Minimum reduce operations to convert a given string into a palindrome

Given a String find the minimum number of reduce operations required to convert a given string into a palindrome. In a reduce operation, we can change character to a immediate lower value. For example b can be converted to a.

Examples :

Input  :  abcd  
Output :  4
We need to reduce c once
and d three times.

Input  : ccc
Output : 0

The idea is simple. We traverse string from left and compare characters of left half with their corresponding characters in right half. We add difference between to characters to result. 

Implementation:

C++

// CPP program to count minimum reduce
// operations to make a palindrome
#include <bits/stdc++.h>
using namespace std;

// Returns count of minimum character
// reduce operations to make palindrome.
int countReduce(string& str)
{
    int n = str.length();
    int res = 0;

    // Compare every character of first half
    // with the corresponding character of
    // second half and add difference to
    // result.
    for (int i = 0; i < n / 2; i++)
        res += abs(str[i] - str[n - i - 1]);

    return res;
}

// Driver code
int main()
{
    string str = "abcd";
    cout << countReduce(str);
    return 0;
}

Java

// Java program to count minimum reduce
// operations to make a palindrome
import java.io.*;

class GFG
{
    // Returns count of minimum character
    // reduce operations to make palindrome.
    static int countReduce(String str)
    {
        int n = str.length();
        int res = 0;

        // Compare every character of first half
        // with the corresponding character of
        // second half and add difference to
        // result.
        for (int i = 0; i < n / 2; i++)
            res += Math.abs(str.charAt(i)
                   - str.charAt(n - i - 1));

        return res;
    }

    // Driver code
    public static void main (String[] args)
    {
        String str = "abcd";
        System.out.println( countReduce(str));

    }
}

// This code is contributed by vt_m.

Python3

# python3 program to count minimum reduce
# operations to make a palindrome

# Returns count of minimum character
# reduce operations to make palindrome.
def countReduce(str):

    n = len(str)
    res = 0

    # Compare every character of first half
    # with the corresponding character of
    # second half and add difference to
    # result.
    for i in range(0, int(n/2)):
        res += abs( int(ord(str[i])) -
               int(ord(str[n - i - 1])) )

    return res

# Driver code
str = "abcd"
print(countReduce(str))

# This code is contributed by Sam007

C#

// C# program to count minimum reduce
// operations to make a palindrome
using System;

class GFG {

    // Returns count of minimum character
    // reduce operations to make palindrome.
    static int countReduce(string str)
    {
        int n = str.Length;
        int res = 0;

        // Compare every character of first
        // half with the corresponding
        // character of second half and
        // add difference to result.
        for (int i = 0; i < n / 2; i++)
            res += Math.Abs(str[i]
                    - str[n - i - 1]);

        return res;
    }

    // Driver code
    public static void Main ()
    {
        string str = "abcd";
        Console.WriteLine( countReduce(str));

    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP program to count minimum
// reduce operations to make a
// palindrome

// Returns count of minimum
// character reduce operations
// to make palindrome.
function countReduce($str)
{
    $n = strlen($str);
    $res = 0;

    // Compare every character
    // of first half with the
    // corresponding character
    // of second half and add
    // difference to result.
    for ($i = 0; $i < $n / 2; $i++)
        $res += abs(ord($str[$i]) -
                    ord($str[($n - $i - 1)]));
    return $res;
}

// Driver code
$str = "abcd";
echo countReduce($str);

// This code is contributed by Sam007
?>

Javascript

<script>
    // Javascript program to count minimum reduce
    // operations to make a palindrome

    // Returns count of minimum character
    // reduce operations to make palindrome.
    function countReduce(str)
    {
        let n = str.length;
        let res = 0;

        // Compare every character of first
        // half with the corresponding
        // character of second half and
        // add difference to result.
        for (let i = 0; i < parseInt(n / 2, 10); i++)
            res += Math.abs(str[i].charCodeAt() - str[n - i - 1].charCodeAt());

        return res;
    }

      let str = "abcd";
      document.write(countReduce(str));

</script>

Output

4

Time Complexity: O(n) where n is the length of the string
Auxiliary Space: O(1)