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 Codeint 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 approachimport 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 Codepublic static void main(String args[]){ String str = "ABCDEFGHI"; addAllVowel(str);}}// This code is contributed by Arnab Kundu |
Python3
# Python3 implementation of the above approachs = "ABCDEFGHI"# converting String to a listS = list(s) # All vowelsx = ["A", "E", "I", "O", "U"] # List to store distinct vowelsy = [] 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 approachusing 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 Codepublic static void Main(string []args){ string str = "ABCDEFGHI"; addAllVowel(str);}}// This code is contributed by rutvik_56 |
Javascript
<script>// JavaScript implementation of the above approachfunction 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 Codelet 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 ss = "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 orderfunction 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.
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