给定长度为N 的字符串str ，任务是从字符串删除频率为素数的所有字符。
例子：
Input: str = “geeksforgeeks”
Output: eeforee
Explanation:
The frequencies of characters is: { g=2, e=4, k=2, s=2, f=1, o=1, r=1}
So, g, k and s are the characters with prime frequencies so, they are removed from the string.
The resultant string is “eeforee”
Input: str = “abcdef”
Output: abcdef
Explanation:
Since all the characters are unique with frequency as 1, so no character is removed.
朴素的方法：解决这个问题的最简单的方法是找到 字符串存在的每个不同字符的频率以及每个字符，检查其频率是否为Prime 。如果发现频率为素数，则跳过该字符。否则，将其附加到形成的新字符串。
时间复杂度： O( N 3/2 )
辅助空间： O(N)
高效的方法：可以通过使用 Eratosthenes 筛来预先计算素数来优化上述方法。请按照以下步骤解决问题：
- 使用Eratosthenes 筛分法，生成最大为N 的所有素数并将它们存储在数组prime[] 中。
- 初始化地图 并存储每个字符的频率。
- 然后，遍历字符串并借助 map 和 prime[] 数组找出哪些字符具有素数频率。
- 忽略所有具有主要频率的字符并将其余字符存储在一个新字符串。
- 完成上述步骤后，打印新的字符串。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to perform the seive of
// eratosthenes algorithm
void SieveOfEratosthenes(bool* prime, int n)
{
// Initialize all entries in
// prime[] as true
for (int i = 0; i <= n; i++) {
prime[i] = true;
}
// Initialize 0 and 1 as non prime
prime[0] = prime[1] = false;
// Traversing the prime array
for (int i = 2; i * i <= n; i++) {
// If i is prime
if (prime[i] == true) {
// All multiples of i must
// be marked false as they
// are non prime
for (int j = 2; i * j <= n; j++) {
prime[i * j] = false;
}
}
}
}
// Function to remove characters which
// have prime frequency in the string
void removePrimeFrequecies(string s)
{
// Length of the string
int n = s.length();
// Create a boolean array prime
bool prime[n + 1];
// Sieve of Eratosthenes
SieveOfEratosthenes(prime, n);
// Stores the frequency of character
unordered_map m;
// Storing the frequecies
for (int i = 0; i < s.length(); i++) {
m[s[i]]++;
}
// New string that will be formed
string new_string = "";
// Removing the characters which
// have prime frequencies
for (int i = 0; i < s.length(); i++) {
// If the character has
// prime frequency then skip
if (prime[m[s[i]]])
continue;
// Else concatenate the
// character to the new string
new_string += s[i];
}
// Print the modified string
cout << new_string;
}
// Driver Code
int main()
{
string str = "geeksforgeeks";
// Function Call
removePrimeFrequecies(str);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to perform the seive of
// eratosthenes algorithm
static void SieveOfEratosthenes(boolean[] prime,
int n)
{
// Initialize all entries in
// prime[] as true
for(int i = 0; i <= n; i++)
{
prime[i] = true;
}
// Initialize 0 and 1 as non prime
prime[0] = prime[1] = false;
// Traversing the prime array
for(int i = 2; i * i <= n; i++)
{
// If i is prime
if (prime[i] == true)
{
// All multiples of i must
// be marked false as they
// are non prime
for(int j = 2; i * j <= n; j++)
{
prime[i * j] = false;
}
}
}
}
// Function to remove characters which
// have prime frequency in the String
static void removePrimeFrequecies(char[] s)
{
// Length of the String
int n = s.length;
// Create a boolean array prime
boolean []prime = new boolean[n + 1];
// Sieve of Eratosthenes
SieveOfEratosthenes(prime, n);
// Stores the frequency of character
HashMap m = new HashMap<>();
// Storing the frequecies
for(int i = 0; i < s.length; i++)
{
if (m.containsKey(s[i]))
{
m.put(s[i], m.get(s[i]) + 1);
}
else
{
m.put(s[i], 1);
}
}
// New String that will be formed
String new_String = "";
// Removing the characters which
// have prime frequencies
for(int i = 0; i < s.length; i++)
{
// If the character has
