给定一个整数N,任务是检查没有任何前导零的N的任何排列是否为2的幂。如果存在给定数字的任何这样的排列,则打印该排列。否则,打印No。
例子:
Input: N = 46
Output: 64
Explanation:
The permutation of 46 which is perfect power of 2 is 64( = 26)
Input: N = 75
Output: No
Explanation:
There is no possible permutation of 75 that results in perfect power of 2.
天真的方法:一个简单的解决方案是生成数字N的所有排列,对于每个排列,检查其是否为2的幂。如果发现是真的,则打印“是” 。否则,打印No。
时间复杂度: O( (log 10 N )!*(log 10 N)),其中N是给定的数字N。
辅助空间: O(1)
高效方法:给定数字的位数和给定数字的任何排列的位数始终是相同的。因此,要优化上述方法,只需检查给定数字的位数是否等于2的任意完美幂。如果确定为真,则打印2的幂。否则,打印No。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
const int TEN = 10;
// Function to update the frequency
// array such that freq[i] stores
// the frequency of digit i to n
void updateFreq(int n, int freq[])
{
// While there are digits
// left to process
while (n) {
int digit = n % TEN;
// Update the frequency of
// the current digit
freq[digit]++;
// Remove the last digit
n /= TEN;
}
}
// Function that returns true if a
// and b are anagrams of each other
bool areAnagrams(int a, int b)
{
// To store the frequencies of
// the digits in a and b
int freqA[TEN] = { 0 };
int freqB[TEN] = { 0 };
// Update the frequency of
// the digits in a
updateFreq(a, freqA);
// Update the frequency of
// the digits in b
updateFreq(b, freqB);
// Match the frequencies of
// the common digits
for (int i = 0; i < TEN; i++) {
// If frequency differs for any
// digit then the numbers are
// not anagrams of each other
if (freqA[i] != freqB[i])
return false;
}
return true;
}
// Function to check if any permutation
// of a number is a power of 2 or not
bool isPowerOf2(int N)
{
// Iterate over all possible perfect
// power of 2
for (int i = 0; i < 32; i++) {
if (areAnagrams(1 << i, N)) {
// Print that number
cout << (1 << i);
return true;
}
}
return false;
}
// Driver Code
int main()
{
// Given Number N
int N = 46;
// Function Call
if (!isPowerOf2(N)) {
cout << "No";
}
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
static int TEN = 10;
// Function to update the frequency
// array such that freq[i] stores
// the frequency of digit i to n
static void updateFreq(int n, int freq[])
{
// While there are digits
// left to process
while (n > 0)
{
int digit = n % TEN;
// Update the frequency of
// the current digit
freq[digit]++;
// Remove the last digit
n /= TEN;
}
}
// Function that returns true if a
// and b are anagrams of each other
static boolean areAnagrams(int a, int b)
{
// To store the frequencies of
// the digits in a and b
int freqA[] = new int[TEN];
int freqB[] = new int[TEN];
// Update the frequency of
// the digits in a
updateFreq(a, freqA);
// Update the frequency of
// the digits in b
updateFreq(b, freqB);
// Match the frequencies of
// the common digits
for(int i = 0; i < TEN; i++)
{
// If frequency differs for any
// digit then the numbers are
// not anagrams of each other
if (freqA[i] != freqB[i])
return false;
}
return true;
}
// Function to check if any permutation
// of a number is a power of 2 or not
static boolean isPowerOf2(int N)
{
// Iterate over all possible perfect
// power of 2
for(int i = 0; i < 32; i++)
{
if (areAnagrams(1 << i, N))
{
// Print that number
System.out.print((1 << i));
return true;
}
}
return false;
}
// Driver Code
public static void main(String[] args)
{
// Given number N
int N = 46;
// Function call
if (!isPowerOf2(N))
{
System.out.print("No");
}
}
}
// This code is contributed by Amit Katiyar
Python3
# Python3 program for the above approach
TEN = 10
# Function to update the frequency
# array such that freq[i] stores
# the frequency of digit i to n
def updateFreq(n, freq):
# While there are digits
# left to process
while (n):
digit = n % TEN
# Update the frequency of
# the current digit
freq[digit] += 1
# Remove the last digit
n //= TEN
# Function that returns true if a
# and b are anagrams of each other
def areAnagrams(a, b):
# To store the frequencies of
# the digits in a and b
freqA = [0] * (TEN)
freqB = [0] * (TEN)
# Update the frequency of
# the digits in a
updateFreq(a, freqA)
# Update the frequency of
# the digits in b
updateFreq(b, freqB)
# Match the frequencies of
# the common digits
for i in range(TEN):
# If frequency differs for any
# digit then the numbers are
# not anagrams of each other
if (freqA[i] != freqB[i]):
return False
return True
# Function to check if any permutation
# of a number is a power of 2 or not
def isPowerOf2(N):
# Iterate over all possible perfect
# power of 2
for i in range(32):
if (areAnagrams(1 << i, N)):
# Print that number
print(1 << i)
return True
return False
# Driver Code
# Given number N
N = 46
# Function call
if (isPowerOf2(N) == 0):
print("No")
# This code is contributed by code_hunt
C#
// C# program for
// the above approach
using System;
class GFG{
static int TEN = 10;
// Function to update the frequency
// array such that freq[i] stores
// the frequency of digit i to n
static void updateFreq(int n,
int []freq)
{
// While there are digits
// left to process
while (n > 0)
{
int digit = n % TEN;
// Update the frequency of
// the current digit
freq[digit]++;
// Remove the last digit
n /= TEN;
}
}
// Function that returns true if a
// and b are anagrams of each other
static bool areAnagrams(int a, int b)
{
// To store the frequencies of
// the digits in a and b
int []freqA = new int[TEN];
int []freqB = new int[TEN];
// Update the frequency of
// the digits in a
updateFreq(a, freqA);
// Update the frequency of
// the digits in b
updateFreq(b, freqB);
// Match the frequencies of
// the common digits
for(int i = 0; i < TEN; i++)
{
// If frequency differs for any
// digit then the numbers are
// not anagrams of each other
if (freqA[i] != freqB[i])
return false;
}
return true;
}
// Function to check if any
// permutation of a number
// is a power of 2 or not
static bool isPowerOf2(int N)
{
// Iterate over all
// possible perfect power of 2
for(int i = 0; i < 32; i++)
{
if (areAnagrams(1 << i, N))
{
// Print that number
Console.Write((1 << i));
return true;
}
}
return false;
}
// Driver Code
public static void Main(String[] args)
{
// Given number N
int N = 46;
// Function call
if (!isPowerOf2(N))
{
Console.Write("No");
}
}
}
// This code is contributed by 29AjayKumar
Javascript
输出:
64
时间复杂度: O((log 10 N) 2 )
辅助空间: O(log 10 N)