Smallest Greater (than S) String of length K whose letters are subset of S
Last Updated :
20 Mar, 2023
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:
- Store the characters of string S in a STL set (ofcourse, in sorted order)
- Copy the string S to string T upto K characters.
- 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.
- Replace this character with that found in the set.
- 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”;
CPP
#include <bits/stdc++.h>
using namespace std;
string findString(string S, int N, int K)
{
char minChar = 'z';
set<char> found;
for (int i = 0; i < N; i++) {
found.insert(S[i]);
if (S[i] < minChar)
minChar = S[i];
}
string T;
if (N < K) {
T = S;
for (int i = 0; i < (K - N); i++)
T += minChar;
}
else {
T = "";
for (int i = 0; i < K; i++)
T += S[i];
int i;
set<char>::iterator it;
for (i = K - 1; i >= 0; i--) {
it = found.find(T[i]);
it++;
if (it != found.end())
break;
}
T[i] = *it;
for (int j = i + 1; j < K; j++)
T[j] = minChar;
}
return T;
}
int main()
{
string S = "abc";
int N = S.length();
int K = 3;
string T = findString(S, N, K);
cout << "Lexicographically Smallest String "
"greater than " << S
<< " is " << T << endl;
return 0;
}
|
Java
import java.util.*;
public class LexicographicallySmallest {
public static String findString(String S, int N, int K) {
char minChar = 'z';
Set<Character> found = new HashSet<>();
for (int i = 0; i < N; i++) {
found.add(S.charAt(i));
if (S.charAt(i) < minChar) {
minChar = S.charAt(i);
}
}
StringBuilder T = new StringBuilder();
if (N < K) {
T.append(S);
for (int i = 0; i < K - N; i++) {
T.append(minChar);
}
}
else {
T = new StringBuilder();
for (int i = 0; i < K; i++) {
T.append(S.charAt(i));
}
int i;
for (i = K - 1; i >= 0; i--) {
if (found.contains(T.charAt(i))) {
char it = T.charAt(i);
it = (char)(it + 1);
if (found.contains(it)) {
break;
}
}
}
char[] charArray = T.toString().toCharArray();
charArray[i] = (char)(charArray[i] + 1);
T = new StringBuilder(String.valueOf(charArray));
for (int j = i + 1; j < K; j++) {
T.setCharAt(j, minChar);
}
}
return T.toString();
}
public static void main(String[] args) {
String S = "abc";
int N = S.length();
int K = 3;
String T = findString(S, N, K);
System.out.println("Lexicographically Smallest String greater than "
+ S + " is " + T);
}
}
|
Python3
def findString(S, N, K):
minChar = 'z'
found = set([])
for i in range(N):
found.add(S[i])
if (S[i] < minChar):
minChar = S[i]
T = ""
if (N < K):
T = S
for i in range((K - N)):
T += minChar
else:
T = ""
for i in range(K):
T += S[i]
T = list(T)
for i in range(K - 1, -1, -1):
if(T[i] in found):
it = T[i]
it = chr(ord(it)+1)
if (it in found):
break
T[i] = it
for j in range(i + 1, K):
T[j] = minChar
return T
if __name__ == "__main__":
S = "abc"
N = len(S)
K = 3
T = findString(S, N, K)
print("Lexicographically Smallest String "
"greater than " + S + " is " + "".join(T))
|
C#
using System;
using System.Collections.Generic;
public class LexicographicallySmallest {
public static string FindString(string S, int N, int K) {
char minChar = 'z';
HashSet < char > found = new HashSet < char > ();
for (int i = 0; i < N; i++) {
found.Add(S[i]);
if (S[i] < minChar) {
minChar = S[i];
}
}
string T = "";
if (N < K) {
T += S;
for (int i = 0; i < K - N; i++) {
T += minChar;
}
}
else {
T = "";
for (int i = 0; i < K; i++) {
T += S[i];
}
int j;
for (j = K - 1; j >= 0; j--) {
if (found.Contains(T[j])) {
char it = T[j];
it = (char)(it + 1);
if (found.Contains(it)) {
break;
}
}
}
char[] charArray = T.ToCharArray();
charArray[j] = (char)(charArray[j] + 1);
T = new string(charArray);
for (int k = j + 1; k < K; k++) {
charArray[k] = minChar;
}
T = new string(charArray);
}
return T;
}
public static void Main(string[] args) {
string S = "abc";
int N = S.Length;
int K = 3;
string T = FindString(S, N, K);
Console.WriteLine("Lexicographically Smallest String greater than " +
S + " is " + T);
}
}
|
Javascript
function findString(S, N, K) {
let minChar = "z";
let found = new Set();
for (let i = 0; i < N; i++) {
found.add(S[i]);
if (S[i] < minChar) {
minChar = S[i];
}
}
let T = "";
if (N < K) {
T = S;
for (let i = 0; i < K - N; i++) {
T += minChar;
}
}
else {
T = "";
for (let i = 0; i < K; i++) {
T += S[i];
}
T = T.split("");
let i;
for (i = K - 1; i >= 0; i--) {
if (found.has(T[i])) {
var it = T[i];
}
it = String.fromCharCode(it.charCodeAt(0) + 1);
if (found.has(it)) {
break;
}
}
T[i] = it;
for (let j = i + 1; j < K; j++) {
T[j] = minChar;
}
}
return T.join("");
}
let S = "abc";
let N = S.length;
let K = 3;
let T = findString(S, N, K);
console.log(
"Lexicographically Smallest String greater than " + S + " is " + T
);
|
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.
Space Complexity : O(K)
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