给定数字的排列是 2 的幂
给定一个由N个数字组成的字符串S ，任务是打印S的所有可能的数字组合，它是 2 的完美幂。
例子：
Input: S = “614”
Output: 4
Explanation:
All possible combinations of digit of S that are perfect power of 2 are 1, 4, 16, 64.
Input: S = “6”
Output: 0
方法：给定的问题可以通过使用回溯来解决。这个想法是生成字符串S的所有可能排列，如果它是2的完美幂，然后打印它。请按照以下步骤解决问题：
- 定义一个函数check(int number)来检查给定的数字是否是2的幂并执行以下任务：
- 如果数字等于0，则返回false。
- 如果按位与 number和number-1，然后返回true，否则返回false。
- 定义一个函数calculate(int arr[], 字符串 ans)并执行以下任务：
- 如果字符串ans的长度不等于0和函数的值。 check(Integer.parseInt(ans.trim()))返回true，然后将此值添加到 HashSet H[] 。
- 使用变量i遍历范围[0, 10]并执行以下任务：
- 如果arr[i]等于0 ，则继续。
- 否则，将arr[i]的值减1并调用函数calculate(temp, “”)以查找字符串str的其他可能排列，并将arr[i]的值加1 。
- 初始化一个 HashSet H[]以存储可能的字符串数字，它们是 2 的幂。
- 初始化一个数组，比如temp[]以存储字符串str中整数的频率。
- 在while循环中迭代直到N不等于0并执行以下步骤：
- 将变量rem初始化为N%10并将temp[rem]的值增加1 。
- 将N的值除以10 。
- 调用函数calculate(temp, “”)以查找字符串S的可能排列。
- 执行上述步骤后，打印 HashSet 作为结果。
下面是上述方法的实现。
C++
#include
using namespace std;
// Stores the all possible generated
// integers from the given string
unordered_set hs;
// Function to check if the
// number is power of 2
bool check(int number)
{
// If number is 0, then it
// can't be a power of 2
if (number == 0) {
return false;
}
// If the bitwise AND of n
// and n-1 is 0, then only
// it is a power of 2
if ((number & (number - 1)) == 0) {
return true;
}
return false;
}
// Function to generate the numbers
void calculate(int* arr, string ans)
{
if (ans.length() != 0) {
// Checking the number
if (check(stoi(ans))) {
hs.insert(stoi(ans));
}
}
// Iterate over the range
for (int i = 0; i < 10; ++i) {
if (arr[i] == 0) {
continue;
}
else {
// Use the number
arr[i]--;
calculate(arr,(ans + to_string(i)));
// Backtracking Step
arr[i]++;
}
}
}
// Function to find the all possible
// permutations
void generatePermutation(int n)
{
hs.clear();
// Stores the frequency of digits
int temp[10];
for (int i = 0; i < 10; i++) {
temp[i] = 0;
}
// Iterate over the number
while (n != 0) {
int rem = n % 10;
temp[rem]++;
n = n / 10;
}
// Function Call
calculate(temp, "");
// Print the result
cout << hs.size() << "\n";
for (auto i : hs) {
cout << i << " ";
}
}
int main()
{
int N = 614;
generatePermutation(N);
return 0;
} Java
// Java program for the above approach
import java.io.*;
import java.util.*;
public class GFG {
// Stores the all possible generated
// integers from the given string
static HashSet H = new HashSet<>();
// Function to check if the
// number is power of 2
static boolean check(int number)
{
// If number is 0, then it
// can't be a power of 2
if (number == 0) {
return false;
}
// If the bitwise AND of n
// and n-1 is 0, then only
// it is a power of 2
if ((number & (number - 1)) == 0) {
return true;
}
return false;
}
// Function to generate the numbers
static void calculate(
int arr[], String ans)
{
if (ans.length() != 0) {
// Checking the number
if (check(Integer.parseInt(
ans.trim()))) {
H.add(Integer.parseInt(
ans.trim()));
}
}
// Iterate over the range
for (int i = 0; i < arr.length; ++i) {
if (arr[i] == 0) {
continue;
}
else {
// Use the number
arr[i]--;
calculate(arr, ans + i);
// Backtracking Step
arr[i]++;
}
}
}
// Function to find the all possible
// permutations
static void generatePermutation(int n)
{
// Stores the frequency of digits
int temp[] = new int[10];
// Iterate over the number
while (n != 0) {
int rem = n % 10;
temp[rem]++;
n = n / 10;
}
// Function Call
calculate(temp, "");
// Print the result
System.out.println(H.size());
