给定一个字符串S 。任务是检查是否可以重新排列字符串,使得每个奇数长度的子串都是回文。
例子:
Input: S = “oiooi”
Output: YES
The string can be rearranged as “oioio”
Input: S = “yuyuo”
Output: NO
方法:
- 第一个观察是如果字符串的所有字符都相同,那么每个奇数长度的子串都是回文,我们不需要重新排列它们。
- 第二个观察结果是,如果不同字符的数量超过2 ,则无法重新排列。
- 现在,如果不同字符的数量正好是2,那么为了让所有奇数长度的子串成为回文,它们的计数差必须小于或等于1 ,如果这满足,那么我们以另一种方式重新排列字符串意味着
- 对于i <— ( 1 到 n – 1 ) 。其中n是字符串的长度。
下面是上述方法的实现:
C++
// C++ implementation of
// the above approach
#include
using namespace std;
// Function to check is it
// possible to rearrange the string
// such that every odd length
// substring is palindrome
bool IsPossible(string s)
{
// Length of the string
int n = s.length();
// To count number of distinct
// character in string
set count;
// To count frequency of
// each character
map map;
for (int i = 0; i < n; i++) {
// Inserting into set
count.insert(s[i]);
// Incrementing the frequency
map[s[i]] += 1;
}
// All characters in
// the string are same
if (count.size() == 1) {
return true;
}
// There are more than 2 different
// character in string
if (count.size() > 2) {
return false;
}
// Currently there is 2 different
// character in string
auto it = count.begin();
// Get the frequencies of the
// characters that present
// in string
int x = 0, y = 0;
x = map[*it];
it++;
y = map[*it];
// Difference between their
// count is less than or
// equal to 1
if (abs(x - y) <= 1) {
return true;
}
return false;
}
// Driver code
int main()
{
string s = "aaaddad";
if (IsPossible(s))
cout << "YES\n";
else
cout << "NO\n";
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to check is it
// possible to rearrange the string
// such that every odd length
// substring is palindrome
static boolean IsPossible(String s)
{
// Length of the string
int n = s.length();
// To count number of distinct
// character in string
HashSet count = new HashSet<>();
// To count frequency of
// each character
HashMap map = new HashMap<>();
for (int i = 0; i < n; i++)
{
// Inserting into set
count.add(s.charAt(i));
// Incrementing the frequency
map.put(s.charAt(i), map.get(s.charAt(i)) ==
null ? 1 : map.get(s.charAt(i)) + 1);
}
// All characters in
// the string are same
if (count.size() == 1)
return true;
// There are more than 2 different
// character in string
if (count.size() > 2)
return false;
String newString = count.toArray().toString();
// Currently there is 2 different
// character in string
int j = 0;
char it = newString.charAt(j);
// Get the frequencies of the
// characters that present
// in string
int x = 0, y = 0;
x = map.get(it) == null ? 0 : map.get(it);
j++;
it = newString.charAt(j);
y = map.get(it) == null ? 0 : map.get(it);
// Difference between their
// count is less than or
// equal to 1
if (Math.abs(x - y) <= 1)
return true;
return false;
}
// Driver Code
public static void main(String[] args)
{
String s = "aaaddad";
if (IsPossible(s))
System.out.println("YES");
else
System.out.println("NO");
}
}
// This code is contributed by
// sanjeev2552
Python3
# Python3 implementation of
# the above approach
# Function to check is it
# possible to rearrange the string
# such that every odd length
# substring is palindrome
def IsPossible(s) :
# Length of the string
n = len(s);
# To count number of distinct
# character in string
count = set();
# To count frequency of
# each character
map = dict.fromkeys(s, 0);
for i in range(n) :
# Inserting into set
count.add(s[i]);
# Incrementing the frequency
map[s[i]] += 1;
# All characters in
# the string are same
if (len(count) == 1) :
return True;
# There are more than 2 different
# character in string
if (len(count) > 2) :
return False;
# Currently there is 2 different
# character in string
j = 0
it = list(count)[j];
# Get the frequencies of the
# characters that present
# in string
x = 0; y = 0;
x = map[it];
j += 1
it = list(count)[j];
y = map[it];
# Difference between their
# count is less than or
# equal to 1
if (abs(x - y) <= 1) :
return True;
return False;
# Driver code
if __name__ == "__main__" :
s = "aaaddad";
if (IsPossible(s)) :
print("YES");
else :
print("NO");
# This code is contributed by AnkitRai01
C#
// C# implementation of the
// above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to check is it
// possible to rearrange the string
// such that every odd length
// substring is palindrome
static bool IsPossible(String s)
{
// Length of the string
int n = s.Length;
// To count number of distinct
// character in string
HashSet count =
new HashSet();
// To count frequency of
// each character
Dictionary map = new Dictionary();
for (int i = 0; i < n; i++)
{
// Inserting into set
count.Add(s[i]);
// Incrementing the frequency
if(map.ContainsKey(s[i]))
map[s[i]] = map[s[i]] + 1;
else
map.Add(s[i], 1);
}
// All characters in
// the string are same
if (count.Count == 1)
return true;
// There are more than 2 different
// character in string
if (count.Count > 2)
return false;
String newString = count.ToString();
// Currently there is 2 different
// character in string
int j = 0;
char it = newString[j];
// Get the frequencies of the
// characters that present
// in string
int x = 0, y = 0;
x = !map.ContainsKey(it) ?
0 : map[it];
j++;
it = newString[j];
y = !map.ContainsKey(it) ?
0 : map[it];
// Difference between their
// count is less than or
// equal to 1
if (Math.Abs(x - y) <= 1)
return true;
return false;
}
// Driver Code
public static void Main(String[] args)
{
String s = "aaaddad";
if (IsPossible(s))
Console.WriteLine("YES");
else
Console.WriteLine("NO");
}
}
// This code is contributed by 29AjayKumar
输出:
YES
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