Given a string str[] of lower-case characters, the task is to make all characters of the string equal in the minimum number of operations such that in each operation either choose a vowel and change it to a consonant or vice-versa.
Examples:
Input: str[] = “neveropen”
Output: 10
Explanation: To make all the characters equal, make the following changes –
- Change ‘o’ to a consonant(let’s say) ‘z’ and then to ‘e’
- Change every other consonant(‘g’, ‘k’, ‘s’, ‘ f’, ‘r’, ) to ‘e’
This results in the string str = “eeeeeeeeeeeee” and the total number of operations performed is 10.
Input: str[] = “coding”
Output: 6
Approach: It can be deduced from the problem statement that in order to change a consonant to a vowel or vice versa, 1 operation is required. In order to change a vowel to a vowel or a consonant to a consonant, 2 operations are required. Because for changing vowel to vowel, first we need to change vowel -> consonant and then consonant -> vowel. Similarly, for changing consonant to consonant, first we need to change consonant -> vowel and then vowel -> consonant . The idea would be to maintain two variables in parallel :-
- Cv = the cost of changing all the characters to the maximum occurring vowel = no. of consonants + ( total number of vowels – frequency of maximum occurring vowel ) * 2
- Cc = the cost of changing all the characters to the maximum occurring consonant = no. of vowels + (total number of consonants – frequency of maximum occurring consonant) * 2
Now the minimum of these 2 i.e., min(Cv, Cc) will give the required minimum number of steps in which we can transform the string. Follow the steps below to solve the problem:
- Initialize the variable ans, vowels and consonants as 0 to store the answer, number of vowels and the number of consonants.
- Initialize 2 variables max_vowels and max_consonants as INT_MIN to find the maximum occurring vowel and maximum occurring consonant in the given string.
- Initialize 2 unordered_map<char, int> freq_vowels[] and freq_consonant[] to store the frequency of vowels and consonants.
- Iterate over the range [0, N) using the variable i and perform the following steps:
- If the current character is a vowel, then increase the count of vowels by 1 and it’s frequency in the map by 1 otherwise do it for the consonant.
- Traverse both the unordered_maps and find the maximum occurring vowel and consonant.
- Using the above formula, calculate the ans.
- After performing the above steps, print the value of ans as the answer.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the minimum number// of steps to make all characters equalint operations(string s){ // Initializing the variables int ans = 0; int vowels = 0, consonants = 0; int max_vowels = INT_MIN; int max_consonants = INT_MIN; // Store the frequency unordered_map<char, int> freq_consonants; unordered_map<char, int> freq_vowels; // Iterate over the string for (int i = 0; i < s.size(); i++) { if (s[i] == 'a' or s[i] == 'e' or s[i] == 'i' or s[i] == 'o' or s[i] == 'u') { // Calculate the total // number of vowels vowels += 1; // Storing frequency of // each vowel freq_vowels[s[i]] += 1; } else { // Count the consonants consonants += 1; // Storing the frequency of // each consonant freq_consonants[s[i]] += 1; } } // Iterate over the 2 maps for (auto it = freq_consonants.begin(); it != freq_consonants.end(); it++) { // Maximum frequency of consonant max_consonants = max( max_consonants, it->second); } for (auto it = freq_vowels.begin(); it != freq_vowels.end(); it++) { // Maximum frequency of vowel max_vowels = max(max_vowels, it->second); } // Find the result ans = min((2 * (vowels - max_vowels) + consonants), (2 * (consonants - max_vowels) + consonants)); return ans;}// Driver Codeint main(){ string S = "neveropen"; cout << operations(S); return 0;} |
Java
// Java program to implement the above approachimport java.io.*;import java.util.*;class GFG { // Function to find the minimum number of steps to make // all characters equal static int operations(String s) { // Initializign the variables int ans = 0; int vowels = 0, consonants = 0; int max_vowels = Integer.MIN_VALUE; int max_consonants = Integer.MIN_VALUE; // Store the frequency HashMap<Character, Integer> freq_consonants = new HashMap<>(); HashMap<Character, Integer> freq_vowels = new HashMap<>(); // Iterate over the string for (int i = 0; i < s.length(); i++) { if (s.charAt(i) == 'a' || s.charAt(i) == 'e' || s.charAt(i) == 'i' || s.charAt(i) == 'o' || s.charAt(i) == 'u') { // Calculate the total // number of vowels vowels += 1; // Storing frequency of // each vowel freq_vowels.put( s.charAt(i), freq_vowels.getOrDefault(s.charAt(i), 0) + 1); } else { // Count the consonants consonants += 1; // Storing the frequency of // each consonant freq_consonants.put( s.charAt(i), freq_consonants.getOrDefault( s.charAt(i), 0) + 1); } } // Iterate over the 2 maps for (Map.Entry<Character, Integer> it : freq_consonants.entrySet()) { // Maximum frequency of consonant max_consonants = Math.max(max_consonants, it.getValue()); } for (Map.Entry<Character, Integer> it : freq_vowels.entrySet()) { // Maximum frequency of vowel max_vowels = Math.max(max_vowels, it.getValue()); } // Find the result ans = Math.min( (2 * (vowels - max_vowels) + consonants), (2 * (consonants - max_vowels) + consonants)); return ans; } public static void main(String[] args) { String S = "neveropen"; System.out.print(operations(S)); }}// This code is contributed by lokeshmvs21. |
