Convert string S into a palindrome string. You can only replace a character with any other character. When you replace character ‘a’ with any other character, it costs 1 unit, similarly for ‘b’ it is 2 units ….. and for ‘z’, it is 26 units. Find the minimum cost required to convert string S into palindrome string.
Examples :
Input : abcdef Output : 6 Explanation: replace 'a', 'b' and 'c' => cost= 1 + 2 + 3 = 6 Input : aba Output : 0
The idea is to start comparing from the two ends of string. Let i be initialized as 0 index and j initialized as length – 1. If characters at two indices are not same, a cost will apply. To make the cost minimum replace the character which is smaller. Then increment i by 1 and decrement j by 1. Iterate till i less than j.
Implementation:
// CPP program to find minimum cost to make // a palindrome. #include <bits/stdc++.h> using namespace std;
// Function to return cost int cost(string str)
{ // length of string
int len = str.length();
// Iterate from both sides of string.
// If not equal, a cost will be there
int res = 0;
for ( int i=0, j=len-1; i < j; i++, j--)
if (str[i] != str[j])
res += min(str[i], str[j]) - 'a' + 1;
return res;
} // Driver code int main()
{ string str = "abcdef" ;
cout << cost(str) << endl;
return 0;
} |
// Java program to find minimum cost to make // a palindrome. import java.io.*;
class GFG
{ // Function to return cost
static int cost(String str)
{
// length of string
int len = str.length();
// Iterate from both sides of string.
// If not equal, a cost will be there
int res = 0 ;
for ( int i = 0 , j = len - 1 ; i < j; i++, j--)
if (str.charAt(i) != str.charAt(j))
res += Math.min(str.charAt(i), str.charAt(j))
- 'a' + 1 ;
return res;
}
// Driver code
public static void main (String[] args)
{
String str = "abcdef" ;
System.out.println(cost(str));
}
} // This code is contributed by vt_m. |
# python program to find minimum # cost to make a palindrome. # Function to return cost def cost(st):
# length of string
l = len (st)
# Iterate from both sides
# of string. If not equal,
# a cost will be there
res = 0
j = l - 1
i = 0
while (i < j):
if (st[i] ! = st[j]):
res + = ( min ( ord (st[i]),
ord (st[j])) -
ord ( 'a' ) + 1 )
i = i + 1
j = j - 1
return res
# Driver code st = "abcdef" ;
print (cost(st))
# This code is contributed by # Sam007 |
// C# program to find minimum cost // to make a palindrome. using System;
class GFG
{ // Function to return cost
static int cost(String str)
{
// length of string
int len = str.Length;
// Iterate from both sides of string.
// If not equal, a cost will be there
int res = 0;
for ( int i = 0, j = len - 1; i < j; i++, j--)
if (str[i] != str[j])
res += Math.Min(str[i], str[j])
- 'a' + 1;
return res;
}
// Driver code
public static void Main ()
{
string str = "abcdef" ;
Console.WriteLine(cost(str));
}
} // This code is contributed by vt_m. |
<?php // PHP program to find minimum // cost to make a palindrome. // Function to return cost function cost( $str )
{ // length of string
$len = strlen ( $str );
// Iterate from both sides
// of string. If not equal,
// a cost will be there
$res = 0;
for ( $i = 0, $j = $len - 1;
$i < $j ; $i ++, $j --)
if ( $str [ $i ] != $str [ $j ])
$res += (min(ord( $str [ $i ]),
ord( $str [ $j ])) -
ord( 'a' ) + 1 );
return $res ;
} // Driver code $str = "abcdef" ;
echo cost( $str );
// This code is contributed by Sam007 ?> |
<script> // Javascript program to find minimum cost
// to make a palindrome.
// Function to return cost
function cost(str)
{
// length of string
let len = str.length;
// Iterate from both sides of string.
// If not equal, a cost will be there
let res = 0;
for (let i = 0, j = len - 1; i < j; i++, j--)
{
if (str[i] != str[j])
{
res += Math.min(str[i].charCodeAt(), str[j].charCodeAt()) - 'a' .charCodeAt() + 1;
}
}
return res;
}
let str = "abcdef" ;
document.write(cost(str));
</script> |
6
Time Complexity : O(|S|) , where |S| is size of given string.
Space Complexity : O(1) , as we are not using any extra space.