// prime frequency then skip
if (prime[m.get(s[i])])
continue;
// Else concatenate the
// character to the new String
new_String += s[i];
}
// Print the modified String
System.out.print(new_String);
}
// Driver Code
public static void main(String[] args)
{
String str = "geeksforgeeks";
// Function Call
removePrimeFrequecies(str.toCharArray());
}
}
// This code is contributed by aashish1995 Python3
# Python3 program for the above approach
# Function to perform the seive of
# eratosthenes algorithm
def SieveOfEratosthenes(prime, n) :
# Initialize all entries in
# prime[] as true
for i in range(n + 1) :
prime[i] = True
# Initialize 0 and 1 as non prime
prime[0] = prime[1] = False
# Traversing the prime array
i = 2
while i*i <= n :
# If i is prime
if (prime[i] == True) :
# All multiples of i must
# be marked false as they
# are non prime
j = 2
while i*j <= n :
prime[i * j] = False
j += 1
i += 1
# Function to remove characters which
# have prime frequency in the String
def removePrimeFrequecies(s) :
# Length of the String
n = len(s)
# Create a bool array prime
prime = [False] * (n + 1)
# Sieve of Eratosthenes
SieveOfEratosthenes(prime, n)
# Stores the frequency of character
m = {}
# Storing the frequecies
for i in range(len(s)) :
if s[i] in m :
m[s[i]]+= 1
else :
m[s[i]] = 1
# New String that will be formed
new_String = ""
# Removing the characters which
# have prime frequencies
for i in range(len(s)) :
# If the character has
# prime frequency then skip
if (prime[m[s[i]]]) :
continue
# Else concatenate the
# character to the new String
new_String += s[i]
# Print the modified String
print(new_String, end = "")
Str = "geeksforgeeks"
# Function Call
removePrimeFrequecies(list(Str))
# This code is contributed by divyesh072019C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to perform the seive of
// eratosthenes algorithm
static void SieveOfEratosthenes(bool[] prime,
int n)
{
// Initialize all entries in
// prime[] as true
for(int i = 0; i <= n; i++)
{
prime[i] = true;
}
// Initialize 0 and 1 as non prime
prime[0] = prime[1] = false;
// Traversing the prime array
for(int i = 2; i * i <= n; i++)
{
// If i is prime
if (prime[i] == true)
{
// All multiples of i must
// be marked false as they
// are non prime
for(int j = 2; i * j <= n; j++)
{
prime[i * j] = false;
}
}
}
}
// Function to remove characters which
// have prime frequency in the String
static void removePrimeFrequecies(char[] s)
{
// Length of the String
int n = s.Length;
// Create a bool array prime
bool []prime = new bool[n + 1];
// Sieve of Eratosthenes
SieveOfEratosthenes(prime, n);
// Stores the frequency of character
Dictionary m = new Dictionary();
// Storing the frequecies
for(int i = 0; i < s.Length; i++)
{
if (m.ContainsKey(s[i]))
{
m[s[i]]++;
}
else
{
m.Add(s[i], 1);
}
}
// New String that will be formed
String new_String = "";
// Removing the characters which
// have prime frequencies
for(int i = 0; i < s.Length; i++)
{
// If the character has
// prime frequency then skip
if (prime[m[s[i]]])
continue;
// Else concatenate the
// character to the new String
new_String += s[i];
}
// Print the modified String
Console.Write(new_String);
}
// Driver Code
public static void Main(String[] args)
{
String str = "geeksforgeeks";
// Function Call
removePrimeFrequecies(str.ToCharArray());
}
}
// This code is contributed by aashish1995 输出：
eeforee时间复杂度： O(N*log (log N))
辅助空间： O(N)
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