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
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++
/* 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; } |
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<String> preProcess(int n) { Vector<String> 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<String> 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<String> 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 = 4125findLargestCube(n) n = 876findLargestCube(n) # This code is contributed # by mohit kumar 29 |
C#
/* 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<String> preProcess(int n) { List<String> preProcessedCubes = new List<String>(); 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<String> 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<String> 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
<?php // PHP code to implement maximum perfect // cube formed after deleting minimum digits // Returns vector of Pre Processed // perfect cubes function preProcess($n) { $preProcessedCubes = array(); for ($i = 1; $i * $i * $i < $n; $i++) { $iThCube = $i * $i * $i; // convert the cube to string and // push into preProcessedCubes vector $cubeString = strval($iThCube); array_push($preProcessedCubes,$cubeString); } return $preProcessedCubes; } // Utility function for findLargestCube(). // Returns the Largest cube number that // can be formed function findLargestCubeUtil($num,$preProcessedCubes) { // reverse the preProcessed cubes so // that we have the largest cube in // the beginning of the vector $preProcessedCubes = array_reverse($preProcessedCubes); $totalCubes = count($preProcessedCubes); // iterate over all cubes for($i = 0; $i < $totalCubes; $i++) { $currCube = $preProcessedCubes[$i]; $digitsInCube = strlen($currCube); $index = 0; $digitsInNumber = strlen($num); for ($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]) $index += 1; if ($digitsInCube == $index) return $currCube; } } // if control reaches here, the its // not possible to form a perfect cube return "Not Possible"; } // wrapper for findLargestCubeUtil() function findLargestCube($n) { // pre process perfect cubes $preProcessedCubes = preProcess($n); $num = strval($n); $ans = findLargestCubeUtil($num, $preProcessedCubes); print("Largest Cube that can be formed from ".$n." is ".$ans."\n"); } // Driver Code $n = 4125; findLargestCube($n); $n = 876; findLargestCube($n) // This code is contributed by mits ?> |
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(N1/3log(N) log(N) is due to the fact that the number of digits in N are Log(N) + 1.
