Given a number n, find two pairs that can represent the number as sum of two cubes. In other words, find two pairs (a, b) and (c, d) such that given number n can be expressed as
n = a^3 + b^3 = c^3 + d^3
where a, b, c and d are four distinct numbers.
Examples:
Input: N = 1729 Output: (1, 12) and (9, 10) Explanation: 1729 = 1^3 + 12^3 = 9^3 + 10^3 Input: N = 4104 Output: (2, 16) and (9, 15) Explanation: 4104 = 2^3 + 16^3 = 9^3 + 15^3 Input: N = 13832 Output: (2, 24) and (18, 20) Explanation: 13832 = 2^3 + 24^3 = 18^3 + 20^3
Any number n that satisfies the constraint will have two distinct pairs (a, b) and (c, d) such that a, b, c and d are all less than n1/3. The idea is very simple. For every distinct pair (x, y) formed by numbers less than the n1/3, if their sum (x3 + y3) is equal to given number, we store them in a hash table using sum as a key. If pairs with sum equal to given number appears again, we simply print both pairs.
1) Create an empty hash map, say s. 2) cubeRoot = n1/3 3) for (int x = 1; x < cubeRoot; x++) for (int y = x + 1; y <= cubeRoot; y++) int sum = x3 + y3; if (sum != n) continue; if sum exists in s, we found two pairs with sum, print the pairs else insert pair(x, y) in s using sum as key
Below is the implementation of above idea –
C++
// C++ program to find pairs that can represent // the given number as sum of two cubes #include <bits/stdc++.h> using namespace std; // Function to find pairs that can represent // the given number as sum of two cubes void findPairs( int n) { // find cube root of n int cubeRoot = pow (n, 1.0/3.0); // create an empty map unordered_map< int , pair< int , int > > s; // Consider all pairs such with values less // than cuberoot for ( int x = 1; x < cubeRoot; x++) { for ( int y = x + 1; y <= cubeRoot; y++) { // find sum of current pair (x, y) int sum = x*x*x + y*y*y; // do nothing if sum is not equal to // given number if (sum != n) continue ; // if sum is seen before, we found two pairs if (s.find(sum) != s.end()) { cout << "(" << s[sum].first << ", " << s[sum].second << ") and (" << x << ", " << y << ")" << endl; } else // if sum is seen for the first time s[sum] = make_pair(x, y); } } } // Driver function int main() { int n = 13832; findPairs(n); return 0; } |
Java
// Java program to find pairs that can represent // the given number as sum of two cubes import java.util.*; class GFG { static class pair { int first, second; public pair( int first, int second) { this .first = first; this .second = second; } } // Function to find pairs that can represent // the given number as sum of two cubes static void findPairs( int n) { // find cube root of n int cubeRoot = ( int ) Math.pow(n, 1.0 / 3.0 ); // create an empty map HashMap<Integer, pair> s = new HashMap<Integer, pair>(); // Consider all pairs such with values less // than cuberoot for ( int x = 1 ; x < cubeRoot; x++) { for ( int y = x + 1 ; y <= cubeRoot; y++) { // find sum of current pair (x, y) int sum = x*x*x + y*y*y; // do nothing if sum is not equal to // given number if (sum != n) continue ; // if sum is seen before, we found two pairs if (s.containsKey(sum)) { System.out.print( "(" + s.get(sum).first+ ", " + s.get(sum).second+ ") and (" + x+ ", " + y+ ")" + "\n" ); } else // if sum is seen for the first time s.put(sum, new pair(x, y)); } } } // Driver code public static void main(String[] args) { int n = 13832 ; findPairs(n); } } // This code is contributed by PrinciRaj1992 |
Python3
# Python3 program to find pairs # that can represent the given # number as sum of two cubes # Function to find pairs that # can represent the given number # as sum of two cubes def findPairs(n): # Find cube root of n cubeRoot = pow (n, 1.0 / 3.0 ); # Create an empty map s = {} # Consider all pairs such with # values less than cuberoot for x in range ( int (cubeRoot)): for y in range (x + 1 , int (cubeRoot) + 1 ): # Find sum of current pair (x, y) sum = x * x * x + y * y * y; # Do nothing if sum is not equal to # given number if ( sum ! = n): continue ; # If sum is seen before, we # found two pairs if sum in s.keys(): print ( "(" + str (s[ sum ][ 0 ]) + ", " + str (s[ sum ][ 1 ]) + ") and (" + str (x) + ", " + str (y) + ")" + "\n" ) else : # If sum is seen for the first time s[ sum ] = [x, y] # Driver code if __name__ = = "__main__" : n = 13832 findPairs(n) # This code is contributed by rutvik_56 |
C#
// C# program to find pairs that can represent // the given number as sum of two cubes using System; using System.Collections.Generic; class GFG { class pair { public int first, second; public pair( int first, int second) { this .first = first; this .second = second; } } // Function to find pairs that can represent // the given number as sum of two cubes static void findPairs( int n) { // find cube root of n int cubeRoot = ( int ) Math.Pow(n, 1.0/3.0); // create an empty map Dictionary< int , pair> s = new Dictionary< int , pair>(); // Consider all pairs such with values less // than cuberoot for ( int x = 1; x < cubeRoot; x++) { for ( int y = x + 1; y <= cubeRoot; y++) { // find sum of current pair (x, y) int sum = x*x*x + y*y*y; // do nothing if sum is not equal to // given number if (sum != n) continue ; // if sum is seen before, we found two pairs if (s.ContainsKey(sum)) { Console.Write( "(" + s[sum].first+ ", " + s[sum].second+ ") and (" + x+ ", " + y+ ")" + "\n" ); } else // if sum is seen for the first time s.Add(sum, new pair(x, y)); } } } // Driver code public static void Main(String[] args) { int n = 13832; findPairs(n); } } // This code is contributed by PrinciRaj1992 |
Output:
(2, 24) and (18, 20)
Time Complexity of above solution is O(n2/3) which is much less than O(n).
Can we solve the above problem in O(n1/3) time? We will be discussing that in next post.
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