给定一个仅包含大写字母的字符串S，任务是找到用所有元音获得一个字符串所需的最少字符替换数，如果我们不能制作所需的字符串，则打印不可能。
例子：
Input: str = "ABCDEFGHI"
Output: AOUDEFGHI
There are already 3 Vowels present in the string
A, E, I we just change B and C to O and U respectively.
Input: str = "ABC"
Output: IMPOSSIBLE方法：由于只有5个元音A，E，I，O，U。因此，如果字符串长度小于5，则总是不可能的。
对于长度大于等于5的字符串，总是可能的。只需遍历每个字符并用字符串中不存在的元音代替。如果当前字符是元音，并且如果不早点访问它，则我们不会将字符更改为元音。如果所有元音从一开始就已经存在，则无需更改任何字符。
下面是上述方法的实现：
C++14
// C++14 implementation of the above approach
#include
using namespace std;
void addAllVowel(string str)
{
// All vowels
char x[] = {'A', 'E', 'I', 'O', 'U'};
// List to store distinct vowels
vector y;
int length = str.length();
// if length of string is less than 5
// then always Impossible
if (length < 5)
cout << "Impossible" << endl;
else
{
// Storing the distinct vowels in the string
// by checking if it in the list of string and not
// in the list of distinct vowels
for (int i = 0; i < length; i++)
{
if (find(x, x + 5, str[i]) != x + 5 and
find(y.begin(), y.end(), str[i]) == y.end())
y.push_back(str[i]);
}
// Storing the vowels which are
// not present in the string
vector z;
for (int i = 0; i < 5; i++)
if (find(y.begin(),
y.end(), x[i]) == y.end())
z.push_back(x[i]);
// No replacement needed condition
if (z.empty())
cout << str << endl;
else
{
int cc = 0;
vector y;
// Replacing the characters to get all Vowels
for (int i = 0; i < length; i++)
{
if (find(x, x + 5, str[i]) != x + 5 and
find(y.begin(), y.end(),
str[i]) == y.end())
y.push_back(str[i]);
else
{
str[i] = z[cc];
cc++;
}
if (cc == z.size()) break;
}
cout << str << endl;
}
}
}
// Driver Code
int main(int argc, char const *argv[])
{
string str = "ABCDEFGHI";
addAllVowel(str);
return 0;
}
// This code is contributed by
// sanjeev2552 Java
// Java implementation of the above approach
import java.util.*;
class GFG
{
static boolean find(char x[], char c)
{
for(int i = 0; i < x.length; i++)
if(x[i] == c)
return true;
return false;
}
static boolean find(Vector v,char c)
{
for(int i = 0; i < v.size(); i++)
if(v.get(i) == c)
return true;
return false;
}
static void addAllVowel(String str)
{
// All vowels
char x[] = {'A', 'E', 'I', 'O', 'U'};
// List to store distinct vowels
Vector y = new Vector<>();
int length = str.length();
// if length of String is less than 5
// then always Impossible
if (length < 5)
System.out.println("Impossible");
else
{
// Storing the distinct vowels in the String
// by checking if it in the list of String and not
// in the list of distinct vowels
for (int i = 0; i < length; i++)
{
if (find(x, str.charAt(i))&&
!find(y, str.charAt(i)))
y.add(str.charAt(i));
}
// Storing the vowels which are
// not present in the String
Vector z = new Vector<>();
for (int i = 0; i < 5; i++)
if (!find(y, x[i]) )
z.add(x[i]);
// No replacement needed condition
if (z.size() == 0)
System.out.println( str );
else
{
int cc = 0;
Vector y1 = new Vector<>();
String ans = "";
// Replacing the characters to get all Vowels
for (int i = 0; i < length; i++)
{
if (find(x, str.charAt(i))&&
!find(y1, str.charAt(i)))
{
ans += str.charAt(i);
y1.add(str.charAt(i));
}
else
{
ans += z.get(cc);
cc++;
}
if (cc == z.size())
{
//copy th rest of the string
for(int j = i + 1; j < length; j++)
ans += str.charAt(j);
break;
}
}
System.out.println(ans);
}
}
}
// Driver Code
public static void main(String args[])
{
String str = "ABCDEFGHI";
addAllVowel(str);
}
}
// This code is contributed by Arnab Kundu Python3
# Python3 implementation of the above approach
s = "ABCDEFGHI"
# converting String to a list
S = list(s)
# All vowels
x = ["A", "E", "I", "O", "U"]
# List to store distinct vowels
y = []
le = len(S)
if (le < 5):
# if length of string is less than 5 then always
# Impossible
print("Impossible")
else:
# Storing the distinct vowels in the string
# by checking if it in the list of string and not
# in the list of distinct vowels
for i in range(le):
if (S[i] in x and S[i] not in y):
y.append(S[i])
# Storing the vowels which are not present in the string
z = []
for i in range(5):
if (x[i] not in y):
z.append(x[i])
if (len(z) == 0):
# No replacement needed condition
print(s)
else:
cc = 0
y = []
# Replacing the characters to get all Vowels
for i in range(le):
if (S[i] in x and S[i] not in y):
y.append(S[i])
else:
S[i] = z[cc]
cc += 1
if (cc == len(z)):
break
print(*S, sep ="")C#
// C# implementation of the above approach
using System;
using System.Collections;
class GFG{
static bool find(char []x, char c)
{
for(int i = 0; i < x.Length; i++)
if (x[i] == c)
return true;
return false;
}
static bool find(ArrayList v, char c)
{
for(int i = 0; i < v.Count; i++)
if ((char)v[i] == c)
return true;
return false;
}
static void addAllVowel(string str)
{
// All vowels
char []x = { 'A', 'E', 'I', 'O', 'U' };
// List to store distinct vowels
ArrayList y = new ArrayList();
int length = str.Length;
// If length of String is less than 5
// then always Impossible
if (length < 5)
Console.Write("Impossible");
else
{
// Storing the distinct vowels in
// the String by checking if it in
// the list of String and not
// in the list of distinct vowels
for(int i = 0; i < length; i++)
{
if (find(x, str[i]) &&
!find(y, str[i]))
y.Add(str[i]);
}
// Storing the vowels which are
// not present in the String
ArrayList z = new ArrayList();
for(int i = 0; i < 5; i++)
if (!find(y, x[i]) )
z.Add(x[i]);
// No replacement needed condition
if (z.Count == 0)
Console.Write(str);
else
{
int cc = 0;
ArrayList y1 = new ArrayList();
string ans = "";
// Replacing the characters to
// get all Vowels
for(int i = 0; i < length; i++)
{
if (find(x, str[i]) &&
!find(y1, str[i]))
{
ans += str[i];
y1.Add(str[i]);
}
else
{
ans += (char)z[cc];
cc++;
}
if (cc == z.Count)
{
// Copy th rest of the string
for(int j = i + 1;
j < length; j++)
ans += str[j];
break;
}
}
Console.Write(ans);
}
}
}
// Driver Code
public static void Main(string []args)
{
string str = "ABCDEFGHI";
addAllVowel(str);
}
}
// This code is contributed by rutvik_56输出：
AOUDEFGHI时间复杂度： O(N)，其中N是字符串的大小
辅助空间： O(N)