Min. operations required so that no two vowels or consonants are adjacent

Given a string S of size N. In one operation you can replace any character of the string with any other character, the task is to find the minimum number of operations required so that no two vowels or consonants are adjacent.

Examples:

Input: N = 4, S = ‘wlrb’
Output: 2
Explanation: We can replace w and r with a.

Input: N = 3, S = ‘owf’
Output: 1
Explanation: We can replace f with o.

Approach: This can be solved with the following idea:

It is always optimal to keep vowels and consonent alternatively. So, keep checking whether at even and odd index consonent and vowel are present there. Keep a count of that.

Below are the steps involved:

• Iterate in string s:
  • If vowel, convert it to ‘1’.
  • If consonebt, convert it to ‘0’.
• First check:
  • If even index is having consonent, increment cnt1.
  • If odd index is having vowel, increment cnt1.
• Second check:
  • If even index is having vowel, increment cnt2.
  • If odd index is having consonent, increment cnt2.
• Return min(cnt1, cnt2).

Below is the implementation of the code:

C++

// C++ Implementation
#include <bits/stdc++.h>
#include <iostream>
using namespace std;

// Function to find whether it is vowel
// or consonent
int check(char crr)
{
    // If vowel
    if (crr == 'a' || crr == 'e' || crr == 'i' || crr == 'o'
        || crr == 'u')
        return 1;
    return 0;
}

// Function to find minimum operations required
int solve(int n, string s)
{

    // Iterate in string
    for (int i = 0; i < n; i++) {

        // Check for vowel and consonent
        if (check(s[i]) == 1)
            s[i] = '1';
        else
            s[i] = '0';
    }

    int cnt1 = 0;

    // Check if vowel at even indexes and
    // consonent at odd indexes
    for (int i = 0; i < n; i++) {

        if (i % 2 == 0 && s[i] != '1')
            cnt1++;
        if (i % 2 == 1 && s[i] != '0')
            cnt1++;
    }
    int cnt2 = 0;

    // Check if vowel at odd indexes and
    // consonent at even indexes
    for (int i = 0; i < n; i++) {

        if (i % 2 == 0 && s[i] != '0')
            cnt2++;
        if (i % 2 == 1 && s[i] != '1')
            cnt2++;
    }

    // Return the minimum number of operations
    return min(cnt1, cnt2);
}

// Driver code
int main()
{

    int n = 4;
    string s = "abcd";

    // Function call
    cout << solve(n, s);

    return 0;
}

Java

import java.util.Scanner;

public class Main {

    // Function to find whether it is vowel
    // or consonant
    static int check(char crr) {
        // If vowel
        if (crr == 'a' || crr == 'e' || crr == 'i' || crr == 'o' || crr == 'u')
            return 1;
        return 0;
    }

    // Function to find minimum operations required
    static int solve(int n, String s) {

        // Iterate in string
        for (int i = 0; i < n; i++) {

            // Check for vowel and consonant
            if (check(s.charAt(i)) == 1)
                s = s.substring(0, i) + '1' + s.substring(i + 1);
            else
                s = s.substring(0, i) + '0' + s.substring(i + 1);
        }

        int cnt1 = 0;

        // Check if vowel at even indexes and
        // consonant at odd indexes
        for (int i = 0; i < n; i++) {

            if (i % 2 == 0 && s.charAt(i) != '1')
                cnt1++;
            if (i % 2 == 1 && s.charAt(i) != '0')
                cnt1++;
        }
        int cnt2 = 0;

        // Check if vowel at odd indexes and
        // consonant at even indexes
        for (int i = 0; i < n; i++) {

            if (i % 2 == 0 && s.charAt(i) != '0')
                cnt2++;
            if (i % 2 == 1 && s.charAt(i) != '1')
                cnt2++;
        }

        // Return the minimum number of operations
        return Math.min(cnt1, cnt2);
    }

    // Driver code
    public static void main(String[] args) {

        int n = 4;
        String s = "abcd";

