Given string str of lower case alphabets, the task is to find the minimum cost to change the input string in a string that contains only vowels. Each consonant is changed to the nearest vowels. The cost is defined as the absolute difference between the ASCII value of consonant and vowel.
Examples:
Input: str = “abcde”
Output: 4
Explanation:
Here, a and e are already the vowels but b, c, and d are consonants. So b, c, and d are changed to nearest vowels as:
1. b –> a = |98 – 97| = 1,
2. c –> a = |99 – 97| = 2 or c –> e = |99 – 101| = 2 ( to minimum the cost),
3. d –> e = |100 – 101| = 1.
Therefore, the minimum cost is 1 + 2 + 1 = 4.Input: str = “aaa”
Output: 0
Explanation:
There is no consonant in the string.
Approach: The idea is to traverse over the string and replace each consonant with their nearest vowel and add the cost as the difference between the consonants and changed vowel. Print the cost after all operations.
Below is the implementation of the above approach:
// C++ program for the above approach #include<bits/stdc++.h> using namespace std;
// Function to find the minimum cost int min_cost(string st)
{ // Store vowels
string vow = "aeiou" ;
int cost = 0;
// Loop for iteration of string
for ( int i = 0; i < st.size(); i++)
{
vector< int > costs;
// Loop to calculate the cost
for ( int j = 0; j < 5; j++)
costs.push_back( abs (st[i] - vow[j]));
// Add minimum cost
cost += *min_element(costs.begin(),
costs.end());
}
return cost;
} // Driver Code int main()
{ // Given String
string str = "abcde" ;
// Function Call
cout << (min_cost(str));
} // This code is contributed by grand_master |
// Java program for the above approach import java.io.*;
import java.util.*;
class GFG{
// Function to find the minimum cost static int min_cost(String st)
{ // Store vowels
String vow = "aeiou" ;
int cost = 0 ;
// Loop for iteration of string
for ( int i = 0 ; i < st.length(); i++)
{
ArrayList<Integer> costs = new ArrayList<>();
// Loop to calculate the cost
for ( int j = 0 ; j < 5 ; j++)
costs.add(Math.abs(st.charAt(i) -
vow.charAt(j)));
// Add minimum cost
int minx = Integer.MAX_VALUE;
for ( int x : costs)
{
if (x < minx)
{
minx = x;
}
}
cost += minx;
}
return cost;
} // Driver Code public static void main(String[] args)
{ // Given string
String str = "abcde" ;
// Function call
System.out.println(min_cost(str));
} } // This code is contributed by rutvik_56 |
# Python3 program for above approach # Function to find the minimum cost def min_cost(st):
# Store vowels
vow = "aeiou"
cost = 0
# Loop for iteration of string
for i in range ( len (st)):
costs = []
# Loop to calculate the cost
for j in range ( 5 ):
costs.append( abs ( ord (st[i]) - ord (vow[j])))
# Add minimum cost
cost + = min (costs)
return cost
# Driver Code # Given String str = "abcde"
# Function Call print (min_cost( str ))
|
// C# program for the above approach using System;
using System.Collections.Generic;
class GFG{
// Function to find the minimum cost static int min_cost( string st)
{ // Store vowels
string vow = "aeiou" ;
int cost = 0;
// Loop for iteration of string
for ( int i = 0; i < st.Length; i++)
{
List< int > costs = new List< int >();
// Loop to calculate the cost
for ( int j = 0; j < 5; j++)
costs.Add(Math.Abs(st[i] -
vow[j]));
// Add minimum cost
int minx = Int32.MaxValue;
foreach ( int x in costs)
{
if (x < minx)
{
minx = x;
}
}
cost += minx;
}
return cost;
} // Driver Code public static void Main()
{ // Given string
string str = "abcde" ;
// Function call
Console.WriteLine(min_cost(str));
} } // This code is contributed by code_hunt |
<script> // JavaScript program for above approach // Function to find the minimum cost function min_cost(st){
// Store vowels
let vow = "aeiou"
let cost = 0
// Loop for iteration of string
for (let i=0;i<st.length;i++){
let costs = []
// Loop to calculate the cost
for (let j = 0; j < 5; j++)
costs.push(Math.abs(st.charCodeAt(i)-vow.charCodeAt(j)))
// Add minimum cost
cost += Math.min(...costs)
}
return cost
} // Driver Code // Given String let str = "abcde"
// Function Call document.write(min_cost(str), "</br>" )
// This code is contributed by shinjanpatra </script> |
4
Time Complexity: O(N), where N is the length of the string.
Auxiliary Space: O(1)