Minimum insertions to form a palindrome with permutations allowed

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
Explanation: geeksforgeeks can be changed as: geeksroforskeeg or geeksorfroskeeg and many more using two insertion.

Input: aabbc
Output: 0
Explanation: aabbc can be changed as: abcba or bacab without any insertion

[Expected Approach] Character Frequency Counting – O(n) Time and O(1) Space

To make a string palindromic, count the occurrences of each character. A palindrome can have at most one character with an odd frequency; the rest must occur an even number of times. The minimum insertions required are one less than the count of characters with odd frequencies. If the string is already a palindrome, no insertions are needed (result = 0).

#include <bits/stdc++.h>
using namespace std;

int minInsertion(string &str)
{
    int n = str.length();
    int res = 0;

    vector<int> count(26,0);

    for (int i = 0; i < n; i++)
        count[str[i] - 'a']++;

    for (int i = 0; i < 26; i++)
        if (count[i] % 2 == 1)
            res++;

    return (res == 0) ? 0 : res - 1;
}

// Driver program
int main()
{
    string str = "geeksforgeeks";
    cout << minInsertion(str);

    return 0;
}

public class GfG {

    static int minInsertion(String str)
    {
        int n = str.length();

        int res = 0;

        int[] count = new int[26];

        for (int i = 0; i < n; i++)
            count[str.charAt(i) - 'a']++;

        for (int i = 0; i < 26; i++) {
            if (count[i] % 2 == 1)
                res++;
        }

        return (res == 0) ? 0 : res - 1;
    }

    // Driver program
    public static void main(String[] args)
    {
        String str = "geeksforgeeks";
        System.out.println(minInsertion(str));
    }
}

import math as mt

def minInsertion(tr1):

   
    n = len(str1)

    res = 0

    count = [0 for i in range(26)]

    for i in range(n):
        count[ord(str1[i]) - ord('a')] += 1

    for i in range(26):
        if (count[i] % 2 == 1):
            res += 1

    if (res == 0):
        return 0
    else:
        return res - 1

# Driver Code
str1 = "geeksforgeeks"
print(minInsertion(str1))

using System;

public class GfG {

    static int minInsertion(String str)
    {

        int n = str.Length;

        int res = 0;

        int[] count = new int[26];

        for (int i = 0; i < n; i++)
            count[str[i] - 'a']++;

        for (int i = 0; i < 26; i++) {
            if (count[i] % 2 == 1)
                res++;
        }

        return (res == 0) ? 0 : res - 1;
    }

    // Driver program
    public static void Main()
    {
        string str = "geeksforgeeks";

        Console.WriteLine(minInsertion(str));
    }
}

function minInsertion(str)
{
    let n = str.length;

    let res = 0;

    let count = new Array(26);
    for (let i = 0; i < count.length; i++) {
        count[i] = 0;
    }

    for (let i = 0; i < n; i++)
        count[str[i].charCodeAt(0) - "a".charCodeAt(0)]++;

    for (let i = 0; i < 26; i++) {
        if (count[i] % 2 == 1)
            res++;
    }

    return (res == 0) ? 0 : res - 1;
}

let str = "geeksforgeeks";
console.log(minInsertion(str));

2

[Alternate Approach] using Bit manipulation – O(n) Time and O(1) Space

Initialize a mask to 0 and toggle bits for each character based on its position in the alphabet. If the mask becomes 0, the string is already a palindrome (return 0). Otherwise, the minimum insertions required are the number of set bits in the mask minus one.

#include <bits/stdc++.h>
using namespace std;

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

import java.util.*;

public class GfG {

    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

def minInsertion(str):
    mask = 0

    for c in str:
        mask ^= (1 << (ord(c) - ord('a')))

    if mask == 0:
        return 0
    count = 0

    while mask:
        count += mask & 1
        mask = mask >> 1

    return count - 1

str = "geeksforgeeks"
print(minInsertion(str))

using System;
using System.Collections.Generic;

class GfG
{
  
  static int minInsertion(string str)
  {
    int mask = 0;

    foreach(char c in str)
      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
  static void Main(string[] args)
  {
    string str = "geeksforgeeks";
    Console.Write(minInsertion(str));
  }
}

function minInsertion(str) {
    let mask = 0;

    for (let c of str) {
        mask ^= (1 << (c.charCodeAt(0) - 'a'.charCodeAt(0)));
    }

    if (mask === 0) {
        return 0;
    }

    let count = 0;

    while (mask) {
        count += mask & 1;
        mask = mask >> 1;
    }

    // return the count minus 1
    return count - 1;
}

// Driver program
let str = "geeksforgeeks";
console.log(minInsertion(str));

2