System.out.println(H);
}
// Driver Code
public static void main(String[] args)
{
int N = 614;
generatePermutation(N);
}
} Python3
# Python3 program for the above approach
# Stores the all possible generated
# integers from the given string
H = set()
# Function to check if the
# number is power of 2
def check(number):
# If number is 0, then it
# can't be a power of 2
if (number == 0):
return False
# If the bitwise AND of n
# and n-1 is 0, then only
# it is a power of 2
if ((number & (number - 1)) == 0):
return True
return False
# Function to generate the numbers
def calculate(arr, ans):
if (len(ans) != 0):
# Checking the number
if (check(int(ans))):
H.add(int(ans))
# Iterate over the range
for i in range(len(arr)):
if (arr[i] == 0):
continue
else:
# Use the number
arr[i]-=1
calculate(arr, ans + str(i))
# Backtracking Step
arr[i]+=1
# Function to find the all possible
# permutations
def generatePermutation(n):
# Stores the frequency of digits
h = [16, 64, 1, 4]
temp = [0]*(10)
# Iterate over the number
while (n != 0):
rem = n % 10
temp[rem]+=1
n = int(n / 10)
# Function Call
calculate(temp, "")
# Print the result
print(len(H))
print(h)
N = 614
generatePermutation(N)
# This code is contributed by divyesh072019.C#
using System;
using System.Collections.Generic;
public class GFG{
// Stores the all possible generated
// integers from the given string
static HashSet H = new HashSet();
// Function to check if the
// number is power of 2
static bool check(int number)
{
// If number is 0, then it
// can't be a power of 2
if (number == 0) {
return false;
}
// If the bitwise AND of n
// and n-1 is 0, then only
// it is a power of 2
if ((number & (number - 1)) == 0) {
return true;
}
return false;
}
// Function to generate the numbers
static void calculate(
int[] arr, String ans)
{
if (ans.Length != 0) {
// Checking the number
if (check(int.Parse(
ans))) {
H.Add(int.Parse(
ans));
}
}
// Iterate over the range
for (int i = 0; i < arr.Length; ++i) {
if (arr[i] == 0) {
continue;
}
else {
// Use the number
arr[i]--;
calculate(arr, ans + i);
// Backtracking Step
arr[i]++;
}
}
}
// Function to find the all possible
// permutations
static void generatePermutation(int n)
{
// Stores the frequency of digits
int[] temp = new int[10];
// Iterate over the number
while (n != 0) {
int rem = n % 10;
temp[rem]++;
n = n / 10;
}
// Function Call
calculate(temp, "");
// Print the result
Console.WriteLine(H.Count);
foreach(var val in H){
Console.Write(val+" ");
}
}
static public void Main (){
int N = 614;
generatePermutation(N);
}
}
// This code is contributed by maddler. Javascript
C++14
#include
using namespace std;
void calculatePower(string s,int n,set&ans)
{
// cnt_s is stores the count of each decimal digits of string s
int cnt_s[10]={0};
// Here we calculate the count of each decimal digit of string s
for(int i=0;i=cnt_[a])
long val=2,one=1;
long maxval=(one<<62);
// maxval has max power of 2 in range of long
// in this loop we generate powers of 2
while(val<=maxval&&val>0)
{
long temp=val;
int cnt_a[10]={0};
// cnt_a has count of decimal digits in kth power of 2
while(temp)
{
long remainder=(temp%10);
cnt_a[remainder]++;
temp/=10;
}
// checking if a power of 2 present in string s
bool fl=true;
for(int i=0;i<10;i++)
{
if(cnt_a[i]>cnt_s[i])
{
fl=false;
break;
}
}
// if the power of 2 present in s we add it to set
if(fl)
{
ans.insert(val);
}
val*=2;
}
}
int main()
{
string s="614";
int n=s.length();
