Find all Ramanujan Numbers that can be formed by numbers upto L (original) (raw)
Last Updated : 15 Jul, 2025
Given a positive integer L, the task is to find all the Ramanujan Numbers that can be generated by any set of quadruples (a, b, c, d), where 0 < a, b, c, d ? L.
Ramanujan Numbers are the numbers that can be expressed as sum of two cubes in two different ways.
Therefore, Ramanujan Number (N) = a3 + b3 = c3 + d3.
Examples:
Input: L = 20
Output: 1729, 4104
Explanation:
The number 1729 can be expressed as 123 + 13 and 103 + 93.
The number 4104 can be expressed as 163 + 23 and 153 + 93.Input: L = 30
Output: 1729, 4104, 13832, 20683
Naive Approach: The simplest approach is to check for all combination of quadruples (a, b, c, d) from the range [1, L] consisting of distinct elements that satisfy the equation a3 + b3 = c3 + d3. For elements found to be satisfying the conditions, store the Ramanujan Numbers as a3 + b3. Finally, after checking for all possible combinations, print all the stored numbers.
Below is the implementation of the above approach:
C++ `
// CPP program for the above approach #include<bits/stdc++.h> using namespace std;
// Function to find Ramanujan numbers // made up of cubes of numbers up to L map<int,vector> ramanujan_On4(int limit) { map<int,vector> dictionary;
// Generate all quadruples a, b, c, d
// of integers from the range [1, L]
for(int a = 0; a < limit; a++)
{
for(int b = 0; b < limit; b++)
{
for(int c = 0; c < limit; c++)
{
for(int d = 0; d < limit; d++)
{
// Condition // 2:
// a, b, c, d are not equal
if ((a != b) and (a != c) and (a != d)
and (b != c) and (b != d)
and (c != d)){
int x = pow(a, 3) + pow(b, 3);
int y = pow(c, 3) + pow(d, 3);
if ((x) == (y))
{
int number = pow(a, 3) + pow(b, 3);
dictionary[number] = {a, b, c, d};
}
}
}
}
}}
// Return all the possible number
return dictionary;}
// Driver Code int main() {
// Given range L int L = 30; map<int, vector> ra1_dict = ramanujan_On4(L);
// Print all the generated numbers for(auto x:ra1_dict) { cout << x.first << ": (";
// sort(x.second.begin(),x.second.end());
for(int i = x.second.size() - 1; i >= 0; i--)
{
if(i == 0)
cout << x.second[i] << ")";
else
cout << x.second[i] << ", ";
}
cout << endl;
}
}
// This code is contributed by SURENDRA_GANGWAR.
Java
// Java program for the above approach import java.util.; import java.lang.;
class GFG{
static Map<Integer, List> ra1_dict;
// Function to find Ramanujan numbers // made up of cubes of numbers up to L static void ramanujan_On4(int limit) {
// Generate all quadruples a, b, c, d
// of integers from the range [1, L]
for(int a = 0; a < limit; a++)
{
for(int b = 0; b < limit; b++)
{
for(int c = 0; c < limit; c++)
{
for(int d = 0; d < limit; d++)
{
// Condition // 2:
// a, b, c, d are not equal
if ((a != b) && (a != c) && (a != d) &&
(b != c) && (b != d) && (c != d))
{
int x = (int)Math.pow(a, 3) +
(int) Math.pow(b, 3);
int y = (int)Math.pow(c, 3) +
(int) Math.pow(d, 3);
if ((x) == (y))
{
int number = (int)Math.pow(a, 3) +
(int) Math.pow(b, 3);
ra1_dict.put(number, new ArrayList<>(
Arrays.asList(a, b, c, d)));
}
}
}
}
}
}}
// Driver code public static void main(String[] args) {
// Given range L
int L = 30;
ra1_dict = new HashMap<>();
ramanujan_On4(L);
// Print all the generated numbers
for(Map.Entry<Integer,
List<Integer>> x: ra1_dict.entrySet())
{
System.out.print(x.getKey() + ": (");
