Given an integer n, the task is to find the minimum number of cubes whose sum equals N.
Examples:
Input: N = 496
Output: 3
43 + 63 + 63 = 496
Note that 13 + 33 + 53 + 73 = 496 but it requires 4 cubes.
Input: N = 15
Output: 8
Naive approach: Write a recursive method that will take every perfect cube less than N say X as part of the summation and then recur for the number of cubes required for the remaining sum N – X. The time complexity of this solution is exponential.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <iostream>using namespace std;// Function to return the minimum// number of cubes whose sum is kint MinOfCubed(int k){ // If k is less than the 2^3 if (k < 8) return k; // Initialize with the maximum // number of cubes required int res = k; for (int i = 1; i <= k; i++) { if ((i * i * i) > k) return res; res = min(res, MinOfCubed(k - (i * i * i)) + 1); } return res;}// Driver codeint main(){ int num = 15; cout << MinOfCubed(num); return 0;} |
Java
// Java implementation of the approachclass GFG { // Function to return the minimum // number of cubes whose sum is k static int MinOfCubed(int k) { // If k is less than the 2^3 if (k < 8) return k; // Initialize with the maximum // number of cubes required int res = k; for (int i = 1; i <= k; i++) { if ((i * i * i) > k) return res; res = Math.min(res, MinOfCubed(k - (i * i * i)) + 1); } return res; } // Driver code public static void main(String[] args) { int num = 15; System.out.println(MinOfCubed(num)); }}// This code has been contributed by 29AjayKumar |
Python3
# Python3 implementation of the approach # Function to return the minimum # number of cubes whose sum is k def MinOfCubed(k): # If k is less than the 2 ^ 3 if (k < 8): return k; # Initialize with the maximum # number of cubes required res = k; for i in range(1, k + 1): if ((i * i * i) > k): return res; res = min(res, MinOfCubed(k - (i * i * i)) + 1); return res; # Driver code num = 15; print(MinOfCubed(num));# This code contributed by PrinciRaj1992 |
C#
// C# implementation of the approachusing System;class GFG { // Function to return the minimum // number of cubes whose sum is k static int MinOfCubed(int k) { // If k is less than the 2^3 if (k < 8) return k; // Initialize with the maximum // number of cubes required int res = k; for (int i = 1; i <= k; i++) { if ((i * i * i) > k) return res; res = Math.Min(res, MinOfCubed(k - (i * i * i)) + 1); } return res; } // Driver code static public void Main() { int num = 15; Console.WriteLine(MinOfCubed(num)); }}// This code has been contributed by ajit. |
PHP
<?php// PHP implementation of the approach // Function to return the minimum // number of cubes whose sum is k function MinOfCubed($k) { // If k is less than the 2^3 if ($k < 8) return $k; // Initialize with the maximum // number of cubes required $res = $k; for ($i = 1; $i <= $k; $i++) { if (($i * $i * $i) > $k) return $res; $res = min($res, MinOfCubed($k - ($i *$i * $i)) + 1); } return $res; } // Driver code $num = 15; echo MinOfCubed($num); // This code is contributed by Ryuga?> |
Javascript
<script> // Javascript implementation of the approach // Function to return the minimum // number of cubes whose sum is k function MinOfCubed(k) { // If k is less than the 2^3 if (k < 8) return k; // Initialize with the maximum // number of cubes required let res = k; for (let i = 1; i <= k; i++) { if ((i * i * i) > k) return res; res = Math.min(res, MinOfCubed(k - (i * i * i)) + 1); } return res; } let num = 15; document.write(MinOfCubed(num));</script> |
Output:
8
Efficient approach: If we draw the complete recursion tree for the above solution, we can see that many sub-problems are solved, again and again, so we can see that this problem has the overlapping sub-problems property. This leads us to solve the problem using the Dynamic Programming paradigm.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the minimum// number of cubes whose sum is kint MinOfCubedDP(int k){ vector<int> DP(k+1); int j = 1, t = 1; DP[0] = 0; for (int i = 1; i <= k; i++) { DP[i] = INT_MAX; // While current perfect cube // is less than current element while (j <= i) { // If i is a perfect cube if (j == i) DP[i] = 1; // i = (i - 1) + 1^3 else if (DP[i] > DP[i - j]) DP[i] = DP[i - j] + 1; // Next perfect cube t++; j = t * t * t; } // Re-initialization for next element t = j = 1; } return DP[k];}// Driver codeint main(){ int num = 15; cout << MinOfCubedDP(num); return 0;} |
