Find the Largest Cube formed by Deleting minimum Digits from a number

Given a number n, the task is to find the largest perfect cube that can be formed by deleting minimum digits(possibly 0) from the number.
X is called a perfect cube if X = Y³ for some Y.

Examples:

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We can generate cubes of all numbers till from 1 to N^1/3 (We don’t consider 0 as 0 is not considered as a perfect cube). We iterate the cubes from largest to the smallest.
Now if we look at the number n given to us, then we know that this number contains only log(n) + 1 digits, thus we can efficiently approach the problem if we treat this number n as a string hereafter.
While iterating on the perfect cubes, we check if the perfect cube is a subsequence of the number n when its represented as a string.If this is the case then the deletions required for changing the number n to the current perfect cube is:

No of deleted digits = No of digits in number n - 
                       Number of digits in current 
                                      perfect cube

Since we want the largest cube number we traverse the array of preprocessed cubes in reverse order.

/* C++ code to implement maximum perfect cube  
   formed after deleting  minimum digits */
#include <bits/stdc++.h> 
using namespace std; 
  
// Returns vector of Pre Processed perfect cubes 
vector<string> preProcess(long long int n) 
{ 
    vector<string> 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<string> 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<string> 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; 
} 

Output:

Largest Cube that can be formed from 4125 is 125
Largest Cube that can be formed from 876 is 8

Time Complexity of the above algorithm is O(N^1/3log(N) log(N) is due to the fact that the number of digits in N are Log(N) + 1.

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sirjan13

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