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 –
CPP
// 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 |
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.
This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. 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.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
Recommended Posts:
- Find Cube Pairs | Set 2 (A n^(1/3) Solution)
- Find initial integral solution of Linear Diophantine equation if finite solution exists
- Count of pairs in an Array whose sum is a Perfect Cube
- Find the diagonal of the Cube
- Find Cube root of a number using Log function
- Find the concentration of a solution using given Mass and Volume
- Find the Largest Cube formed by Deleting minimum Digits from a number
- Percentage increase in volume of the cube if a side of cube is increased by a given percentage
- Given GCD G and LCM L, find number of possible pairs (a, b)
- Find all Pairs possible from the given Array
- Find the maximum possible value of a[i] % a[j] over all pairs of i and j
- Find the number of pairs such that their gcd is equals to 1
- Find the count of even odd pairs in a given Array
- Find the sum of all possible pairs in an array of N elements
- Find number of pairs (x, y) in an Array such that x^y > y^x | Set 2
- Find minimum GCD of all pairs in an array
- Find the GCD of LCM of all unique pairs in an Array
- Find the minimum value of the given expression over all pairs of the array
- Find the number of ordered pairs such that a * p + b * q = N, where p and q are primes
- Find unique pairs such that each element is less than or equal to N
Improved By : princiraj1992
