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# Minimum insertions to form a palindrome with permutations allowed

• Difficulty Level : Easy
• Last Updated : 27 Dec, 2022

Given a string of lowercase letters. Find minimum characters to be inserted in the string so that it can become palindrome. We can change the positions of characters in the string.

Examples:

```Input : geeksforgeeks
Output : 2
geeksforgeeks can be changed as:
geeksroforskeeg
geeksorfroskeeg
and many more

Input : aabbc
Output : 0
aabbc can be changed as:
abcba
bacab```

Method 1: A palindromic string can have one odd character only when the length of the string is odd otherwise all characters occur an even number of times. So, we have to find characters that occur at odd times in a string.

The idea is to count the occurrence of each character in a string. As a palindromic string can have one character which occurs odd times, so the number of insertion will be one less than the count of characters that occur at odd times. And if the string is already palindrome, we do not need to add any character, so the result will be 0.

Implementation:

## C++

 `// CPP program to find minimum number``// of insertions to make a string``// palindrome``#include ``using` `namespace` `std;` `// Function will return number of``// characters to be added``int` `minInsertion(string str)``{``    ``// To store string length``    ``int` `n = str.length();` `    ``// To store number of characters``    ``// occurring odd number of times``    ``int` `res = 0;` `    ``// To store count of each``    ``// character``    ``int` `count = { 0 };` `    ``// To store occurrence of each``    ``// character``    ``for` `(``int` `i = 0; i < n; i++)``        ``count[str[i] - ``'a'``]++;` `    ``// To count characters with odd``    ``// occurrence``    ``for` `(``int` `i = 0; i < 26; i++)``        ``if` `(count[i] % 2 == 1)``            ``res++;` `    ``// As one character can be odd return``    ``// res - 1 but if string is already``    ``// palindrome return 0``    ``return` `(res == 0) ? 0 : res - 1;``}` `// Driver program``int` `main()``{``    ``string str = ``"geeksforgeeks"``;``    ``cout << minInsertion(str);` `    ``return` `0;``}`

## Java

 `// Java program to find minimum number``// of insertions to make a string``// palindrome``public` `class` `Palindrome {` `    ``// Function will return number of``    ``// characters to be added``    ``static` `int` `minInsertion(String str)``    ``{``        ``// To store string length``        ``int` `n = str.length();` `        ``// To store number of characters``        ``// occurring odd number of times``        ``int` `res = ``0``;` `        ``// To store count of each``        ``// character``        ``int``[] count = ``new` `int``[``26``];` `        ``// To store occurrence of each``        ``// character``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``count[str.charAt(i) - ``'a'``]++;` `        ``// To count characters with odd``        ``// occurrence``        ``for` `(``int` `i = ``0``; i < ``26``; i++) {``            ``if` `(count[i] % ``2` `== ``1``)``                ``res++;``        ``}` `        ``// As one character can be odd return``        ``// res - 1 but if string is already``        ``// palindrome return 0``        ``return` `(res == ``0``) ? ``0` `: res - ``1``;``    ``}` `    ``// Driver program``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String str = ``"geeksforgeeks"``;``        ``System.out.println(minInsertion(str));``    ``}``}`

