# Minimum number of characters to be replaced to make a given string Palindrome

Given a string str, the task is to find the minimum number of characters to be replaced to make a given string palindrome. Replacing a character means replacing it with any single character at the same position. We are not allowed to remove or add any characters.

If there are multiple answers, print the lexicographically smallest string.

Examples:

```Input: str = "geeks"
Output: 2
geeks can be converted to geeeg to make it palindrome
by replacing minimum characters.

Input: str = "ameba"
Output: 1
We can get "abeba" or "amema" with only 1 change.
Among those two, "abeba" is lexicographically smallest.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Run a loop from 0 up to (length)/2-1 and check if a character at ith index i.e. s[i]!=s[length-i-1] then we will replace the alphabetically larger character with the one which is alphabetically smaller among them and continue the same process until all the elements gets traversed.

Below is the implementation of the above approach:

## C++

 `// C++ Implementation of the above approach  ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the minimum number  ` `// character change required  ` `  `  `void` `change(string s) ` `{ ` ` `  `    ``// Finding the length of the string  ` `    ``int` `n = s.length();  ` ` `  `    ``// To store the number of replacement operations  ` `    ``int` `cc = 0; ` ` `  `    ``for``(``int` `i=0;i

## Java

 `// Java Implementation of the above approach  ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to find the minimum number  ` `// character change required  ` `static` `void` `change(String s) ` `{ ` ` `  `    ``// Finding the length of the string  ` `    ``int` `n = s.length();  ` ` `  `    ``// To store the number of replacement operations  ` `    ``int` `cc = ``0``; ` ` `  `    ``for``(``int` `i = ``0``; i < n/``2``; i++) ` `    ``{ ` ` `  `        ``// If the characters at location  ` `        ``// i and n-i-1 are same then  ` `        ``// no change is required  ` `        ``if``(s.charAt(i) == s.charAt(n - i - ``1``))  ` `            ``continue``; ` ` `  `        ``// Counting one change operation  ` `        ``cc += ``1``; ` ` `  `        ``// Changing the character with higher  ` `        ``// ascii value with lower ascii value  ` `        ``if``(s.charAt(i) < s.charAt(n - i - ``1``))  ` `            ``s = s.replace(s.charAt(n - i - ``1``),s.charAt(i)); ` `        ``else` `            ``s = s.replace(s.charAt(n),s.charAt(n - i - ``1``)); ` `    ``} ` `    ``System.out.println(``"Minimum characters to be replaced = "``+(cc)) ; ` `    ``System.out.println(s);  ` `} ` ` `  `// Driver code  ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``String s = ``"geeks"``; ` `    ``change((s)); ` ` `  `} ` `} ` ` `  `// This code is contributed by ` `// Surendra_Gangwar `

## Python

 `# Python Implementation of the above approach ` ` `  `# Function to find the minimum number  ` `# character change required ` `import` `math as ma ` `def` `change(s): ` ` `  `    ``# Finding the length of the string ` `    ``n ``=` `len``(s) ` ` `  `    ``# To store the number of replacement operations ` `    ``cc ``=` `0` ` `  `    ``for` `i ``in` `range``(n``/``/``2``): ` ` `  `        ``# If the characters at location  ` `        ``# i and n-i-1 are same then  ` `        ``# no change is required ` `        ``if``(s[i]``=``=` `s[n``-``i``-``1``]): ` `            ``continue` ` `  `        ``# Counting one change operation ` `        ``cc``+``=` `1` ` `  `        ``# Changing the character with higher  ` `        ``# ascii value with lower ascii value ` `        ``if``(s[i]

## C#

 `// C# Implementation of the above approach  ` `using` `System;  ` `     `  `class` `GFG ` `{ ` ` `  `// Function to find the minimum number  ` `// character change required  ` `static` `void` `change(String s) ` `{ ` ` `  `    ``// Finding the length of the string  ` `    ``int` `n = s.Length;  ` ` `  `    ``// To store the number of  ` `    ``//replacement operations  ` `    ``int` `cc = 0; ` ` `  `    ``for``(``int` `i = 0; i < n / 2; i++) ` `    ``{ ` ` `  `        ``// If the characters at location  ` `        ``// i and n-i-1 are same then  ` `        ``// no change is required  ` `        ``if``(s[i] == s[n - i - 1])  ` `            ``continue``; ` ` `  `        ``// Counting one change operation  ` `        ``cc += 1; ` ` `  `        ``// Changing the character with higher  ` `        ``// ascii value with lower ascii value  ` `        ``if``(s[i] < s[n - i - 1])  ` `            ``s = s.Replace(s[n - i - 1], s[i]); ` `        ``else` `            ``s = s.Replace(s[n], s[n - i - 1]); ` `    ``} ` `    ``Console.WriteLine(``"Minimum characters "` `+  ` `                      ``"to be replaced = "` `+ (cc)); ` `    ``Console.WriteLine(s);  ` `} ` ` `  `// Driver code  ` `public` `static` `void` `Main(String []args) ` `{ ` `    ``String s = ``"geeks"``; ` `    ``change((s)); ` `} ` `} ` ` `  `// This code contributed by Rajput-Ji `

## PHP

 ` `

Output:

```Minimum characters to be replaced =  2
geeeg
```

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