给定一个包含小写英文字母的字符串“str”，任务是找出字符串中字符的频率是否在卢卡斯序列中。您可以以任何方式自由排列频率数，以形成卢卡斯序列。如果可能，则打印YES否则打印NO 。
卢卡斯序列。
注意：必须使用所有频率来检查它们是否在 Lucas 序列中。
例子：
Input: str = “gggeek”
Output: YES
frequency of ‘g’ = 3
frequency of ‘e’ = 2
frequency of ‘k’ = 1
These frequency can be arranged to form first 3 terms of Lucas sequence, {2, 1, 3}.
Input: str = “geeksforgeeks”
Output: NO
方法：
- 使用 map STL 将每个字符的频率存储在向量中。之后对向量进行排序。
- 将第一个向量的第一个和第二个元素分别更改为 ‘2’ 和 ‘1’，因为 Lucas 序列将 ‘2’ 和 ‘1’ 作为前两个数字。但是，只有在向量中存在“1”和“2”时才必须进行更改。如果它不存在，则频率永远不会在 Lucas 序列中并输出 NO。
- 然后，制作另一个向量。设第一个向量的大小为 n。
- 在第二个向量中插入第一个“n”卢卡斯数字。
- 然后，比较两个向量中的每个元素。如果两个向量相同，则输出“YES”，否则输出“NO”。
下面是上述方法的实现：
C++
// C++ implementation of
// the approach
#include
using namespace std;
// function that checks
// if the frequencies
// are in Lucas sequence.
string lucas_sequence(string s, int n)
{
// map is used to store
// character frequencies
map m;
for (int i = 0; i < n; i++) {
if (m.find(s[i]) == m.end())
m[s[i]] = 1;
else
m[s[i]]++;
}
vector v1, v2;
map::iterator it;
// frequencies are extracted from
// map and stored in vector v1
for (it = m.begin(); it != m.end(); it++)
v1.push_back((*it).second);
// vector v1 elements are sorted,
// but first and second element are
// changed to '2' and '1' respectively,
// only if '1' and '2' are present in the vector.
sort(v1.begin(), v1.end());
if (v1[0] == 1 && v1[1] == 2) {
v1[0] = 2;
v1[1] = 1;
}
else
return "NO";
// a and b are first and
// second terms of
// Lucas sequence
int a = 2, b = 1;
int c;
v2.push_back(a);
v2.push_back(b);
// v2 contains Lucas sequence
for (int i = 0; i < v1.size() - 2; i++) {
v2.push_back(a + b);
c = a + b;
a = b;
b = c;
}
int flag = 1;
// both vectors are compared
for (int i = 0; i < v1.size(); i++) {
if (v1[i] != v2[i]) {
flag = 0;
break;
}
}
if (flag == 1)
return "YES";
else
return "NO";
}
// Driver code
int main()
{
string s = "oooeeeeqkk";
int n = s.length();
cout << lucas_sequence(s, n);
return 0;
} Java
// Java implementation of the approach
import java.util.Collections;
import java.util.HashMap;
import java.util.Vector;
class GFG
{
// function that checks
// if the frequencies
// are in Lucas sequence.
static String lucas_sequence(String s,
int n)
{
// map is used to store
// character frequencies
HashMap m = new HashMap<>();
for (int i = 0; i < n; i++)
m.put(s.charAt(i),
m.get(s.charAt(i)) == null ? 1 :
m.get(s.charAt(i)) + 1);
Vector v1 = new Vector<>();
Vector v2 = new Vector<>();
// frequencies are extracted from
// map and stored in vector v1
for (HashMap.Entry entry : m.entrySet())
v1.add(entry.getValue());
// vector v1 elements are sorted,
// but first and second element are
// changed to '2' and '1' respectively,
// only if '1' and '2' are present in the vector.
