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

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:

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# Python 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 ="")

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Output:

AOUDEFGHI


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