// set-ans has all possible powers of 2 that are present in string
setans;
calculatePower(s,n,ans);
cout<<"THE TOTAL POSSIBLE COMBINATIONS THAT ARE POWER OF 2 ARE: "< C
#include
#include
void calculatePower(char s[],int n)
{
// cnt_s is stores the count of each decimal digits of string s
int cnt_s[10]={0};
// Here we calculate the count of each decimal digit of string s
for(int i=0;i=cnt_[a])
long val=2,one=1;
long maxval=(one<<62);
// maxval has max power of 2 in range of long
// in this loop we generate power of 2
while(val<=maxval&&val>0)
{
long temp=val;
int cnt_a[10]={0};
// cnt_a has count of decimal digits in kth power of 2
while(temp)
{
long remainder=(temp%10);
cnt_a[remainder]++;
temp/=10;
}
// checking if a power of 2 present in string s
int fl=1;
for(int i=0;i<10;i++)
{
if(cnt_a[i]>cnt_s[i])
{
fl=0;
break;
}
}
if(fl)
{
printf("%ld is present in string\n",val);
}
val*=2;
}
}
int main()
{
char s[]="614";
int n=strlen(s);
calculatePower(s,n);
} Java
import java.util.*;
public class Main
{
public static void calculatePower(String s,int n)
{
ArrayList ans=new ArrayList();
// cnt_s is stores the count of each decimal digits of string s
int cnt_s[]=new int[10];
// Here we calculate the count of each decimal digit of string s
for(int i=0;i=cnt_[a])
long value=2,one=1;
long maxvalue=(one<<62);
// maxval has max power of 2 in range of long
// in this loop we generate powers of 2
while(value<=maxvalue && value>=0)
{
long temp=value;
int cnt_a[]=new int[10];
// cnt_a has count of decimal digits in kth power of 2
while(temp>0)
{
long remainder=(temp%10);
cnt_a[(int)remainder]++;
temp/=10;
}
// checking if a power of 2 present in string s
boolean fl=true;
for(int i=0;i<10;i++)
{
if(cnt_a[i]>cnt_s[i])
{
fl=false;
break;
}
}
// if the power of 2 present in s we add it to ans
if(fl)
{
ans.add(value);
}
value*=2;
}
System.out.println("THE TOTAL POSSIBLE COMBINATIONS THAT ARE POWER OF 2 ARE: "+ans.size());
for(Long i:ans)
{
System.out.println(i);
}
}
public static void main(String[] args) {
String s="614";
int n=s.length();
calculatePower(s,n);
}
} Python3
def calculatePower(s,n):
# cnt_s is stores the count of each decimal digits of string s
cnt_s=[0]*10
# Here we calculate the count of each decimal digit of string s
for i in s:
cnt_s[int(i)]+=1
# our main logic is to generate powers of 2 and store the count of decimal digit of powers of 2 in cnt_a i.e we check for every i in range(0,10): if(cnt_s[i]>=cnt_[a])
one=1
value=2
maxvalue=(one<<62)
ans=[]
# maxval has max power of 2 in range of long in this loop we generate powers of 2
while(value<=maxvalue):
# cnt_a has count of decimal digits in kth power of 2
cnt_a=[0]*10
fl=True
temp=value
while(temp>0):
remainder=temp%10
cnt_a[remainder]+=1
temp=int(temp/10)
# checking if a power of 2 present in string s
for i in range(0,10):
if(cnt_a[i]>cnt_s[i]):
fl=False
break
# if the power of 2 present in s we add it to ans
if(fl):
ans.append(value)
value*=2
print("THE TOTAL POSSIBLE COMBINATIONS THAT ARE POWER OF 2 ARE: ",len(ans))
for i in ans:
print(i)
s="614"
n=len(s)
calculatePower(s,n)C#
using System;
using System.Collections.Generic;
public class GFG {
public static void calculatePower(string s, int n)
{
List ans = new List();
// cnt_s is stores the count of each decimal digits
// of string s
int[] cnt_s = new int[10];
// Here we calculate the count of each decimal digit
// of string s
for (int i = 0; i < n; i++) {
