Sub-string that contains all lowercase alphabets after performing the given operation

Given a string str containing lower case alphabets and character ‘?’. The task is to check if it is possible to make str good or not.
A string is called good if it contains a sub-string of length 26 which has every character of lower case alphabets in it. 
The task is to check if it is possible to make the string good by replacing ‘?’ characters with any lower case alphabet. If it is possible then print the modified string otherwise print -1.

Examples:

Input: str = “abcdefghijkl?nopqrstuvwxy?” 
Output: abcdefghijklmnopqrstuvwxyz 
Replace first ‘?’ with ‘m’ and second with ‘z’.

Input: str = “abcdefghijklmnopqrstuvwxyz??????” 
Output: abcdefghijklmnopqrstuvwxyzaaaaaa 
Given string already has a sub-string which contains all the 26 lower case alphabets. 

Approach:

If the length of the string is less than 26 then print -1. The Task is to make a sub-string of length 26 that has all the lowercase characters. Thus, the simplest way is to iterate through all sub-strings of length 26 then for each sub-string count the number of occurrences of each alphabet, ignoring the question marks. After that, if there exists a character that occurs twice or more than this sub-string cannot contain all letters of the alphabet, and we process the next sub-string. Otherwise, we can fill in the question marks with the letters that have not appeared in the sub-string and obtain a sub-string of length 26 which contains all letters of the alphabet.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function that returns true if every
// lowercase character appears atmost once
bool valid(int cnt[])
{
    // every character frequency must be not
    // greater than one
    for (int i = 0; i < 26; i++) {
        if (cnt[i] >= 2)
            return false;
    }

    return true;
}

// Function that returns the modified
// good string if possible
string getGoodString(string s, int n)
{
    // If the length of the string is less than n
    if (n < 26)
        return "-1";

    // Sub-strings of length 26
    for (int i = 25; i < n; i++) {

        // To store frequency of each character
        int cnt[26] = { 0 };

        // Get the frequency of each character
        // in the current sub-string
        for (int j = i; j >= i - 25; j--) {
            cnt[s[j] - 'a']++;
        }

        // Check if we can get sub-string containing all
        // the 26 characters
        if (valid(cnt)) {

            // Find which character is missing
            int cur = 0;
            while (cnt[cur] > 0)
                cur++;

            for (int j = i - 25; j <= i; j++) {

                // Fill with missing characters
                if (s[j] == '?') {
                    s[j] = cur + 'a';
                    cur++;

                    // Find the next missing character
                    while (cnt[cur] > 0)
                        cur++;
                }
            }

            // Return the modified good string
            return s;
        }
    }

    return "-1";
}

// Driver code
int main()
{
    string s = "abcdefghijkl?nopqrstuvwxy?";
    int n = s.length();

    cout << getGoodString(s, n);

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG
{

// Function that returns true if every
// lowercase character appears atmost once
static boolean valid(int []cnt)
{
    // every character frequency must be not
    // greater than one
    for (int i = 0; i < 26; i++)
    {
        if (cnt[i] >= 2)
            return false;
    }

    return true;
}

// Function that returns the modified
// good string if possible
static String getGoodString(String ss, int n)
{
    char[] s=ss.toCharArray();

    // If the length of the string is less than n
    if (n < 26)
        return "-1";

    // To store frequency of each character
        int[] cnt = new int[27];

    // Sub-strings of length 26
    for (int i = 25; i < n; i++)
    {



        // Get the frequency of each character
        // in the current sub-string
        for (int j = i; j >= i - 25; j--)
        {
            if (s[j] != '?')
            cnt[((int)s[j] - (int)'a')]++;
        }

        // Check if we can get sub-string containing all
        // the 26 characters
        if (valid(cnt))
        {

            // Find which character is missing
            int cur = 0;
            while (cnt[cur] > 0)
                cur++;

            for (int j = i - 25; j <= i; j++)
            {

                // Fill with missing characters
                if (s[j] == '?')
                {
                    s[j] = (char)(cur + (int)('a'));
                    cur++;

                    // Find the next missing character
                    while (cnt[cur] > 0)
                        cur++;
                }
            }

            // Return the modified good string
            return new String(s);
        }
    }

    return "-1";
}

// Driver code
public static void main (String[] args)
{
    String s = "abcdefghijkl?nopqrstuvwxy?";
    int n = s.length();

