Next word that does not contain a palindrome and has characters from first k

Given a string and a limit k, find lexicographically next word which contains characters from a set of first K letters of the English alphabet and does not contain a palindrome as it’s substring of length more than one. It may be assumed that the input string does not contain a palindromic substring.

Examples:

Input : s = "cba" 
        k = 4
Output : cbd

Input : s = "cba"
        k = 3
Output : -1
we can't form such word

Approach : Our aim is to make lexicographically next word, so, we need to move from right to left. While traversing from right to left change last alphabet from right. While incrementing the last alphabet make sure that formed string contains first k letters and does not contain any substring as a palindrome.

Below is the implementation of the above approach:

C++

// CPP program to find lexicographically
// next word which contains first K
// letters of the English alphabet
// and does not contain a palindrome
// as it's substring of length more
// than one.
#include <bits/stdc++.h>
using namespace std;

// function to return lexicographically
// next word
void findNextWord(string s, int m)
{
    // we made m as m+97 that means
    // our required string contains
    // not more than m+97(as per ASCII
    // value) in it.
    m += 97;
    int n = s.length();
    int i = s.length() - 1;

    // increment last alphabet to make
    // next lexicographically next word.
    s[i]++;

    while (i >= 0 && i <= n - 1) {

        // if i-th alphabet not in first
        // k letters then make it as "a"
        // and then increase (i-1)th letter
        if (s[i] >= m) {
            s[i] = 'a';
            s[--i]++;
        }

        // to check whether formed string
        // palindrome or not.
        else if (s[i] == s[i - 1] ||
                 s[i] == s[i - 2])
            s[i]++;

        // increment i.
        else
            i++;
    }

    // if i less than or equals to one
    // that means we not formed such word.
    if (i <= -1)
        cout << "-1";
    else
        cout << s;
}

// Driver code for above function.
int main()
{
    string str = "abcd";
    int k = 4;
    findNextWord(str, k);
    return 0;
}

Java

// Java program to find lexicographically
// next word which contains first K
// letters of the English alphabet
// and does not contain a palindrome
// as it's substring of length more
// than one.

public class GFG
{

    // function to return lexicographically
    // next word
    static void findNextWord(char[] s, int m)
    {
        // we made m as m+97 that means
        // our required string contains
        // not more than m+97(as per ASCII
        // value) in it.
        m += 97;
        int n = s.length;
        int i = s.length - 1;

        // increment last alphabet to make
        // next lexicographically next word.
        s[i]++;

        while (i >= 0 && i <= n - 1)
        {

            // if i-th alphabet not in first
            // k letters then make it as "a"
            // and then increase (i-1)th letter
            if (s[i] >= m)
            {
                s[i] = 'a';
                s[--i]++;
            }
            // to check whether formed string
            // palindrome or not.
            else if (s[i] == s[i - 1]
                    || s[i] == s[i - 2])
            {
                s[i]++;
            }
            // increment i.
            else
            {
                i++;
            }
        }

        // if i less than or equals to one
        // that means we not formed such word.
        if (i <= -1)
        {
            System.out.println("-1");
        }
        else
        {
            System.out.println(s);
        }
    }

    // Driver code
    public static void main(String[] args)
    {
        char[] str = "abcd".toCharArray();
        int k = 4;
        findNextWord(str, k);
    }
}

// This code contributed by Rajput-Ji

Python3

# Python3 program to find lexicographically
# next word which contains first K
# letters of the English alphabet and
# does not contain a palindrome as it's
# substring of length more than one.

# Function to return lexicographically next word
def findNextWord(s, m):

    # we made m as m+97 that means
    # our required string contains
    # not more than m+97(as per ASCII
    # value) in it.
    m += 97
    n = len(s)
    i = len(s) - 1

    # increment last alphabet to make
    # next lexicographically next word.
    s[i] = chr(ord(s[i]) + 1)

    while i >= 0 and i <= n - 1:

        # if i-th alphabet not in first
        # k letters then make it as "a"
        # and then increase (i-1)th letter
        if ord(s[i]) >= m:
            s[i] = 'a'
            i -= 1
            s[i] = chr(ord(s[i]) + 1)

        # to check whether formed string
        # palindrome or not.
        elif s[i] == s[i - 1] or s[i] == s[i - 2]:
            s[i] = chr(ord(s[i]) + 1)

        # increment i.
        else:
            i += 1

    # if i less than or equals to one
    # that means we not formed such word.
    if i <= -1:
        print("-1")
    else:
        print(''.join(s))

# Driver code
if __name__ == "__main__":

    string = "abcd"
    k = 4
    findNextWord(list(string), k)

# This code is contributed by Rituraj Jain

C#

// C# program to find lexicographically
// next word which contains first K
// letters of the English alphabet
// and does not contain a palindrome
// as it's substring of length more
// than one.
using System;

class GFG
{

    // function to return lexicographically
    // next word
    static void findNextWord(char[] s, int m)
    {
        // we made m as m+97 that means
        // our required string contains
        // not more than m+97(as per ASCII
        // value) in it.
        m += 97;
        int n = s.Length;
        int i = s.Length - 1;

        // increment last alphabet to make
        // next lexicographically next word.
        s[i]++;

        while (i >= 0 && i <= n - 1)
        {

            // if i-th alphabet not in first
            // k letters then make it as "a"
            // and then increase (i-1)th letter
            if (s[i] >= m)
            {
                s[i] = 'a';
                s[--i]++;
            }
            // to check whether formed string
            // palindrome or not.
            else if (s[i] == s[i - 1] ||
                     s[i] == s[i - 2])
            {
                s[i]++;
            }

            // increment i.
            else
            {
                i++;
            }
        }

        // if i less than or equals to one
        // that means we not formed such word.
        if (i <= -1)
        {
            Console.WriteLine("-1");
        }
        else
        {
            Console.WriteLine(s);
        }
    }

    // Driver code
    public static void Main(String[] args)
    {
        char[] str = "abcd".ToCharArray();
        int k = 4;
        findNextWord(str, k);
    }
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// Javascript program to find lexicographically
// next word which contains first K
// letters of the English alphabet
// and does not contain a palindrome
// as it's substring of length more
// than one.

// function to return lexicographically
// next word
function findNextWord(s, m)
{
    // we made m as m+97 that means
    // our required string contains
    // not more than m+97(as per ASCII
    // value) in it.
    m += 97;
    var n = s.length;
    var i = s.length - 1;

    // increment last alphabet to make
    // next lexicographically next word.
    s[i] = String.fromCharCode(s[i].charCodeAt(0)+1);

    while (i >= 0 && i <= n - 1) {

        // if i-th alphabet not in first
        // k letters then make it as "a"
        // and then increase (i-1)th letter

        if (s[i].charCodeAt(0) >= m) {

            s[i] = 'a';
            s[i-1] = String.fromCharCode(s[i-1].charCodeAt(0)+1);
            i--;
        }

        // to check whether formed string
        // palindrome or not.
        else if (s[i] == s[i - 1] ||
                 s[i] == s[i - 2])
            s[i] = String.fromCharCode(s[i].charCodeAt(0)+1);

        // increment i.
        else
            i++;
    }

    // if i less than or equals to one
    // that means we not formed such word.
    if (i <= -1)
        document.write( "-1");
    else
        document.write( s.join(''));
}

// Driver code for above function.
var str = "abcd".split('');
var k = 4;
findNextWord(str, k);

// This code is contributed by noob2000.
</script>

Output

abda

Time Complexity: O(N), where N represents the length of the given string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.