进行回文排列的最小去除
给定一个字符串S，我们必须找到可以删除的最少字符，以使字符串S 的任何排列成为回文。
简而言之，问题指出：通过删除最小数量的字符（包括尽可能删除 0 个字符）以任何方式重新排列字符串，从而使字符串成为回文。
注意：我们只考虑小字母。
例子 ：
Input : geeksforgeeks
Output : 2
Explanation : if we remove 2 characters lets
say 'f' and 'r', we remain with "geeksogeeks"
which can be re-arranged like "skeegogeeks"
to make it a palindrome. Removal of less than
2 character wouldn't make this string a
palindrome.
Input : shubham
Output : 4
If we remove any 4 characters except 'h' (let's
say 's', 'b', 'a', 'm'), we remain with "huh"
which is a palindrome.一种朴素的方法会检查字符串的每个排列是否有回文，如果没有找到，则删除一个字符并再次检查。这种方法非常复杂，需要很多时间。
一种有效的方法是注意到我们不需要打印最少的字符，只需要打印最小的数字。因此，一个有效的想法是关键：可以有两种类型的回文，偶数长度和奇数长度回文。我们可以推论出一个偶数长度的回文，每个字符必须出现偶数次（即每个字符出现的频率是偶数）。类似地，奇数回文必须使每个字符出现偶数次，除了一个字符出现奇数次。
从这些事实来看，问题其实很简单。我们检查每个字符的频率，然后计算出现奇数次的字符。然后结果是奇数频率字符减1的总数。
C++
// CPP Program to find minimum number of removal to
// make any permutation of the string a palindrome
#include
using namespace std;
#define MAX_CHAR 26
// function to find minimum removal of characters
int minRemoval(string str) {
// hash to store frequency of each character
int hash[MAX_CHAR];
// to set hash array to zeros
memset(hash, 0, sizeof(hash));
// count frequency of each character
for (int i = 0; str[i]; i++)
hash[str[i] - 'a']++;
// count the odd frequency characters
int count = 0;
for (int i = 0; i < MAX_CHAR; i++)
if (hash[i] % 2)
count++;
// if count is -1 return 0
// otherwise return count
return (count == 0) ? 0 : count-1;
}
// Driver's Code
int main() {
string str = "geeksforgeeks";
cout << minRemoval(str) << endl;
return 0;
} Java
// Java Program to find minimum number of removal to
// make any permutation of the string a palindrome
import java.util.Arrays;
class GFG {
static final int MAX_CHAR = 26;
// function to find minimum removal of characters
static int minRemoval(String str) {
// hash to store frequency of each character
int hash[] = new int[MAX_CHAR];
// to set hash array to zeros
Arrays.fill(hash, 0);
// count frequency of each character
for (int i = 0; i < str.length(); i++)
hash[str.charAt(i) - 'a']++;
// count the odd frequency characters
int count = 0;
for (int i = 0; i < MAX_CHAR; i++)
if (hash[i] % 2 == 1)
count++;
// if count is -1 return 0
// otherwise return count
return (count == 0) ? 0 : count - 1;
}
// Driver code
public static void main(String[] args) {
String str = "geeksforgeeks";
System.out.println(minRemoval(str));
}
}
// This code is contributed by Anant Agarwal.Python
# Python Program to find minimum number of
# removal to make any permutation of the
# string a palindrome
# function to find minimum removal of
# characters
def minRemoval(strr):
# hash to store frequency of each character
# to set hash array to zeros
hash = [0] * 26
# count frequency of each character
for char in strr:
hash[ord(char)-ord('a')] = hash[ord(char)-ord('a')] + 1
# count the odd frequency characters
count = 0
for i in range(26):
if hash[i]% 2:
count = count + 1
# if count is 0, return 0
# otherwise return count
return 0 if count == 0 else count-1
# Driver's Code
if __name__ == "__main__":
strr = "geeksforgeeks";
# minRemoval to find minimum characters to remove
print(minRemoval(strr))C#
// C# Program to find minimum number of
// removal to make any permutation of
// the string a palindrome
using System;
class GFG {
static int MAX_CHAR = 26;
// function to find minimum removal
// of characters
static int minRemoval(string str) {
// hash to store frequency of
// each character
int []hash = new int[MAX_CHAR];
// to set hash array to zeros
for(int i = 0; i < MAX_CHAR; i++)
hash[i] = 0;
// count frequency of each character
for (int i = 0; i < str.Length; i++)
hash[str[i] - 'a']++;
// count the odd frequency characters
int count = 0;
for (int i = 0; i < MAX_CHAR; i++)
if (hash[i] % 2 == 1)
count++;
// if count is -1 return 0
// otherwise return count
return (count == 0) ? 0 : count - 1;
}
// Driver code
public static void Main() {
string str = "geeksforgeeks";
Console.Write(minRemoval(str));
}
}
// This code is contributed by nitin mittalPHP
Javascript
输出 ：
2