Java
// Java implementation of the approachclass GFG { // Function to return the minimum // number of cubes whose sum is k static int MinOfCubedDP(int k) { int[] DP = new int[k + 1]; int j = 1, t = 1; DP[0] = 0; for (int i = 1; i <= k; i++) { DP[i] = Integer.MAX_VALUE; // While current perfect cube // is less than current element while (j <= i) { // If i is a perfect cube if (j == i) DP[i] = 1; // i = (i - 1) + 1^3 else if (DP[i] > DP[i - j]) DP[i] = DP[i - j] + 1; // Next perfect cube t++; j = t * t * t; } // Re-initialization for next element t = j = 1; } return DP[k]; } // Driver code public static void main(String[] args) { int num = 15; System.out.println(MinOfCubedDP(num)); }}/* This code contributed by PrinciRaj1992 */ |
Python3
# Python implementation of the approachimport sys# Function to return the minimum# number of cubes whose sum is kdef MinOfCubedDP(k): DP = [0] * (k + 1); j = 1; t = 1; DP[0] = 0; for i in range(1, k + 1): DP[i] = sys.maxsize; # While current perfect cube # is less than current element while (j <= i): # If i is a perfect cube if (j == i): DP[i] = 1; # i = (i - 1) + 1^3 elif (DP[i] > DP[i - j]): DP[i] = DP[i - j] + 1; # Next perfect cube t += 1; j = t * t * t; # Re-initialization for next element t = j = 1; return DP[k];# Driver codenum = 15;print(MinOfCubedDP(num));# This code contributed by Rajput-Ji |
C#
// C# implementation of the approachusing System;class GFG { // Function to return the minimum // number of cubes whose sum is k static int MinOfCubedDP(int k) { int[] DP = new int[k + 1]; int j = 1, t = 1; DP[0] = 0; for (int i = 1; i <= k; i++) { DP[i] = int.MaxValue; // While current perfect cube // is less than current element while (j <= i) { // If i is a perfect cube if (j == i) DP[i] = 1; // i = (i - 1) + 1^3 else if (DP[i] > DP[i - j]) DP[i] = DP[i - j] + 1; // Next perfect cube t++; j = t * t * t; } // Re-initialization for next element t = j = 1; } return DP[k]; } // Driver code public static void Main() { int num = 15; Console.WriteLine(MinOfCubedDP(num)); }}/* This code contributed by Code_Mech */ |
PHP
<?php// PHP implementation of the approach// Function to return the minimum// number of cubes whose sum is kfunction MinOfCubedDP($k){ $DP = array($k + 1); $j = 1; $t = 1; $DP[0] = 0; for ($i = 1; $i <= $k; $i++) { $DP[$i] = PHP_INT_MAX; // While current perfect cube // is less than current element while ($j <= $i) { // If i is a perfect cube if ($j == $i) $DP[$i] = 1; // i = (i - 1) + 1^3 else if ($DP[$i] > $DP[$i - $j]) $DP[$i] = $DP[$i - $j] + 1; // Next perfect cube $t++; $j = $t * $t * $t; } // Re-initialization for next element $t = $j = 1; } return $DP[$k];}// Driver code$num = 15;echo(MinOfCubedDP($num));// This code contributed by Code_Mech ?> |
Javascript
<script> // Javascript implementation of the approach // Function to return the minimum // number of cubes whose sum is k function MinOfCubedDP(k) { let DP = new Array(k + 1); DP.fill(0); let j = 1, t = 1; DP[0] = 0; for (let i = 1; i <= k; i++) { DP[i] = Number.MAX_VALUE; // While current perfect cube // is less than current element while (j <= i) { // If i is a perfect cube if (j == i) DP[i] = 1; // i = (i - 1) + 1^3 else if (DP[i] > DP[i - j]) DP[i] = DP[i - j] + 1; // Next perfect cube t++; j = t * t * t; } // Re-initialization for next element t = j = 1; } return DP[k]; } let num = 15; document.write(MinOfCubedDP(num));</script> |
Output:
8
At the first glance, it seems that the algorithm works in polynomial time because we have two nested loops, the outer loop takes O(n), and the inner loop takes O(n^(1/3)). So the overall algorithm takes O(n*n^(1/3)). But what is the complexity as a function of input length? To represent a number in size of n we need m=log(n) – (log of base 2) bits. In this case n=2^m. If we set 2^m as n in the formula O(n*n^(1/3)), we see that the time complexity is still exponential. This algorithm is called Pseudo-polynomial.
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