Python3
# Python 3 program for the above approachimport sysfrom collections import defaultdict# Function to find the minimum number# of steps to make all characters equaldef operations(s): # Initializing the variables ans = 0 vowels = 0 consonants = 0 max_vowels = -sys.maxsize-1 max_consonants = -sys.maxsize-1 # Store the frequency freq_consonants = defaultdict(int) freq_vowels = defaultdict(int) # Iterate over the string for i in range(len(s)): if (s[i] == 'a' or s[i] == 'e' or s[i] == 'i' or s[i] == 'o' or s[i] == 'u'): # Calculate the total # number of vowels vowels += 1 # Storing frequency of # each vowel freq_vowels[s[i]] += 1 else: # Count the consonants consonants += 1 # Storing the frequency of # each consonant freq_consonants[s[i]] += 1 # Iterate over the 2 maps for it in freq_consonants: # Maximum frequency of consonant max_consonants = max( max_consonants, freq_consonants[it]) for it in freq_vowels: # Maximum frequency of vowel max_vowels = max(max_vowels, freq_vowels[it]) # Find the result ans = min((2 * (vowels - max_vowels) + consonants), (2 * (consonants - max_vowels) + consonants)) return ans# Driver Codeif __name__ == "__main__": S = "neveropen" print(operations(S)) # This code is contributed by ukasp. |
Javascript
<script> // JavaScript Program to implement // the above approach // Function to find the minimum number // of steps to make all characters equal function operations(s) { // Initializing the variables let ans = 0; let vowels = 0, consonants = 0; let max_vowels = Number.MIN_VALUE; let max_consonants = Number.MIN_VALUE; // Store the frequency let freq_consonants = new Map(); let freq_vowels = new Map(); // Iterate over the string for (let i = 0; i < s.length; i++) { if (s.charAt(i) == 'a' || s.charAt(i) == 'e' || s.charAt(i) == 'i' || s.charAt(i) == 'o' || s.charAt(i) == 'u') { // Calculate the total // number of vowels vowels += 1; // Storing frequency of // each vowel if (freq_vowels.has(s.charAt(i))) { freq_vowels.set(s.charAt(i), freq_vowels.get(s.charAt(i)) + 1) } else { freq_vowels.set(s.charAt(i), 1); } } else { // Count the consonants consonants += 1; // Storing the frequency of // each consonant if (freq_consonants.has(s.charAt(i))) { freq_consonants.set(s.charAt(i), freq_consonants.get(s.charAt(i)) + 1) } else { freq_consonants.set(s.charAt(i), 1); } } } // Iterate over the 2 maps for (let [key, value] of freq_consonants) { // Maximum frequency of consonant max_consonants = Math.max( max_consonants, value); } for (let [key1, value1] of freq_vowels) { // Maximum frequency of vowel max_vowels = Math.max(max_vowels, value1); } // Find the result ans = Math.min((2 * (vowels - max_vowels) + consonants), (2 * (consonants - max_vowels) + consonants)); return ans; } // Driver Code let S = "neveropen"; document.write(operations(S)); // This code is contributed by Potta Lokesh </script> |
C#
using System;using System.Collections.Generic;class Program { // Function to find the minimum number // of steps to make all characters equal static int Operations(string s) { // Initializing the variables int ans = 0; int vowels = 0, consonants = 0; int maxVowels = int.MinValue; int maxConsonants = int.MinValue; // Store the frequency var freqConsonants = new Dictionary<char, int>(); var freqVowels = new Dictionary<char, int>(); // Iterate over the string for (int i = 0; i < s.Length; i++) { if (s[i] == 'a' || s[i] == 'e' || s[i] == 'i' || s[i] == 'o' || s[i] == 'u') { // Calculate the total // number of vowels vowels += 1; // Storing frequency of // each vowel if (!freqVowels.ContainsKey(s[i])) { freqVowels[s[i]] = 1; } else { freqVowels[s[i]] += 1; } } else { // Count the consonants consonants += 1; // Storing the frequency of // each consonant if (!freqConsonants.ContainsKey(s[i])) { freqConsonants[s[i]] = 1; } else { freqConsonants[s[i]] += 1; } } } // Iterate over the 2 dictionaries foreach(var pair in freqConsonants) { // Maximum frequency of consonant maxConsonants = Math.Max(maxConsonants, pair.Value); } foreach(var pair in freqVowels) { // Maximum frequency of vowel maxVowels = Math.Max(maxVowels, pair.Value); } // Find the result ans = Math.Min( (2 * (vowels - maxVowels) + consonants), (2 * (consonants - maxVowels) + consonants)); return ans; } // Driver Code static void Main() { string S = "neveropen"; Console.WriteLine(Operations(S)); }} |
Output:
10
Time Complexity: O(N)
Auxiliary Space: O(1) as it is using constant space for variables and map to store the frequency of characters
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