// sort(x.second.begin(),x.second.end());
for(int i = x.getValue().size() - 1; i >= 0; i--)
{
if (i == 0)
System.out.print(x.getValue().get(i) + ")");
else
System.out.print(x.getValue().get(i) + ", ");
}
System.out.println();
}} }
// This code is contributed by offbeat
Python3
Python program for the above approach
import time
Function to find Ramanujan numbers
made up of cubes of numbers up to L
def ramanujan_On4(limit): dictionary = dict()
# Generate all quadruples a, b, c, d
# of integers from the range [1, L]
for a in range(0, limit):
for b in range(0, limit):
for c in range(0, limit):
for d in range(0, limit):
# Condition # 2:
# a, b, c, d are not equal
if ((a != b) and (a != c) and (a != d)
and (b != c) and (b != d)
and (c != d)):
x = a ** 3 + b ** 3
y = c ** 3 + d ** 3
if (x) == (y):
number = a ** 3 + b ** 3
dictionary[number] = a, b, c, d
# Return all the possible number
return dictionaryDriver Code
Given range L
L = 30 ra1_dict = ramanujan_On4(L)
Print all the generated numbers
for i in sorted(ra1_dict): print(f'{i}: {ra1_dict[i]}', end ='\n')
C#
// C# code to implement the approach using System; using System.Collections.Generic;
class GFG {
static Dictionary<int, List> ra1_dict;
// Function to find Ramanujan numbers // made up of cubes of numbers up to limit static void Ramanujan_On4(int limit) {
// Generate all quadruples a, b, c, d
// of integers from the range [1, limit]
for (int a = 0; a < limit; a++) {
for (int b = 0; b < limit; b++) {
for (int c = 0; c < limit; c++) {
for (int d = 0; d < limit; d++) {
// Condition 2:
// a, b, c, d are not equal
if ((a != b) && (a != c) && (a != d) &&
(b != c) && (b != d) && (c != d)) {
int x = (int)Math.Pow(a, 3) +
(int)Math.Pow(b, 3);
int y = (int)Math.Pow(c, 3) +
(int)Math.Pow(d, 3);
if (x == y) {
int number = (int)Math.Pow(a, 3) +
(int)Math.Pow(b, 3);
ra1_dict[number] = new List<int>{a, b, c, d};
}
}
}
}
}
}}
// Driver code static void Main(string[] args) {
// Given range L
int L = 30;
ra1_dict = new Dictionary<int, List<int>>();
// Function call
Ramanujan_On4(L);
// Print all the generated numbers
foreach (var x in ra1_dict) {
Console.Write(x.Key + ": (");
// sort(x.second.begin(),x.second.end());
for (int i = x.Value.Count - 1; i >= 0; i--) {
if (i == 0)
Console.Write(x.Value[i] + ")");
else
Console.Write(x.Value[i] + ", ");
}
Console.WriteLine();
}} }
// This code is contributed by phasing17
JavaScript
`
Output:
1729: (9, 10, 1, 12) 4104: (9, 15, 2, 16) 13832: (18, 20, 2, 24) 20683: (19, 24, 10, 27)
Time Complexity: O(L4)
Auxiliary Space: O(1)
Efficient Approach: The above approach can also be optimized by using Hashing. Follow the steps below to solve the problem:
- Initialize an array, say ans[], to stores all the possible Ramanujan Numbers that satisfy the given conditions.
- Precompute and store the cubes of all numbers from the range [1, L] in an auxiliary array arr[].
- Initialize a HashMap, say M, that stores the sum of all possible combinations of a pair of cubes generated from the array arr[].
- Now, generate all possible pairs(i, j) of the array arr[] and if the sum of pairs doesn't exist in the array, then mark the occurrence of the current sum of pairs in the Map. Otherwise, add the current sum to the array ans[] as it is one of the Ramanujan Numbers.
- After completing the above steps, print the numbers stored in the array ans[].
