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Rearrange a string to maximize the minimum distance between any pair of vowels
  • Last Updated : 05 Feb, 2021
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Given a string str, the task is to rearrange the characters of the given string such that the minimum distance between any pair of vowels is maximum possible.

Examples:

Input: str = “aacbbc”
Output: abcbca
Explanation: Maximized distance between the only pair of vowels is 4.

Input: str = “aaaabbbcc”
Output: ababacbac

Approach: Follow the below steps to solve the problem:



  • Iterate over the characters of the string and count the number of vowels and consonants present in the string str, say Nv and Nc respectively.
  • Now, calculate the maximum number of consonants that can be put between every pair of vowels using the following formula:
     

M = rounded down[Nc / (Nv – 1)]

  •  
  • Now, construct the resultant string by placing all vowels and then inserting M consonants between each adjacent vowel of the pair.
  • After constructing the resultant string, if any consonant is still remaining, then simply insert them at the end of the string.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to rearrange the string
// such that the minimum distance
// between any of vowels is maximum.
string solution(string s)
{
 
    // Store vowels and consonants
    vector<char> vowel, consonant;
 
    // Iterate over the characters
    // of string
    for (auto i : s) {
 
        // If current character is a vowel
        if (i == 'a' || i == 'e'
            || i == 'i' || i == 'o'
            || i == 'u') {
 
            vowel.push_back(i);
        }
 
        // If current character is
        // a consonant
        else {
 
            consonant.push_back(i);
        }
    }
 
    // Stores count of vowels and
    // consonants respectively
    int Nc, Nv;
    Nv = vowel.size();
    Nc = consonant.size();
 
    int M = Nc / (Nv - 1);
 
    // Stores the resultant string
    string ans = "";
 
    // Stores count of consonants
    // appended into ans
    int consotnant_till = 0;
 
    for (auto i : vowel) {
 
        // Append vowel to ans
        ans += i;
        int temp = 0;
 
        // Append consonants
        for (int j = consotnant_till;
             j < min(Nc, consotnant_till + M);
             j++) {
 
            // Append consonant to ans
            ans += consonant[j];
 
            // Update temp
            temp++;
        }
 
        // Remove the taken
        // elements of consonant
        consotnant_till += temp;
    }
 
    // Return final ans
    return ans;
}
 
// Driver Code
int main()
{
    string str = "aaaabbbcc";
 
    // Function Call
    cout << solution(str);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
class GFG
{
 
// Function to rearrange the String
// such that the minimum distance
// between any of vowels is maximum.
static String solution(String s)
{
 
    // Store vowels and consonants
    Vector<Character> vowel = new Vector<Character>();
    Vector<Character> consonant = new Vector<Character>();
 
    // Iterate over the characters
    // of String
    for (char i : s.toCharArray())
    {
 
        // If current character is a vowel
        if (i == 'a' || i == 'e'
            || i == 'i' || i == 'o'
            || i == 'u')
        {
 
            vowel.add(i);
        }
 
        // If current character is
        // a consonant
        else
        {
 
            consonant.add(i);
        }
    }
 
    // Stores count of vowels and
    // consonants respectively
    int Nc, Nv;
    Nv = vowel.size();
    Nc = consonant.size();
    int M = Nc / (Nv - 1);
 
    // Stores the resultant String
    String ans = "";
 
    // Stores count of consonants
    // appended into ans
    int consotnant_till = 0;
    for (char i : vowel)
    {
 
        // Append vowel to ans
        ans += i;
        int temp = 0;
 
        // Append consonants
        for (int j = consotnant_till;
             j < Math.min(Nc, consotnant_till + M);
             j++) {
 
            // Append consonant to ans
            ans += consonant.get(j);
 
            // Update temp
            temp++;
        }
 
        // Remove the taken
        // elements of consonant
        consotnant_till += temp;
    }
 
    // Return final ans
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
    String str = "aaaabbbcc";
 
    // Function Call
    System.out.print(solution(str));
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python Program for the above approach
 
# Function to rearrange the string
# such that the minimum distance
# between any of vowels is maximum.
def solution(S):
   
    # store vowels and consonants
    vowels = []
    consonants = []
     
    # Iterate over the
    # characters of string
    for i in S:
       
        # if current character is a vowel
        if (i == 'a' or i == 'e' or i == 'i' or i == 'o' or i == 'u'):
            vowels.append(i)
             
        # if current character is consonant
        else:
            consonants.append(i)
             
    # store count of vowels and
    # consonants respectively
    Nc = len(consonants)
    Nv = len(vowels)
    M = Nc // (Nv - 1)
 
    # store the resultant string
    ans = ""
 
    # store count of consonants
    # append into ans
    consonant_till = 0
      
    for i in vowels:
        # Append vowel to ans
        ans += i
        temp = 0
 
        # Append consonants
        for j in range(consonant_till, min(Nc, consonant_till + M)):
           
            # Appendconsonant to ans
            ans += consonants[j]
             
            # update temp
            temp += 1
             
        # Remove the taken
        # elements of consonant
        consonant_till += temp
     
    # return final answer
    return ans
 
# Driver code
S = "aaaabbbcc"
print(solution(S))
 
# This code is contributed by Virusbuddah

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
 
// Function to rearrange the String
// such that the minimum distance
// between any of vowels is maximum.
static String solution(string s)
{
 
    // Store vowels and consonants
    List<char> vowel = new List<char>();
    List<char> consonant = new List<char>();
 
    // Iterate over the characters
    // of String
    foreach (char i in s.ToCharArray())
    {
 
        // If current character is a vowel
        if (i == 'a' || i == 'e'
            || i == 'i' || i == 'o'
            || i == 'u')
        {
 
            vowel.Add(i);
        }
 
        // If current character is
        // a consonant
        else
        {
 
            consonant.Add(i);
        }
    }
 
    // Stores count of vowels and
    // consonants respectively
    int Nc, Nv;
    Nv = vowel.Count;
    Nc = consonant.Count;
    int M = Nc / (Nv - 1);
 
    // Stores the resultant String
    string ans = "";
 
    // Stores count of consonants
    // appended into ans
    int consotnant_till = 0;
    foreach (char i in vowel)
    {
 
        // Append vowel to ans
        ans += i;
        int temp = 0;
 
        // Append consonants
        for (int j = consotnant_till;
             j < Math.Min(Nc, consotnant_till + M);
             j++) {
 
            // Append consonant to ans
            ans += consonant[j];
 
            // Update temp
            temp++;
        }
 
        // Remove the taken
        // elements of consonant
        consotnant_till += temp;
    }
 
    // Return final ans
    return ans;
}
 
// Driver Code
static public void Main()
{
    String str = "aaaabbbcc";
 
    // Function Call
    Console.WriteLine(solution(str));
    }
}
 
// This code is contributed by sanjoy_62.
Output: 
abababac

 

Time Complexity: O(N)
Auxiliary Space: O(N)

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