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.


Input : 4125
Output : 125
125 = 53. We can form 125 by deleting digit 4 from 4125

Input : 876
Output :8
8 = 23. We can form 8 by deleting digits 7 and 6 from 876

We can generate cubes of all numbers till from 1 to N1/3 (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);
    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]) 
            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;

    n = 876;

    return 0;

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(N1/3log(N) log(N) is due to the fact that the number of digits in N are Log(N) + 1.

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Recommended Posts:

4.5 Average Difficulty : 4.5/5.0
Based on 13 vote(s)