Below is the implementation of the above approach:
C++ `
#include <bits/stdc++.h> using namespace std;
map<int, vector > ramanujan_On2(int limit) { // Stores the cubes from 1 to L int cubes[limit + 1]; for (int i = 0; i <= limit; i++) { cubes[i] = i * i * i; }
// Stores the sum of pairs of cubes
map<int, vector<int> > dict_sum_pairs;
// Stores the Ramanujan Numbers
map<int, vector<int> > dict_ramnujan_nums;
// Generate all pairs (a, b)
// from the range [0, L]
for (int a = 0; a <= limit; a++) {
for (int b = a + 1; b <= limit; b++) {
int a3 = cubes[a];
int b3 = cubes[b];
// Find the sum of pairs
int sum_pairs = a3 + b3;
// Add to Map
if (dict_sum_pairs.count(sum_pairs)) {
// If the current sum_pair is already in
// the sum_pair Map, then store store
// all numbers in ramnujan_no map
int c = dict_sum_pairs[sum_pairs][0];
int d = dict_sum_pairs[sum_pairs][1];
dict_ramnujan_nums[sum_pairs]
= { a, b, c, d };
}
else {
// otherwise add in sum_pair map
dict_sum_pairs[sum_pairs] = { a, b };
}
}
}
// Return possible ramanujan numbers
return dict_ramnujan_nums;}
int main() { // Given range L int L = 30;
map<int, vector<int> > r_dict
= ramanujan_On2(L);
// Print all the generated numbers
for (auto& e : r_dict) {
cout << e.first << ": (";
for (int i = 0; i < e.second.size(); i++) {
if (i == e.second.size() - 1)
cout << e.second[i] << ")";
else
cout << e.second[i] << ", ";
}
cout << endl;
}
return 0;}
Java
// Java program for the above approach import java.io.; import java.util.;
class GFG {
// Function to find Ramanujan numbers static HashMap<Integer, ArrayList > ramanujan_On2(int limit) {
// Stores the cubes from 1 to L
int[] cubes = new int[limit + 1];
for (int i = 0; i <= limit; i++) {
cubes[i] = i * i * i;
}
// Stores the sum of pairs of cubes
HashMap<Integer, ArrayList<Integer> > dict_sum_pairs
= new HashMap<>();
// Stores the Ramanujan Numbers
HashMap<Integer, ArrayList<Integer> >
dict_ramnujan_nums = new HashMap<>();
// Generate all pairs (a, b)
// from the range [0, L]
for (int a = 0; a <= limit; a++) {
for (int b = a + 1; b <= limit; b++) {
int a3 = cubes[a];
int b3 = cubes[b];
// Find the sum of pairs
int sum_pairs = a3 + b3;
// Add to Map
if (dict_sum_pairs.containsKey(sum_pairs))
{
// If the current sum_pair is already in
// the sum_pair Map, then store store
// all numbers in ramnujan_no map
int c
= dict_sum_pairs.get(sum_pairs).get(
0);
int d
= dict_sum_pairs.get(sum_pairs).get(
1);
dict_ramnujan_nums.put(
sum_pairs,
new ArrayList<>(
Arrays.asList(a, b, c, d)));
}
else
{
// otherwise add in sum_pair map
dict_sum_pairs.put(
sum_pairs,
new ArrayList<>(
Arrays.asList(a, b)));
}
}
}
// Return possible ramanujan numbers
return dict_ramnujan_nums;}
// Driver code public static void main(String[] args) {
// Given range L
int L = 30;
HashMap<Integer, ArrayList<Integer> > r_dict
= ramanujan_On2(L);
// Print all the generated numbers
for (Map.Entry<Integer, ArrayList<Integer> > e :
(Iterable<
Map.Entry<Integer, ArrayList<Integer> > >)
r_dict.entrySet()
.stream()
.sorted(
Map.Entry
.comparingByKey())::iterator) {
System.out.print(e.getKey() + ": (");
for (int i = 0; i <= e.getValue().size() - 1;
i++) {
if (i == e.getValue().size() - 1)
System.out.print(e.getValue().get(i)
+ ")");
else
System.out.print(e.getValue().get(i)
+ ", ");
}
System.out.println();
}} }
// This code is contributed by Karandeep1234
Python3
Python program for the above approach
from array import * import time
Function to find Ramanujan numbers
made up of cubes of numbers up to L
def ramanujan_On2(limit): cubes = array('i', [])
# Stores the sum of pairs of cubes
dict_sum_pairs = dict()
# Stores the Ramanujan Numbers
dict_ramnujan_nums = dict()
sum_pairs = 0
# Stores the cubes from 1 to L
for i in range(0, limit):
cubes.append(i ** 3)
# Generate all pairs (a, b)
# from the range [0, L]
for a in range(0, limit):
