给定数字n，任务是找到可以通过从数字中删除最小数字(可能为0)而形成的最大正方体。
如果对于某些Y，X = Y 3 ，则X称为理想立方体。
例子：
Input : 4125
Output : 125
Explanation
125 = 53. We can form 125 by deleting digit 4 from 4125
Input : 876
Output :8
Explanation
8 = 23. We can form 8 by deleting digits 7 and 6 from 876
我们可以生成从1到N 1/3的所有数字的多维数据集(我们不认为0，因为0被认为不是理想的多维数据集)。我们从最大到最小迭代多维数据集。
现在，如果我们看一下给定给我们的数字n，那么我们知道该数字仅包含log(n)+ 1个数字，因此，如果此后将此数字n视为字符串，则可以有效地解决该问题。
在对理想立方体进行迭代时，我们检查理想立方体是否为数字n的子序列(如果将其表示为字符串)。如果是这种情况，则将数字n更改为当前理想立方体所需的删除为：
No of deleted digits = No of digits in number n -
Number of digits in current
perfect cube
由于我们想要最大的多维数据集数，因此我们以相反的顺序遍历预处理多维数据集的数组。
C++
/* C++ code to implement maximum perfect cube
formed after deleting minimum digits */
#include
using namespace std;
// Returns vector of Pre Processed perfect cubes
vector preProcess(long long int n)
{
vector preProcessedCubes;
for (int i = 1; i * i * i <= n; i++) {
long long int iThCube = i * i * i;
// convert the cube to string and push into
// preProcessedCubes vector
string cubeString = to_string(iThCube);
preProcessedCubes.push_back(cubeString);
}
return preProcessedCubes;
}
/* Utility function for findLargestCube().
Returns the Largest cube number that can be formed */
string findLargestCubeUtil(string num,
vector preProcessedCubes)
{
// reverse the preProcessed cubes so that we
// have the largest cube in the beginning
// of the vector
reverse(preProcessedCubes.begin(), preProcessedCubes.end());
int totalCubes = preProcessedCubes.size();
// iterate over all cubes
for (int i = 0; i < totalCubes; i++) {
string currCube = preProcessedCubes[i];
int digitsInCube = currCube.length();
int index = 0;
int digitsInNumber = num.length();
for (int j = 0; j < digitsInNumber; j++) {
// check if the current digit of the cube
// matches with that of the number num
if (num[j] == currCube[index])
index++;
if (digitsInCube == index)
return currCube;
}
}
// if control reaches here, the its
// not possible to form a perfect cube
return "Not Possible";
}
// wrapper for findLargestCubeUtil()
void findLargestCube(long long int n)
{
// pre process perfect cubes
vector preProcessedCubes = preProcess(n);
// convert number n to string
string num = to_string(n);
string ans = findLargestCubeUtil(num, preProcessedCubes);
cout << "Largest Cube that can be formed from "
<< n << " is " << ans << endl;
}
// Driver Code
int main()
{
long long int n;
n = 4125;
findLargestCube(n);
n = 876;
findLargestCube(n);
return 0;
} Java
/* Java code to implement maximum perfect cube
formed after deleting minimum digits */
import java.util.*;
class GFG
{
// Returns vector of Pre Processed perfect cubes
static Vector preProcess(int n)
{
Vector preProcessedCubes = new Vector<>();
for (int i = 1; i * i * i <= n; i++)
{
int iThCube = i * i * i;
// convert the cube to String and push into
// preProcessedCubes vector
String cubeString = String.valueOf(iThCube);
preProcessedCubes.add(cubeString);
}
return preProcessedCubes;
}
/* Utility function for findLargestCube().
