegikmyy”. Then iterating in reverse direction, we first have ‘y’ but there is no greater character than ‘y’ in the set so move ahead. Again we have ‘y’, move ahead now we have ‘m’ for which there is a greater character ‘y’ in the set. Replace ‘m’ with ‘y’ and then after ‘m’ in forward direction replace all characters with minimum character i.e. ‘b’. Hence the string T becomes T = “bcegikybb”;

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/* CPP Program to find the lexicographically
   smallest string T greater than S whose
   letters are subset of letters of S */
#include <bits/stdc++.h>
  
using namespace std;
  
// function to find the lexicographically
// smallest string greater than S
string findString(string S, int N, int K)
{
    // stores minimum character
    char minChar = 'z';
  
    // stores unique characters of
    // string in sorted order
    set<char> found;
  
    // Loop through to find the minimum character
    // and stores characters in the set
    for (int i = 0; i < N; i++) {
        found.insert(S[i]);
        if (S[i] < minChar)
            minChar = S[i];
    }
  
    string T;
  
    // Case 1: If N < K
    if (N < K) {
  
        // copy the string S upto N characters
        T = S;
  
        // append minChar to make the remaining 
        // characters
        for (int i = 0; i < (K - N); i++)
            T += minChar;
    }
  
    // Case 2 :  If N >= K
    else {
        T = "";
  
        // copy the string S upto K characters
        for (int i = 0; i < K; i++)
            T += S[i];
  
        int i;
  
        // an iterator to the set
        set<char>::iterator it;
  
        // iterating in reverse direction
        for (i = K - 1; i >= 0; i--) {
  
            // find the current character in the set
            it = found.find(T[i]);
  
            // increment the iterator
            it++;
  
            // check if some bigger character exists in set
            if (it != found.end())
                break;
        }
  
        // Replace current character with that found in set
        T[i] = *it;
  
        // Replace all characters after with minChar
        for (int j = i + 1; j < K; j++)
            T[j] = minChar;
    }
  
    return T;
}
  
// Driver Code to test the above function
int main()
{
    string S = "abc";
    int N = S.length();
  
    // Length of required string
    int K = 3;
  
    string T = findString(S, N, K);
    cout << "Lexicographically Smallest String "
            "greater than " << S
         << " is " << T << endl;
    return 0;
}

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Output:

Lexicographically Smallest String greater than abc is aca

Time Complexity: O(N + K), where N is the length of given string and K is the length of required string.



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