Given a string S, the task is to find the minimum cost to convert all the vowels {a, e, i, o, u} in the given string to any one of them. The cost of converting a vowel is given by the change in ASCII value.
Examples:
Input: S = “neveropen”
Output: 10
Explanation:
Count of e’s = 4
Count of o’s = 1
Conversion from ‘o’ to ‘e’ costs abs(‘o’ – ‘e’) = 10.
Hence, the output is 10.
Input: S = “coding”
Output: 6
Explanation:
Conversion from ‘o’ to ‘i’ costs abs(‘o’ – ‘i’) = 6.
Hence, the output is 10.
Approach:
The idea is to calculate the separate cost for converting every vowel in the string to one of the vowels {a, e, i, o, u} and choose the vowel which gives the minimum cost. Follow the steps below:
- Initialize 5 variables cA, cE, cI, cO, and cU with a cost of 0 which stores the cost of vowels {a, e, i, o, u} respectively.
- Iterate through the string, and for each vowel find the cost to convert it into all the other vowels and store it in the variables cA, cE, cI, cO, and cU respectively.
- The minimum cost to convert all the vowels into a single one is given by min(cA, cE, cI, cO, cU).
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function that return true if the // given character is a vowel bool isVowel(char ch) { if (ch == 'a' or ch == 'e' or ch == 'i' or ch == 'o' or ch == 'u') return true; else return false; } // Function to return the minimum // cost to convert all the vowels // of a string to a single one int minCost(string S) { // Stores count of // respective vowels int cA = 0; int cE = 0; int cI = 0; int cO = 0; int cU = 0; // Iterate through the string for (int i = 0; i < S.size(); i++) { // If a vowel is encountered if (isVowel(S[i])) { // Calculate the cost cA += abs(S[i] - 'a'); cE += abs(S[i] - 'e'); cI += abs(S[i] - 'i'); cO += abs(S[i] - 'o'); cU += abs(S[i] - 'u'); } } // Return the minimum cost return min(min(min(min(cA, cE), cI), cO), cU); } // Driver Code int main() { string S = "neveropen"; cout << minCost(S) << endl; return 0; } |
Java
// Java program for the above approach import java.util.*;class GFG{// Function that return true if the// given character is a vowelstatic boolean isVowel(char ch){ if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u') return true; else return false;}// Function to return the minimum// cost to convert all the vowels// of a string to a single onestatic int minCost(String S){ // Stores count of // respective vowels int cA = 0; int cE = 0; int cI = 0; int cO = 0; int cU = 0; // Iterate through the string for(int i = 0; i < S.length(); i++) { // If a vowel is encountered if (isVowel(S.charAt(i))) { // Calculate the cost cA += Math.abs(S.charAt(i) - 'a'); cE += Math.abs(S.charAt(i) - 'e'); cI += Math.abs(S.charAt(i) - 'i'); cO += Math.abs(S.charAt(i) - 'o'); cU += Math.abs(S.charAt(i) - 'u'); } } // Return the minimum cost return Math.min(Math.min(Math.min(Math.min(cA, cE), cI), cO), cU);}// Driver codepublic static void main(String[] args){ String S = "neveropen"; System.out.println(minCost(S));}}// This code is contributed by offbeat |
Python3
# Python3 program for the above approach # Function that return true if the # given character is a vowel def isVowel(ch): if (ch == 'a' or ch == 'e' or ch == 'i' or ch == 'o' or ch == 'u'): return True; else: return False; # Function to return the minimum # cost to convert all the vowels # of a string to a single one def minCost(S): # Stores count of # respective vowels cA = 0; cE = 0; cI = 0; cO = 0; cU = 0; # Iterate through the string for i in range(len(S)): # If a vowel is encountered if (isVowel(S[i])): # Calculate the cost cA += abs(ord(S[i]) - ord('a')); cE += abs(ord(S[i]) - ord('e')); cI += abs(ord(S[i]) - ord('i')); cO += abs(ord(S[i]) - ord('o')); cU += abs(ord(S[i]) - ord('u')); # Return the minimum cost return min(min(min(min(cA, cE), cI), cO), cU); # Driver codeS = "neveropen"; print(minCost(S)) # This code is contributed by grand_master |
C#
// C# program for the above approach using System;class GFG{// Function that return true if the// given character is a vowelstatic bool isVowel(char ch){ if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u') return true; else return false;}// Function to return the minimum// cost to convert all the vowels// of a string to a single onestatic int minCost(String S){ // Stores count of // respective vowels int cA = 0; int cE = 0; int cI = 0; int cO = 0; int cU = 0; // Iterate through the string for(int i = 0; i < S.Length; i++) { // If a vowel is encountered if (isVowel(S[i])) { // Calculate the cost cA += Math.Abs(S[i] - 'a'); cE += Math.Abs(S[i] - 'e'); cI += Math.Abs(S[i] - 'i'); cO += Math.Abs(S[i] - 'o'); cU += Math.Abs(S[i] - 'u'); } } // Return the minimum cost return Math.Min(Math.Min( Math.Min(Math.Min(cA, cE), cI), cO), cU);}// Driver codepublic static void Main(String[] args){ String S = "neveropen"; Console.WriteLine(minCost(S));}}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript program f|| the above approach // Function that return true if the // given character is a vowel function isVowel(ch) { if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u') return true; else return false; } // Function to return the minimum // cost to convert all the vowels // of a string to a single one function minCost(S) { // St||es count of // respective vowels var cA = 0; var cE = 0; var cI = 0; var cO = 0; var cU = 0; // Iterate through the string for(var i = 0; i < S.length; i++) { // If a vowel is encountered if (isVowel(S[i])) { // Calculate the cost cA += Math.abs(S.charCodeAt(i) - 'a'.charCodeAt(0)); cE += Math.abs(S.charCodeAt(i) - 'e'.charCodeAt(0)); cI += Math.abs(S.charCodeAt(i) - 'i'.charCodeAt(0)); cO += Math.abs(S.charCodeAt(i) - 'o'.charCodeAt(0)); cU += Math.abs(S.charCodeAt(i) - 'u'.charCodeAt(0)); } } // Return the Math.minimum cost return Math.min(Math.min(Math.min(Math.min(cA, cE), cI), cO), cU); } // Driver Code var S = "neveropen"; document.write(minCost(S)); // This code is contributed by importantly.</script> |
Output:
10
Time Complexity: O(N)
Auxiliary Space: O(1)
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