cnt_s[s[i] - '0']++;
}
// our main logic is to generate powers of 2 and
// store the count of decimal digit of powers of 2
// in cnt_a i.e we check for every i in range(0,10):
// if(cnt_s[i]>=cnt_[a])
long value = 2, one = 1;
long maxvalue = (one << 62);
// maxval has max power of 2 in range of long
// in this loop we generate powers of 2
while (value <= maxvalue && value >= 0) {
long temp = value;
int[] cnt_a = new int[10];
// cnt_a has count of decimal digits in kth
// power of 2
while (temp > 0) {
long remainder = (temp % 10);
cnt_a[(int)remainder]++;
temp /= 10;
}
// checking if a power of 2 present in string s
bool fl = true;
for (int i = 0; i < 10; i++) {
if (cnt_a[i] > cnt_s[i]) {
fl = false;
break;
}
}
// if the power of 2 present in s we add it to
// ans
if (fl) {
ans.Add(value);
}
value *= 2;
}
Console.WriteLine(
"THE TOTAL POSSIBLE COMBINATIONS THAT ARE POWER OF 2 ARE: "
+ ans.Count);
foreach(long i in ans) { Console.WriteLine(i); }
}
// Driver code
public static void Main(string[] args)
{
string s = "614";
int n = s.Length;
calculatePower(s, n);
}
}
// This code is contributed by ukasp. 输出：
4
[16, 64, 1, 4]时间复杂度： O(N*9 N )
辅助空间： O(1)
高效方法：这个想法是最初生成 2 的所有幂并将它们存储在适当的数据结构中（在本例中为“set OF CPP”）并检查字符串中是否存在 2 的幂
- 这里的困难是检查字符串中是否存在两个的幂。
- 检查字符串的所有可能排列并验证该字符串是否为或字符串的幂为 2 这种方法不推荐，因为时间复杂度约为 O(n!)
- 一种有效的方法是使用一个计数数组，该数组存储十进制数字的计数，并验证 2 的幂的每个十进制数字的计数是否小于或等于字符串中存在的十进制数字的计数。
C++14
#include
using namespace std;
void calculatePower(string s,int n,set&ans)
{
// cnt_s is stores the count of each decimal digits of string s
int cnt_s[10]={0};
// Here we calculate the count of each decimal digit of string s
for(int i=0;i=cnt_[a])
long val=2,one=1;
long maxval=(one<<62);
// maxval has max power of 2 in range of long
// in this loop we generate powers of 2
while(val<=maxval&&val>0)
{
long temp=val;
int cnt_a[10]={0};
// cnt_a has count of decimal digits in kth power of 2
while(temp)
{
long remainder=(temp%10);
cnt_a[remainder]++;
temp/=10;
}
// checking if a power of 2 present in string s
bool fl=true;
for(int i=0;i<10;i++)
{
if(cnt_a[i]>cnt_s[i])
{
fl=false;
break;
}
}
// if the power of 2 present in s we add it to set
if(fl)
{
ans.insert(val);
}
val*=2;
}
}
int main()
{
string s="614";
int n=s.length();
// set-ans has all possible powers of 2 that are present in string
setans;
calculatePower(s,n,ans);
cout<<"THE TOTAL POSSIBLE COMBINATIONS THAT ARE POWER OF 2 ARE: "< C
#include
#include
void calculatePower(char s[],int n)
{
// cnt_s is stores the count of each decimal digits of string s
int cnt_s[10]={0};
// Here we calculate the count of each decimal digit of string s
for(int i=0;i=cnt_[a])
long val=2,one=1;
long maxval=(one<<62);
// maxval has max power of 2 in range of long
// in this loop we generate power of 2
while(val<=maxval&&val>0)
{
long temp=val;
int cnt_a[10]={0};
// cnt_a has count of decimal digits in kth power of 2
while(temp)
{
long remainder=(temp%10);
cnt_a[remainder]++;
temp/=10;
}
// checking if a power of 2 present in string s
int fl=1;
for(int i=0;i<10;i++)
{
if(cnt_a[i]>cnt_s[i])
{
fl=0;
break;
}
}
if(fl)
{
printf("%ld is present in string\n",val);
}
val*=2;
}
}
int main()
{
char s[]="614";
int n=strlen(s);
calculatePower(s,n);
}
Java
import java.util.*;
public class Main
{
public static void calculatePower(String s,int n)
{
ArrayList ans=new ArrayList();
// cnt_s is stores the count of each decimal digits of string s
int cnt_s[]=new int[10];