for b in range(a + 1, limit):
a3, b3 = cubes[a], cubes[b]
# Find the sum of pairs
sum_pairs = a3 + b3
# Append to dictionary
if sum_pairs in dict_sum_pairs:
# If the current sum is in
# the dictionary, then store
# the current number
c, d = dict_sum_pairs.get(sum_pairs)
dict_ramnujan_nums[sum_pairs] = a, b, c, d
# Otherwise append the current
# sum pairs to the sum pairs
# dictionary
else:
dict_sum_pairs[sum_pairs] = a, b
# Return the possible Ramanujan
# Numbers
return dict_ramnujan_numsDriver Code
Given range L
L = 30 r_dict = ramanujan_On2(L)
Print all the numbers
for d in sorted(r_dict): print(f'{d}: {r_dict[d]}', end ='\n')
C#
// C# program for the above approach using System; using System.Linq; using System.Collections.Generic;
class GFG {
// Function to find Ramanujan numbers static Dictionary<int, List > Ramanujan_On2(int limit) {
// Stores the cubes from 1 to L
int[] cubes = new int[limit + 1];
for (int i = 0; i <= limit; i++) {
cubes[i] = i * i * i;
}
// Stores the sum of pairs of cubes
Dictionary<int, List<int> > dict_sum_pairs
= new Dictionary<int, List<int> >();
// Stores the Ramanujan Numbers
Dictionary<int, List<int> > dict_ramanujan_nums
= new Dictionary<int, List<int> >();
// Generate all pairs (a, b)
// from the range [0, L]
for (int a = 0; a <= limit; a++) {
for (int b = a + 1; b <= limit; b++) {
int a3 = cubes[a];
int b3 = cubes[b];
// Find the sum of pairs
int sum_pairs = a3 + b3;
// Add to Map
if (dict_sum_pairs.ContainsKey(sum_pairs)) {
// If the current sum_pair is already in
// the sum_pair Map, then store store
// all numbers in ramnujan_no map
int c = dict_sum_pairs[sum_pairs][0];
int d = dict_sum_pairs[sum_pairs][1];
dict_ramanujan_nums[sum_pairs]
= new List<int>(
new int[] { a, b, c, d });
}
else {
// otherwise add in sum_pair map
dict_sum_pairs[sum_pairs]
= new List<int>(new int[] { a, b });
}
}
}
// Return possible ramanujan numbers
return dict_ramanujan_nums;}
// Driver code static void Main(string[] args) {
// Given range L
int L = 30;
Dictionary<int, List<int> > r_dict
= Ramanujan_On2(L);
// Print all the generated numbers
foreach(KeyValuePair<int, List<int> > e in
r_dict.OrderBy(key => key.Key))
{
Console.Write(e.Key + ": (");
for (int i = 0; i <= e.Value.Count - 1; i++) {
if (i == e.Value.Count - 1)
Console.Write(e.Value[i] + ")");
else
Console.Write(e.Value[i] + ", ");
}
Console.WriteLine();
}} }
// This code is contributed by phasing17
` JavaScript ``
// JavaScript equivalent of the Python code
// Function to find Ramanujan numbers // made up of cubes of numbers up to L function ramanujan_On2(limit) { let cubes = [];
// Stores the sum of pairs of cubes let dict_sum_pairs = {};
// Stores the Ramanujan Numbers let dict_ramanujan_nums = {}; let sum_pairs = 0;
// Stores the cubes from 1 to L for (let i = 0; i < limit; i++) { cubes.push(i ** 3); }
// Generate all pairs (a, b) // from the range [0, L] for (let a = 0; a < limit; a++) { for (let b = a + 1; b < limit; b++) { let a3 = cubes[a]; let b3 = cubes[b];
// Find the sum of pairs
sum_pairs = a3 + b3;
// Append to dictionary
if (dict_sum_pairs[sum_pairs]) {
// If the current sum is in
// the dictionary, then store
// the current number
let [c, d] = dict_sum_pairs[sum_pairs];
dict_ramanujan_nums[sum_pairs] = [a, b, c, d];
} else {
// Otherwise append the current
// sum pairs to the sum pairs
// dictionary
dict_sum_pairs[sum_pairs] = [a, b];
}
}}
// Return the possible Ramanujan // Numbers return dict_ramanujan_nums; }
// Driver Code
// Given range L const L = 30; const r_dict = ramanujan_On2(L);
// Print all the numbers
for (let d in r_dict) {
console.log(${d}: ${r_dict[d]});
}
// This code is contributed by phasing17.
``
Output:
1729: (9, 10, 1, 12) 4104: (9, 15, 2, 16) 13832: (18, 20, 2, 24) 20683: (19, 24, 10, 27)
Time Complexity: O(L2)
Auxiliary Space: O(L2)