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 = Y^{3} for some Y.

Examples:

Input : 4125 Output : 125Explanation125 = 5^{3}. We can form 125 by deleting digit 4 from 4125 Input : 876 Output :8Explanation8 = 2^{3}. We can form 8 by deleting digits 7 and 6 from 876

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

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