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Modify the string such that it contains all vowels at least once

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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
Explanation: 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 exist 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 an 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


Javascript




<script>
 
// JavaScript implementation of the above approach
 
function find(x,c)
{
    for(let i = 0; i < x.length; i++)
        if(x[i] == c)
            return true;
    return false;
}
 
function addAllVowel(str)
{
    // All vowels
    let x = ['A', 'E', 'I', 'O', 'U'];
  
    // List to store distinct vowels
    let y = [];
    let length = str.length;
  
    // if length of String is less than 5
    // then always Impossible
    if (length < 5)
        document.write("Impossible<br>");
    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 (let i = 0; i < length; i++)
        {
            if (find(x, str[i])&&
                !find(y, str[i]))
                y.push(str[i]);
        }
  
        // Storing the vowels which are
        // not present in the String
        let z = [];
        for (let i = 0; i < 5; i++)
            if (!find(y, x[i]) )
                z.push(x[i]);
  
        // No replacement needed condition
        if (z.length == 0)
            document.write( str+"<br>" );
        else
        {
            let cc = 0;
            let y1 = [];
               let ans = "";
          
            // Replacing the characters to get all Vowels
            for (let i = 0; i < length; i++)
            {
            if (find(x, str[i])&&
                !find(y1, str[i]))
                {
                    ans += str[i];
                    y1.push(str[i]);
                }
                else
                {
                    ans += z[cc];
                    cc++;
                }
                if (cc == z.length)
                {
                    //copy th rest of the string
                    for(let j = i + 1; j < length; j++)
                        ans += str[j];
                        break;
                }
            }
            document.write(ans);
        }
    }
}
 
// Driver Code
let str = "ABCDEFGHI";
addAllVowel(str);
 
 
// This code is contributed by avanitrachhadiya2155
 
</script>


Output

AOUDEFGHI

Complexity Analysis:

  • Time Complexity: O(N), where N is the size of the string
  • Auxiliary Space: O(1) because a constant size vector is being used to store vowels

Another approach is to use a dictionary to store the count of each vowel in the string. This can be done by iterating through the string and incrementing the count of each vowel in the dictionary. Then, we can iterate through the dictionary and check if any of the vowel counts are zero. If a vowel count is zero, we can replace a non-vowel character in the string with that vowel. This can be done by iterating through the string again and replacing the first non-vowel character that is encountered with the missing vowel.

Here is an example of this approach in Python:

C++




#include <iostream>
#include <map>
 
using namespace std;
 
string modifyString(string s)
{
    map<char, int> vowels;
    vowels['A'] = 0;
    vowels['E'] = 0;
    vowels['I'] = 0;
    vowels['O'] = 0;
    vowels['U'] = 0;
 
    for (int i = 0; i < s.length(); i++) {
        char c = s[i];
        if (vowels.count(c)) {
            vowels++;
        }
    }
 
    for (int i = 0; i < s.length(); i++) {
        char c = s[i];
        if (!vowels.count(c)) {
            for (auto& kv : vowels) {
                if (kv.second == 0) {
                    s[i] = kv.first;
                    kv.second = 1;
                    break;
                }
            }
        }
    }
 
    return s;
}
 
int main()
{
    string s = "ABCDEFGHI";
    string modified_s = modifyString(s);
    cout << modified_s << endl;
    return 0;
}


Java




import java.util.*;
public class Main {
     
   public static String modifyString(String s) {
    Map<Character, Integer> vowels = new HashMap<Character, Integer>();
    vowels.put('A', 0);
    vowels.put('E', 0);
    vowels.put('I', 0);
    vowels.put('O', 0);
    vowels.put('U', 0);
 
    for (int i = 0; i < s.length(); i++) {
        char c = s.charAt(i);
        if (vowels.containsKey(c)) {
            vowels.put(c, vowels.get(c) + 1);
        }
    }
 
    for (int i = 0; i < s.length(); i++) {
        char c = s.charAt(i);
        if (!vowels.containsKey(c)) {
            for (char v : vowels.keySet()) {
                if (vowels.get(v) == 0) {
                    s = s.substring(0, i) + v + s.substring(i+1);
                    vowels.put(v, 1);
                    break;
                }
            }
        }
    }
 
    return s;
}
 
public static void main(String[] args) {
    String s = "ABCDEFGHI";
    String modified_s = modifyString(s);
    System.out.println(modified_s);
}
 
}


Python3




def modify_string(s):
    vowels = {'A': 0, 'E': 0, 'I': 0, 'O': 0, 'U': 0}
    for c in s:
        if c in vowels:
            vowels += 1
    for c in s:
        if c in vowels:
            continue
        for v in vowels:
            if vowels[v] == 0:
                s = s.replace(c, v, 1)
                vowels[v] = 1
                break
    return s
 
s = "ABCDEFGHI"
modified_s = modify_string(s)
print(modified_s)
#This code is contributed by Edula Vinay Kumar Reddy


Javascript




// JS code to replace the non-vowels in a string with vowels in order
function modifyString(s)
{
 
    // create a dictionary for vowels with initial value 0
    let vowels = {'A': 0, 'E': 0, 'I': 0, 'O': 0, 'U': 0};
     
    // loop through the string and count the vowels
    for (let i = 0; i < s.length; i++) {
        let c = s[i];
        if (c in vowels) {
            vowels += 1;
        }
    }
     
    // loop through the string again
    for (let i = 0; i < s.length; i++) {
        let c = s[i];
        if (c in vowels) {
            continue;
        }
        // loop through the vowels
        for (let v in vowels) {
            if (vowels[v] == 0) {
                // replace the non-vowel with the next unused vowel
                s = s.replace(c, v);
                vowels[v] = 1;
                break;
            }
        }
    }
     
    return s;
}
 
let s = "ABCDEFGHI";
let modified_s = modifyString(s);
console.log(modified_s);
 
// This code is contributed by phasing17


C#




using System;
using System.Collections.Generic;
 
class Program
{
    static string ModifyString(string s)
    {
        Dictionary<char, int> vowels = new Dictionary<char, int>();
        vowels['A'] = 0;
        vowels['E'] = 0;
        vowels['I'] = 0;
        vowels['O'] = 0;
        vowels['U'] = 0;
 
        for (int i = 0; i < s.Length; i++)
        {
            char c = s[i];
            if (vowels.ContainsKey(c))
            {
                vowels++;
            }
        }
 
        for (int i = 0; i < s.Length; i++)
        {
            char c = s[i];
            if (!vowels.ContainsKey(c))
            {
                foreach (var kv in vowels)
                {
                    if (kv.Value == 0)
                    {
                        s = s.Remove(i, 1).Insert(i, kv.Key.ToString());
                        vowels[kv.Key] = 1;
                        break;
                    }
                }
            }
        }
 
        return s;
    }
 
    static void Main()
    {
        string s = "ABCDEFGHI";
        string modified_s = ModifyString(s);
        Console.WriteLine(modified_s);
    }
}


Output

AOUDEFGHI

 Time complexity: O(N) as we iterate through the string twice. 
 Auxiliary space: O(1) as we only use a fixed-size dictionary to store the vowel counts.



Last Updated : 07 Mar, 2023
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