// Here we calculate the count of each decimal digit of string s
for(int i=0;i=cnt_[a])
long value=2,one=1;
long maxvalue=(one<<62);
// maxval has max power of 2 in range of long
// in this loop we generate powers of 2
while(value<=maxvalue && value>=0)
{
long temp=value;
int cnt_a[]=new int[10];
// cnt_a has count of decimal digits in kth power of 2
while(temp>0)
{
long remainder=(temp%10);
cnt_a[(int)remainder]++;
temp/=10;
}
// checking if a power of 2 present in string s
boolean fl=true;
for(int i=0;i<10;i++)
{
if(cnt_a[i]>cnt_s[i])
{
fl=false;
break;
}
}
// if the power of 2 present in s we add it to ans
if(fl)
{
ans.add(value);
}
value*=2;
}
System.out.println("THE TOTAL POSSIBLE COMBINATIONS THAT ARE POWER OF 2 ARE: "+ans.size());
for(Long i:ans)
{
System.out.println(i);
}
}
public static void main(String[] args) {
String s="614";
int n=s.length();
calculatePower(s,n);
}
}
Python3
def calculatePower(s,n):
# cnt_s is stores the count of each decimal digits of string s
cnt_s=[0]*10
# Here we calculate the count of each decimal digit of string s
for i in s:
cnt_s[int(i)]+=1
# our main logic is to generate powers of 2 and store the count of decimal digit of powers of 2 in cnt_a i.e we check for every i in range(0,10): if(cnt_s[i]>=cnt_[a])
one=1
value=2
maxvalue=(one<<62)
ans=[]
# maxval has max power of 2 in range of long in this loop we generate powers of 2
while(value<=maxvalue):
# cnt_a has count of decimal digits in kth power of 2
cnt_a=[0]*10
fl=True
temp=value
while(temp>0):
remainder=temp%10
cnt_a[remainder]+=1
temp=int(temp/10)
# checking if a power of 2 present in string s
for i in range(0,10):
if(cnt_a[i]>cnt_s[i]):
fl=False
break
# if the power of 2 present in s we add it to ans
if(fl):
ans.append(value)
value*=2
print("THE TOTAL POSSIBLE COMBINATIONS THAT ARE POWER OF 2 ARE: ",len(ans))
for i in ans:
print(i)
s="614"
n=len(s)
calculatePower(s,n)
C＃
using System;
using System.Collections.Generic;
public class GFG {
public static void calculatePower(string s, int n)
{
List ans = new List();
// cnt_s is stores the count of each decimal digits
// of string s
int[] cnt_s = new int[10];
// Here we calculate the count of each decimal digit
// of string s
for (int i = 0; i < n; i++) {
cnt_s[s[i] - '0']++;
}
// our main logic is to generate powers of 2 and
// store the count of decimal digit of powers of 2
// in cnt_a i.e we check for every i in range(0,10):
// if(cnt_s[i]>=cnt_[a])
long value = 2, one = 1;
long maxvalue = (one << 62);
// maxval has max power of 2 in range of long
// in this loop we generate powers of 2
while (value <= maxvalue && value >= 0) {
long temp = value;
int[] cnt_a = new int[10];
// cnt_a has count of decimal digits in kth
// power of 2
while (temp > 0) {
long remainder = (temp % 10);
cnt_a[(int)remainder]++;
temp /= 10;
}
// checking if a power of 2 present in string s
bool fl = true;
for (int i = 0; i < 10; i++) {
if (cnt_a[i] > cnt_s[i]) {
fl = false;
break;
}
}
// if the power of 2 present in s we add it to
// ans
if (fl) {
ans.Add(value);
}
value *= 2;
}
Console.WriteLine(
"THE TOTAL POSSIBLE COMBINATIONS THAT ARE POWER OF 2 ARE: "
+ ans.Count);
foreach(long i in ans) { Console.WriteLine(i); }
}
// Driver code
public static void Main(string[] args)
{
string s = "614";
int n = s.Length;
calculatePower(s, n);
}
}
// This code is contributed by ukasp.
时间复杂度：O(log(2^62) * N) 其中 N 是字符串的长度，log(2^62) 因为 2^62 是 Long 范围内的最大有效功率
辅助空间：O(10+10)