        // Function call
        System.out.println(solve(n, s));
    }
}

// This code is contributed by akshitaguprzj3

Python3

# Python Implementation:
def check(crr):
    # If vowel
    if crr == 'a' or crr == 'e' or crr == 'i' or crr == 'o' or crr == 'u':
        return 1
    return 0

def solve(n, s):
    # Convert vowels to '1' and consonants to '0'
    s = [str(check(c)) for c in s]

    cnt1 = 0
    cnt2 = 0

    # Check if vowel at even indexes and consonant at odd indexes
    for i in range(n):
        if i % 2 == 0 and s[i] != '1':
            cnt1 += 1
        if i % 2 == 1 and s[i] != '0':
            cnt1 += 1

    # Check if vowel at odd indexes and consonant at even indexes
    for i in range(n):
        if i % 2 == 0 and s[i] != '0':
            cnt2 += 1
        if i % 2 == 1 and s[i] != '1':
            cnt2 += 1

    # Return the minimum number of operations
    return min(cnt1, cnt2)

n = 4
s = "abcd"

# Function call
print(solve(n, s))

# This code is contributed by Sakshi

C#

using System;

class Program
{
    // Function to find whether it is a vowel or consonant
    static int Check(char crr)
    {
        // If vowel
        if (crr == 'a' || crr == 'e' || crr == 'i' || crr == 'o' || crr == 'u')
            return 1;
        return 0;
    }

    // Function to find minimum operations required
    static int Solve(int n, string s)
    {
        // Iterate in the string
        for (int i = 0; i < n; i++)
        {
            // Check for vowel and consonant
            if (Check(s[i]) == 1)
                s = s.Substring(0, i) + '1' + s.Substring(i + 1);
            else
                s = s.Substring(0, i) + '0' + s.Substring(i + 1);
        }

        int cnt1 = 0;

        // Check if vowel at even indexes and consonant at odd indexes
        for (int i = 0; i < n; i++)
        {
            if (i % 2 == 0 && s[i] != '1')
                cnt1++;
            if (i % 2 == 1 && s[i] != '0')
                cnt1++;
        }

        int cnt2 = 0;

        // Check if vowel at odd indexes and consonant at even indexes
        for (int i = 0; i < n; i++)
        {
            if (i % 2 == 0 && s[i] != '0')
                cnt2++;
            if (i % 2 == 1 && s[i] != '1')
                cnt2++;
        }

        // Return the minimum number of operations
        return Math.Min(cnt1, cnt2);
    }

    // Driver code
    static void Main()
    {
        int n = 4;
        string s = "abcd";

        // Function call
        Console.WriteLine(Solve(n, s));
    }
}

Javascript

function check(crr) {
    // If vowel
    if (crr === 'a' || crr === 'e' || crr === 'i' || crr === 'o' || crr === 'u') {
        return 1;
    }
    return 0;
}

function solve(n, s) {
    // Convert vowels to '1' and consonants to '0'
    s = s.split('').map(c => check(c).toString());

    let cnt1 = 0;
    let cnt2 = 0;

    // Check if vowel at even indexes and consonant at odd indexes
    for (let i = 0; i < n; i++) {
        if (i % 2 === 0 && s[i] !== '1') {
            cnt1 += 1;
        }
        if (i % 2 === 1 && s[i] !== '0') {
            cnt1 += 1;
        }
    }

    // Check if vowel at odd indexes and consonant at even indexes
    for (let i = 0; i < n; i++) {
        if (i % 2 === 0 && s[i] !== '0') {
            cnt2 += 1;
        }
        if (i % 2 === 1 && s[i] !== '1') {
            cnt2 += 1;
        }
    }

    // Return the minimum number of operations
    return Math.min(cnt1, cnt2);
}

const n = 4;
const s = "abcd";

// Function call
console.log(solve(n, s));

Output

1

Time Complexity: O(N)
Auxiliary Space: O(1)