Skip to content
Related Articles

Related Articles

Modify the string such that it contains all vowels at least once
  • Last Updated : 04 Mar, 2021

Given a string S containing only Uppercase letters, the task is to find the minimum number of replacement of characters needed to get a string with all vowels and if we cannot make the required string then print Impossible.

Examples

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

Approach: Since there are only 5 vowels A, E, I, O, U. So, If the string length is less than 5 it is always impossible. 
For a string of length greater than equal to 5, it is always possible. Just iterate over each character and replace it with the vowel that doesn’t exists in the string. If the current character is a vowel and if it is not visited earlier then we will not change the character to the vowel. If all the vowels are already present from early then no need to change any character.

Below is the implementation of the above approach: 

C++14




// C++14 implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
void addAllVowel(string str)
{
    // All vowels
    char x[] = {'A', 'E', 'I', 'O', 'U'};
 
    // List to store distinct vowels
    vector<char> 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<char> 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<char> 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<Character> 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<Character> 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<Character> 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<Character> 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
Output: 
AOUDEFGHI

 

Time Complexity: O(N), where N is the size of the string

Auxiliary Space: O(N)

Attention geek! Strengthen your foundations with the Python Programming Foundation Course and learn the basics.

To begin with, your interview preparations Enhance your Data Structures concepts with the Python DS Course.




My Personal Notes arrow_drop_up
Recommended Articles
Page :