给定一个包含小写字母和字符‘?’的字符串str 。任务是检查是否有可能使str变好。
如果字符串包含长度为26的子字符串，且其中包含小写字母的每个字符，则该字符串称为良。
任务是检查是否可以通过替换‘?’来使字符串好。与任何小写字母字符。如果可能，则打印修改后的字符串，否则打印-1 。
例子：
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.
方法：如果字符串的长度小于26，则打印-1 。任务是制作一个长度为26的子字符串，其中包含所有小写字符。因此，最简单的方法是遍历所有长度为26的子字符串，然后对每个子字符串计数每个字母的出现次数，而忽略问号。此后，如果存在一个出现两次或两次以上的字符，则此子字符串不能包含字母表中的所有字母，因此我们将处理下一个子字符串。否则，我们可以用未出现在子字符串中的字母填充问号，并获得长度为26的子字符串，其中包含字母表中的所有字母。
下面是上述方法的实现：
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[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_jnuPython3
# 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 JainC#
// 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_jnuPHP
= 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
?>输出：
abcdefghijklmnopqrstuvwxyz