给定一个由N个字符串组成的数组arr [] ，任务是在修改每个字符串，通过删除所有非2的幂次幂的字符，以升序对数组进行排序，然后以降序对修改后的字符串进行排序。
例子：
Input: arr[] = {“aaacbb”, “geeks”, “aaa”}
Output: cbb skgee
Explanation:
Following are the modified strings in the array arr[]:
- For the string “aaacbb”: The frequency of a is not the power of 2. Therefore, removing ‘a’ modifies the string to “cbb”. Now, sorting the string “cbb” in increasing order of frequency modifies the string to “cbb”.
- For the string “aaacbb”: The frequency of every character is a power of 2. Now, sorting the string “geeks” in increasing order of frequency modifies the string to “skgee”.
- For the string “aaacbb”: The frequency of a is not the power of 2. Therefore, removing ‘a’ modifies the string to “”.
Therefore, sorting the above strings in increasing order gives {“cbb”, “skgee”}.
Input: S[] = {“c”, “a”, “b”}
Output: a b c
方法：可以通过使用哈希存储每个字符串的所有字符的频率，然后执行给定的操作来解决给定的问题。请按照以下步骤解决问题：
- 遍历给定的字符串数组arr [] ，并对每个字符串执行以下操作：
- 将每个字符的频率存储在地图中。
- 创建一个空字符串，例如T ，以存储修改后的字符串。
- 现在遍历该地图，并将频率为2的幂的字符附加到字符串T上。
- 对字符串T进行升序排序，然后将其添加到字符串res []的数组中。
- 对数组res []进行升序排序。
- 完成上述步骤后，打印在阵列RES []作为结果的字符串。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to check if N is power of
// 2 or not
bool isPowerOfTwo(int n)
{
// Base Case
if (n == 0)
return false;
// Return true if N is power of 2
return (ceil(log2(n))
== floor(log2(n)));
}
// Function to print array of strings
// in ascending order
void printArray(vector res)
{
// Sort strings in ascending order
sort(res.begin(), res.end());
// Print the array
for (int i = 0; i < res.size(); i++) {
cout << res[i] << " ";
}
}
// Function to sort the strings after
// modifying each string according to
// the given conditions
void sortedStrings(string S[], int N)
{
// Store the frequency of each
// characters of the string
unordered_map freq;
// Stores the required
// array of strings
vector res;
// Traverse the array of strings
for (int i = 0; i < N; i++) {
// Temporary string
string st = "";
// Stores frequency of each
// alphabet of the string
for (int j = 0;
j < S[i].size(); j++) {
// Update frequency of S[i][j]
freq[S[i][j]]++;
}
// Traverse the map freq
for (auto i : freq) {
// Check if the frequency
// of i.first is a power of 2
if (isPowerOfTwo(i.second)) {
// Update string st
for (int j = 0;
j < i.second; j++) {
st += i.first;
}
}
}
// Clear the map
freq.clear();
// Null string
if (st.size() == 0)
continue;
// Sort the string in
// descending order
sort(st.begin(), st.end(),
greater());
// Update res
res.push_back(st);
}
// Print the array of strings
printArray(res);
}
// Driver Code
int main()
{
string arr[] = { "aaacbb", "geeks", "aaa" };
int N = sizeof(arr) / sizeof(arr[0]);
sortedStrings(arr, N);
return 0;
} Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG{
// Function to check if N is power of
// 2 or not
static boolean isPowerOfTwo(int n)
{
// Base Case
if (n == 0)
return false;
// Return true if N is power of 2
return (Math.ceil(Math.log(n) / Math.log(2)) ==
Math.floor(Math.log(n) / Math.log(2)));
}
// Function to print array of strings
// in ascending order
static void printArray(ArrayList res)
{
// Sort strings in ascending order
Collections.sort(res);
// Print the array
for(int i = 0; i < res.size(); i++)
{
System.out.print(res.get(i) + " ");
}
}
// Function to sort the strings after
// modifying each string according to
// the given conditions
static void sortedStrings(String S[], int N)
{
// Store the frequency of each
// characters of the string
HashMap freq = new HashMap<>();
// Stores the required
// array of strings
ArrayList res = new ArrayList<>();
// Traverse the array of strings
for(int i = 0; i < N; i++)
{
// Temporary string
String st = "";
// Stores frequency of each
// alphabet of the string
for(int j = 0; j < S[i].length(); j++)
{
// Update frequency of S[i][j]
freq.put(S[i].charAt(j),
freq.getOrDefault(S[i].charAt(j), 0) + 1);
}
// Traverse the map freq
for(char ch : freq.keySet())
{
// Check if the frequency
// of i.first is a power of 2
if (isPowerOfTwo(freq.get(ch)))
{
// Update string st
for(int j = 0; j < freq.get(ch); j++)
{
st += ch;
}
}
}
// Clear the map
freq.clear();
// Null string
if (st.length() == 0)
continue;
// Sort the string in
// descending order
char myCharArr[] = st.toCharArray();
Arrays.sort(myCharArr);
String ns = "";
for(int j = myCharArr.length - 1; j >= 0; --j)
ns += myCharArr[j];
// Update res
res.add(ns);
}
// Print the array of strings
printArray(res);
}
// Driver Code
public static void main(String[] args)
{
String arr[] = { "aaacbb", "geeks", "aaa" };
int N = arr.length;
sortedStrings(arr, N);
}
}
// This code is contributed by Kingash 输出：
cbb skgee时间复杂度： O(N * log N + M * log M)，其中M是S []中字符串的最大长度
辅助空间： O(N)