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# 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 ``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 = { 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``;``        ` `    ``// 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

 `= 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

 ``

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 ``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)

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