    System.out.println(getGoodString(s, n));
}
}

// This code is contributed by chandan_jnu

Python3

# Python3 implementation of the approach

# Function that returns true if every
# lowercase character appears atmost once
def valid(cnt):

    # Every character frequency must
    # be not greater than one
    for i in range(0, 26):
        if cnt[i] >= 2:
            return False

    return True

# Function that returns the modified
# good string if possible
def getGoodString(s, n):

    # If the length of the string is
    # less than n
    if n < 26:
        return "-1"

    # Sub-strings of length 26
    for i in range(25, n):

        # To store frequency of each character
        cnt = [0] * 26

        # Get the frequency of each character
        # in the current sub-string
        for j in range(i, i - 26, -1):
            if s[j] != '?':
                cnt[ord(s[j]) - ord('a')] += 1

        # Check if we can get sub-string
        # containing the 26 characters all
        if valid(cnt):

            # Find which character is missing
            cur = 0
            while cur < 26 and cnt[cur] > 0:
                cur += 1

            for j in range(i - 25, i + 1):

                # Fill with missing characters
                if s[j] == '?':
                    s[j] = chr(cur + ord('a'))
                    cur += 1

                    # Find the next missing character
                    while cur < 26 and cnt[cur] > 0:
                        cur += 1

            # Return the modified good string
            return ''.join(s)

    return "-1"

# Driver code
if __name__ == "__main__":

    s = "abcdefghijkl?nopqrstuvwxy?"
    n = len(s)

    print(getGoodString(list(s), n))

# This code is contributed by Rituraj Jain

C#

// C# implementation of the approach
using System;

class GFG
{

// Function that returns true if every
// lowercase character appears atmost once
static bool valid(int []cnt)
{
    // every character frequency must be not
    // greater than one
    for (int i = 0; i < 26; i++)
    {
        if (cnt[i] >= 2)
            return false;
    }

    return true;
}

// Function that returns the modified
// good string if possible
static string getGoodString(string ss, int n)
{
    char[] s = ss.ToCharArray();

    // If the length of the string is less than n
    if (n < 26)
        return "-1";

    // To store frequency of each character
        int[] cnt = new int[27];

    // Sub-strings of length 26
    for (int i = 25; i < n; i++)
    {



        // Get the frequency of each character
        // in the current sub-string
        for (int j = i; j >= i - 25; j--)
        {
            if (s[j] != '?')
            cnt[((int)s[j] - (int)'a')]++;
        }

        // Check if we can get sub-string containing all
        // the 26 characters
        if (valid(cnt))
        {

            // Find which character is missing
            int cur = 0;
            while (cnt[cur] > 0)
                cur++;

            for (int j = i - 25; j <= i; j++)
            {

                // Fill with missing characters
                if (s[j] == '?')
                {
                    s[j] = (char)(cur + (int)('a'));
                    cur++;

                    // Find the next missing character
                    while (cnt[cur] > 0)
                        cur++;
                }
            }

            // Return the modified good string
            return new String(s);
        }
    }

    return "-1";
}

// Driver code
static void Main()
{
    string s = "abcdefghijkl?nopqrstuvwxy?";
    int n = s.Length;

    Console.WriteLine(getGoodString(s, n));
}
}

// This code is contributed by chandan_jnu

PHP

<?php
// PHP implementation of the approach

// Function that returns true if every
// lowercase character appears atmost once
function valid(&$cnt)
{
    // every character frequency must
    // be not greater than one
    for ($i = 0; $i < 26; $i++)
    {
        if ($cnt[$i] >= 2)
            return false;
    }

    return true;
}

// Function that returns the modified
// good string if possible
function getGoodString($s, $n)
{
    // If the length of the string is
    // less than n
    if ($n < 26)
        return "-1";

    // Sub-strings of length 26
    for ($i = 25; $i < $n; $i++)
    {

        // To store frequency of each character
        $cnt = array_fill(0, 26, NULL);

        // Get the frequency of each character
        // in the current sub-string
        for ($j = $i; $j >= $i - 25; $j--)
        {
            if ($s[$j] != '?')
            $cnt[ord($s[$j]) - ord('a')]++;
        }

        // Check if we can get sub-string
        // containing all the 26 characters
        if (valid($cnt))
        {