## Python3

 `# Python3 program to find minimum number``# of insertions to make a string``# palindrome``import` `math as mt` `# Function will return number of``# characters to be added``def` `minInsertion(tr1):` `    ``# To store string length``    ``n ``=` `len``(str1)` `    ``# To store number of characters``    ``# occurring odd number of times``    ``res ``=` `0` `    ``# To store count of each``    ``# character``    ``count ``=` `[``0` `for` `i ``in` `range``(``26``)]` `    ``# To store occurrence of each``    ``# character``    ``for` `i ``in` `range``(n):``        ``count[``ord``(str1[i]) ``-` `ord``(``'a'``)] ``+``=` `1` `    ``# To count characters with odd``    ``# occurrence``    ``for` `i ``in` `range``(``26``):``        ``if` `(count[i] ``%` `2` `=``=` `1``):``            ``res ``+``=` `1` `    ``# As one character can be odd return``    ``# res - 1 but if string is already``    ``# palindrome return 0``    ``if` `(res ``=``=` `0``):``        ``return` `0``    ``else``:``        ``return` `res ``-` `1` `# Driver Code``str1 ``=` `"geeksforgeeks"``print``(minInsertion(str1))` `# This code is contributed by``# Mohit kumar 29`

## C#

 `// C# program to find minimum number``// of insertions to make a string``// palindrome``using` `System;` `public` `class` `GFG {` `    ``// Function will return number of``    ``// characters to be added``    ``static` `int` `minInsertion(String str)``    ``{``        ` `        ``// To store string length``        ``int` `n = str.Length;` `        ``// To store number of characters``        ``// occurring odd number of times``        ``int` `res = 0;` `        ``// To store count of each``        ``// character``        ``int``[] count = ``new` `int``;` `        ``// To store occurrence of each``        ``// character``        ``for` `(``int` `i = 0; i < n; i++)``            ``count[str[i] - ``'a'``]++;` `        ``// To count characters with odd``        ``// occurrence``        ``for` `(``int` `i = 0; i < 26; i++) {``            ``if` `(count[i] % 2 == 1)``                ``res++;``        ``}` `        ``// As one character can be odd``        ``// return res - 1 but if string``        ``// is already palindrome``        ``// return 0``        ``return` `(res == 0) ? 0 : res - 1;``    ``}` `    ``// Driver program``    ``public` `static` `void` `Main()``    ``{``        ``string` `str = ``"geeksforgeeks"``;``        ` `        ``Console.WriteLine(minInsertion(str));``    ``}``}` `// This code is contributed by vt_m.`

## PHP

 ``

## Javascript

 ``

Output

`2`

Time Complexity: O(n)
Auxiliary Space: O(1)

Method 2: An approach using Bit manipulation:

• Create a mask and initialise it to zero.
• For each character in string str, toggle the bit into the mask with its corresponding position in the alphabet.
• Check if mask is equal to zero, and return 0.
• Otherwise, return number of setbit in mask – 1.

Below is the implementation of the above approach:

## C++

 `// CPP program to find minimum number``// of insertions to make a string``// palindrome``#include ``using` `namespace` `std;` `// Function will return number of``// characters to be added``int` `minInsertion(string str)``{``    ``long` `long` `mask = 0;` `    ``for` `(``auto` `c : str)``        ``mask ^= (1 << (c - ``'a'``));` `    ``if` `(mask == 0)``        ``return` `0;``    ``int` `count = 0;` `    ``while` `(mask) {``        ``count += mask & 1;``        ``mask = mask >> 1;``    ``}` `    ``return` `count - 1;``}` `// Driver program``int` `main()``{``    ``string str = ``"geeksforgeeks"``;``    ``cout << minInsertion(str);` `    ``return` `0;``}` `// This code is contributed by hkdass001`

## Java

 `// Java program to find minimum number``// of insertions to make a string``// palindrome``import` `java.util.*;` `public` `class` `GFG {` `    ``// Function will return number of``    ``// characters to be added``    ``static` `int` `minInsertion(String str)``    ``{``        ``long` `mask = ``0``;` `        ``for` `(``char` `c : str.toCharArray())``            ``mask ^= (``1` `<< (c - ``'a'``));` `        ``if` `(mask == ``0``)``            ``return` `0``;``        ``int` `count = ``0``;` `        ``while` `(mask != ``0``) {``            ``count += mask & ``1``;``            ``mask = mask >> ``1``;``        ``}` `        ``return` `count - ``1``;``    ``}` `    ``// Driver program``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String str = ``"geeksforgeeks"``;``        ``System.out.println(minInsertion(str));``    ``}``}` `// This code is contributed by Karandeep1234`

Output

`2`

Time Complexity: O(n)
Auxiliary Space: O(1)

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