Find Cube Pairs | Set 1 (A n^(2/3) Solution)
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 cubesvoid 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 functionint main(){ int n = 13832; findPairs(n); return 0;} |
Java
// Java program to find pairs that can represent// the given number as sum of two cubesimport 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 cubesstatic 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 codepublic 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 cubesdef 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 codeif __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 cubesusing 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 cubesstatic 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 codepublic 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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