# Minimum removal to make palindrome permutation in C++

## Problem statement

Given a string S, we have to find minimum characters that we can remove to make any permutation of the string S a palindrome

## Example

If str = “abcdba” then we remove to 1 character i.e. either ‘c’ or ‘d’.

## Algorithms

1. There can be two types of a palindrome, even length, and odd length palindromes
2. We can deduce the fact that an even length palindrome must have every character occurring even number of times
3.An odd palindrome must have every character occurring even number of times except one character occurring odd number of time
4. Check the frequency of every character and those characters occurring odd number of times are then counted. Then the result is total count of odd frequency characters’ minus 1

## Example

Live Demo

#include <bits/stdc++.h>
#define MAX 26
using namespace std;
int minCharactersRemoved(string str) {
int hash[MAX] = {0};
for (int i = 0; str[i]; ++i) {
hash[str[i] - 'a']++;
}
int cnt = 0;
for (int i = 0; i < MAX; ++i) {
if (hash[i] & 1) {
++cnt;
}
}
return (cnt == 0) ? 0 : (cnt - 1);
}
int main(){
string str = "abcdba";
cout << "Minimum characters to be removed = " <<
minCharactersRemoved(str) << endl;
return 0;
}

When you compile and execute above program. It generates following output

## Output

Minimum characters to be removed = 1

Updated on: 23-Dec-2019

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