Collections.sort(v1);
if (v1.elementAt(0) == 1 &&
v1.elementAt(1) == 2)
{
v1.set(0, 2);
v1.set(1, 1);
}
else
return "NO";
// a and b are first and
// second terms of
// Lucas sequence
int a = 2, b = 1;
int c;
v2.add(a);
v2.add(b);
// v2 contains Lucas sequence
for (int i = 0; i < v1.size() - 2; i++)
{
v2.add(a + b);
c = a + b;
a = b;
b = c;
}
int flag = 1;
// both vectors are compared
for (int i = 0; i < v1.size(); i++)
{
if (v1.elementAt(i) != v2.elementAt(i))
{
flag = 0;
break;
}
}
if (flag == 1)
return "YES";
else
return "NO";
}
// Driver Code
public static void main(String[] args)
{
String s = "oooeeeeqkk";
int n = s.length();
System.out.println(lucas_sequence(s, n));
}
}
// This code is contributed by
// sanjeev2552 Python3
# Python3 implementation of the approach
from collections import defaultdict
# Function that checks if the
# frequencies are in Lucas sequence.
def lucas_sequence(s, n):
# map is used to store
# character frequencies
m = defaultdict(lambda:0)
for i in range(0, n):
m[s[i]] += 1
v1, v2 = [], []
# frequencies are extracted from
# map and stored in vector v1
for it in m:
v1.append(m[it])
# vector v1 elements are sorted, but
# first and second element are changed
# to '2' and '1' respectively, only if
# '1' and '2' are present in the vector.
v1.sort()
if v1[0] == 1 and v1[1] == 2:
v1[0], v1[1] = 2, 1
else:
return "NO"
# a and b are first and second terms
# of Lucas sequence
a, b = 2, 1
v2.append(a)
v2.append(b)
# v2 contains Lucas sequence
for i in range(0, len(v1) - 2):
v2.append(a + b)
a, b = b, a + b
flag = 1
# both vectors are compared
for i in range(0, len(v1)):
if v1[i] != v2[i]:
flag = 0
break
if flag == 1:
return "YES"
else:
return "NO"
# Driver code
if __name__ == "__main__":
s = "oooeeeeqkk"
n = len(s)
print(lucas_sequence(s, n))
# This code is contributed by Rituraj JainC#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// function that checks
// if the frequencies
// are in Lucas sequence.
static String lucas_sequence(String s,
int n)
{
// map is used to store
// character frequencies
Dictionary m = new Dictionary();
for (int i = 0; i < n; i++)
{
if(m.ContainsKey(s[i]))
{
m[s[i]] = m[s[i]] + 1;
}
else
{
m.Add(s[i], 1);
}
}
List v1 = new List();
List v2 = new List();
// frequencies are extracted from
// map and stored in vector v1
foreach(KeyValuePair entry in m)
v1.Add(entry.Value);
// vector v1 elements are sorted,
// but first and second element are
// changed to '2' and '1' respectively,
// only if '1' and '2' are present in the vector.
v1.Sort();
if (v1[0] == 1 &&
v1[1] == 2)
{
v1[0] = 2;
v1[1] = 1;
}
else
return "NO";
// a and b are first and
// second terms of
// Lucas sequence
int a = 2, b = 1;
int c;
v2.Add(a);
v2.Add(b);
// v2 contains Lucas sequence
for (int i = 0; i < v1.Count - 2; i++)
{
v2.Add(a + b);
c = a + b;
a = b;
b = c;
}
int flag = 1;
// both vectors are compared
for (int i = 0; i < v1.Count; i++)
{
if (v1[i] != v2[i])
{
flag = 0;
break;
}
}
if (flag == 1)
return "YES";
else
return "NO";
}
// Driver Code
public static void Main(String[] args)
{
String s = "oooeeeeqkk";
int n = s.Length;
Console.WriteLine(lucas_sequence(s, n));
}
}
// This code is contributed by Rajput-Ji 输出：
YES
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