            // Find which character is missing
            $cur = 0;
            while ($cur < 26 && $cnt[$cur] > 0)
                $cur++;

            for ($j = $i - 25; $j <= $i; $j++)
            {

                // Fill with missing characters
                if ($s[$j] == '?')
                {
                    $s[$j] = chr($cur + ord('a'));
                    $cur++;

                    // Find the next missing character
                    while ($cur < 26 && $cnt[$cur] > 0)
                        $cur++;
                }
            }

            // Return the modified good string
            return $s;
        }
    }

    return "-1";
}

// Driver code
$s = "abcdefghijkl?nopqrstuvwxy?";
$n = strlen($s);
echo getGoodString($s, $n);

// This code is contributed by ita_c
?>

Javascript

<script>
      // JavaScript implementation of the approach
      // Function that returns true if every
      // lowercase character appears atmost once
      function valid(cnt) {
        // every character frequency must be not
        // greater than one
        for (var i = 0; i < 26; i++) {
          if (cnt[i] >= 2) return false;
        }

        return true;
      }

      // Function that returns the modified
      // good string if possible
      function getGoodString(ss, n) {
        var s = ss.split("");
        // If the length of the string is less than n
        if (n < 26) return "-1";

        // Sub-strings of length 26
        for (var i = 25; i < n; i++) {
          // To store frequency of each character
          var cnt = new Array(26).fill(0);

          // Get the frequency of each character
          // in the current sub-string
          for (var j = i; j >= i - 25; j--) {
            cnt[s[j].charCodeAt(0) - "a".charCodeAt(0)]++;
          }

          // Check if we can get sub-string containing all
          // the 26 characters
          if (valid(cnt)) {
            // Find which character is missing
            var cur = 0;
            while (cnt[cur] > 0) cur++;

            for (var j = i - 25; j <= i; j++) {
              // Fill with missing characters
              if (s[j] === "?") {
                s[j] = String.fromCharCode(cur + "a".charCodeAt(0));
                cur++;

                // Find the next missing character
                while (cnt[cur] > 0) cur++;
              }
            }

            // Return the modified good string
            return s.join("");
          }
        }

        return "-1";
      }

      // Driver code
      var s = "abcdefghijkl?nopqrstuvwxy?";
      var n = s.length;
      document.write(getGoodString(s, n));
    </script>

Output

abcdefghijklmnopqrstuvwxyz

Complexity Analysis:

• Time Complexity: O(N²), where N is the size of the given string.
• Auxiliary Space: O(1)

Approach – 2 :

Here is an alternate approach to the problem:

Algorithm:

1. Traverse the string from left to right.
2. For each position i in the string, check if there exists a substring of length 26 starting from position i that contains all lowercase English alphabets.
3. If such a substring is found, replace all the ‘?’ characters in the substring with the missing characters such that each lowercase English alphabet appears exactly once in the substring.
4. If no such substring is found, return -1.
   

Code:

C++

#include <bits/stdc++.h>
using namespace std;

string getGoodString(string s) {
    int n = s.size();
    vector<int> freq(26, 0);

    for (int i = 0; i < n - 25; i++) {
        bool found = true;
        fill(freq.begin(), freq.end(), 0);

        for (int j = i; j < i + 26; j++) {
            if (s[j] != '?') {
                freq[s[j] - 'a']++;
                if (freq[s[j] - 'a'] > 1) {
                    found = false;
                    break;
                }
            }
        }

        if (found) {
            int idx = 0;
            for (int j = i; j < i + 26; j++) {
                if (s[j] == '?') {
                    while (freq[idx]) {
                        idx++;
                    }
                    s[j] = idx + 'a';
                    freq[idx] = 1;
                }
            }

            for (int j = 0; j < n; j++) {
                if (s[j] == '?') {
                    s[j] = 'a';
                }
            }
            return s;
        }
    }

    return "-1";
}

int main() {
    string s = "abcdefghijkl?nopqrstuvwxy?";
    cout << getGoodString(s) << endl; // Output: "abcdefghijklmnopqrstuvwxyz"
    return 0;
}

Output

abcdefghijklmnopqrstuvwxyz

Complexity Analysis:

• Time Complexity: O(N2)
• Auxiliary Space: O(1)