Returns the Largest cube number that can be formed */
static String findLargestCubeUtil(String num,
Vector preProcessedCubes)
{
// reverse the preProcessed cubes so that we
// have the largest cube in the beginning
// of the vector
Collections.reverse(preProcessedCubes);
int totalCubes = preProcessedCubes.size();
// iterate over all cubes
for (int i = 0; i < totalCubes; i++)
{
String currCube = preProcessedCubes.get(i);
int digitsInCube = currCube.length();
int index = 0;
int digitsInNumber = num.length();
for (int j = 0; j < digitsInNumber; j++)
{
// check if the current digit of the cube
// matches with that of the number num
if (num.charAt(j) == currCube.charAt(index))
{
index++;
}
if (digitsInCube == index)
{
return currCube;
}
}
}
// if control reaches here, the its
// not possible to form a perfect cube
return "Not Possible";
}
// wrapper for findLargestCubeUtil()
static void findLargestCube(int n)
{
// pre process perfect cubes
Vector preProcessedCubes = preProcess(n);
// convert number n to String
String num = String.valueOf(n);
String ans = findLargestCubeUtil(num, preProcessedCubes);
System.out.println("Largest Cube that can be formed from "
+ n + " is " + ans);
}
// Driver Code
public static void main(String[] args)
{
int n;
n = 4125;
findLargestCube(n);
n = 876;
findLargestCube(n);
}
}
// This code has been contributed by 29AjayKumar Python3
# Python3 code to implement maximum perfect
# cube formed after deleting minimum digits
import math as mt
# Returns vector of Pre Processed
# perfect cubes
def preProcess(n):
preProcessedCubes = list()
for i in range(1, mt.ceil(n**(1. / 3.))):
iThCube = i**3
# convert the cube to string and
# push into preProcessedCubes vector
cubeString = str(iThCube)
preProcessedCubes.append(cubeString)
return preProcessedCubes
# Utility function for findLargestCube().
# Returns the Largest cube number that
# can be formed
def findLargestCubeUtil(num,preProcessedCubes):
# reverse the preProcessed cubes so
# that we have the largest cube in
# the beginning of the vector
preProcessedCubes = preProcessedCubes[::-1]
totalCubes = len(preProcessedCubes)
# iterate over all cubes
for i in range(totalCubes):
currCube = preProcessedCubes[i]
digitsInCube = len(currCube)
index = 0
digitsInNumber = len(num)
for j in range(digitsInNumber):
# check if the current digit of the cube
# matches with that of the number num
if (num[j] == currCube[index]):
index += 1
if (digitsInCube == index):
return currCube
# if control reaches here, the its
# not possible to form a perfect cube
return "Not Possible"
# wrapper for findLargestCubeUtil()
def findLargestCube(n):
# pre process perfect cubes
preProcessedCubes = preProcess(n)
num = str(n)
ans = findLargestCubeUtil(num, preProcessedCubes)
print("Largest Cube that can be formed from",
n, "is", ans)
# Driver Code
n = 4125
findLargestCube(n)
n = 876
findLargestCube(n)
# This code is contributed
# by mohit kumar 29C#
/* C# code to implement maximum perfect cube
formed after deleting minimum digits */
using System;
using System.Collections.Generic;
class GFG
{
// Returns vector of Pre Processed perfect cubes
static List preProcess(int n)
{
List preProcessedCubes = new List();
for (int i = 1; i * i * i <= n; i++)
{
int iThCube = i * i * i;
// convert the cube to String and push into
// preProcessedCubes vector
String cubeString = String.Join("",iThCube);
preProcessedCubes.Add(cubeString);
}
return preProcessedCubes;
}
/* Utility function for findLargestCube().
Returns the Largest cube number that can be formed */
static String findLargestCubeUtil(String num,
List preProcessedCubes)
{
// reverse the preProcessed cubes so that we
// have the largest cube in the beginning
// of the vector
preProcessedCubes.Reverse();
int totalCubes = preProcessedCubes.Count;
// iterate over all cubes
for (int i = 0; i < totalCubes; i++)
{
String currCube = preProcessedCubes[i];
int digitsInCube = currCube.Length;
int index = 0;
int digitsInNumber = num.Length;
for (int j = 0; j < digitsInNumber; j++)
{
// check if the current digit of the cube
// matches with that of the number num
if (num[j] == currCube[index])
{
index++;
}
if (digitsInCube == index)
{
return currCube;
}
}
}
// if control reaches here, the its
// not possible to form a perfect cube
return "Not Possible";
}
// wrapper for findLargestCubeUtil()
static void findLargestCube(int n)
{
// pre process perfect cubes
List preProcessedCubes = preProcess(n);
// convert number n to String
String num = String.Join("",n);
String ans = findLargestCubeUtil(num, preProcessedCubes);
Console.WriteLine("Largest Cube that can be formed from "
+ n + " is " + ans);
}
// Driver Code
public static void Main(String[] args)
{
int n;
n = 4125;
findLargestCube(n);
n = 876;
findLargestCube(n);
}
}
// This code contributed by Rajput-Ji PHP
输出：
Largest Cube that can be formed from 4125 is 125
Largest Cube that can be formed from 876 is 8
上面算法的时间复杂度为O(N 1/3 log(N)log(N)，这是因为N中的位数是Log(N)+ 1。