CareerCup

I am just writing my thoughts on this:
use hash map and get all the numbers.

for (i=0;i<10;i++)
		for (j=0;j<10;j++)
			hashmap[x,a,b] = a^3+b^3;
	for (i=0;i<10;i++)
		for (j=0;j<10;j++)
			if c^3+d^3 == hashpmap[x]
				printf("a, b, c, d")

for (i=0;i<10;i++)
		for (j=0;j<10;j++)
			hashmap[x, a, b] = a^3+b^3;
for (i=0;i<10;i++)
		for (j=0;j<10;j++)
			if (hashmap[x] == c^3+d^3)
				printf("a b c d");

Using a hashmap:

//MehrdadAP

void solve(int N)
{

	unordered_map<long long int,vector< pair<int,int> > > umap;

	for (int i=0;i<=N;i++)
		for (int j=i;j<=N;j++){
			long long int tmp = i*1LL*i*aLL*i+j*1LL*j*1LL*j;
			umap[tmp].push_back({i,j});
		}

	for (auto it:umap){
		if (it.second.size()==1) continue;
		for (auto x:it.second)
			cout << "(" << x.first << "," << x.second << ")" << " " ;
		cout << endl;
	}
}

Use a hash table to keep the sums of all possible pairs (i,j).

def a3b3c3d3(n):
  l=[pow(i,3) for i in range(n)]
  d={}
  for i in range(n):
    for j in range(i, n):
      s=l[i]+l[j]
      d.setdefault(l[i]+l[j], []).append((i,j))

  for l in d.values():
    if len(l)>=2:
      first=True
      for i,j in l:
        if (first):
          first=False
        else:
          sys.stdout.write("=")
        sys.stdout.write("{0}^3+{1}^3".format(i,j))
      sys.stdout.write("\n")

public static void main(String args[])
	{
		Map<Integer,Set<String>> list=new HashMap<Integer,Set<String>>();
		int n=(int) Math.pow(10,2);
		int timecomplexity=0;
		for(int a=0;a<n;a++){
			for(int b=0;b<n;b++){
				Set<String> l;
				timecomplexity++;
				int result=(int) (Math.pow(a, 3)+Math.pow(b, 3));
				if(list.containsKey(result)){
					l=list.get(result);
					l.add(a+","+b);
					list.put(result, l);
				}
				else{
					l=new TreeSet<String>();
					l.add(a+","+b);
					list.put(result, l);
				}
			}
		}
for(Map.Entry<Integer,Set<String>> entry:list.entrySet()){
			Set<String> l=entry.getValue();
			System.out.print(entry.getKey()+": ");
			for(String s:l){
				System.out.print(s+" ; ");
			}
			System.out.println();
		}

public static void main(String args[])
	{
		Map<Integer,Set<String>> list=new HashMap<Integer,Set<String>>();
		int n=(int) Math.pow(10,2);
		for(int a=0;a<n;a++){
			for(int b=0;b<n;b++){
				Set<String> l;
				int result=(int) (Math.pow(a, 3)+Math.pow(b, 3));
				if(list.containsKey(result)){
					l=list.get(result);
					l.add(a+","+b);
					list.put(result, l);
				}
				else{
					l=new TreeSet<String>();
					l.add(a+","+b);
					list.put(result, l);
				}
			}
		}for(Map.Entry<Integer,Set<String>> entry:list.entrySet()){
			Set<String> l=entry.getValue();
			System.out.print(entry.getKey()+": ");
			for(String s:l){
				System.out.print(s+" ; ");
			}
			System.out.println();
		}
}

Time Complexity = O(n^2)
The idea is to maintain a hash map of integer and pair-of-numbers as shown below.

public class Main {
	
	public static class Pair{
		int a,b;
		Pair(int a, int b){
			this.a = a;
			this.b = b;
		}
		public String toString(){
			return "("+this.a+", "+this.b+")";
		}
	}
	public static void main(String[] args) {
		int n = 100000;
		printTaxiCabNumbers(n);
	}

	private static void printTaxiCabNumbers(int n) {
		HashMap<Integer, Pair> map = new HashMap<Integer, Pair>();
		for(int i=0; i<n; i++){
			for(int j=i+1; j<n; j++){
				int m = (int) (Math.pow(i, 3)+Math.pow(j, 3));
				if(map.containsKey(m)){
					System.out.println(new Pair(i,j).toString()+" and "+map.get(m).toString());
				}else{
					map.put((int) (m), new Pair(i,j));
				}
			}
		}
	}
}

Time Complexity = O(n)
The idea is to maintain a HashMap of sum and Integer-pairs.

public class Main {
	
	public static class Pair{
		int a,b;
		Pair(int a, int b){
			this.a = a;
			this.b = b;
		}
		public String toString(){
			return "("+this.a+", "+this.b+")";
		}
	}
	public static void main(String[] args) {
		int n = 100000;
		printTaxiCabNumbers(n);
	}

	private static void printTaxiCabNumbers(int n) {
		HashMap<Integer, Pair> map = new HashMap<Integer, Pair>();
		for(int i=0; i<n; i++){
			for(int j=i+1; j<n; j++){
				int m = (int) (Math.pow(i, 3)+Math.pow(j, 3));
				if(map.containsKey(m)){
					System.out.println(new Pair(i,j).toString()+" and "+map.get(m).toString());
				}else{
					map.put((int) (m), new Pair(i,j));
				}
			}
		}
	}
}

must it be in collection? or I can write solution while searching solutions.

public class AllSolutions {
    public static void main(String[] args){
        long maxint = (int)Math.pow(10,5);
        long powOfI=0;
        long powOfJ =0;
        for(int i =0;i<=maxint;i++){
            powOfI=(long)Math.pow(i,3);
            for (int j =i;j<=maxint;j++){
                powOfJ=(long)Math.pow(j,3);
                System.out.println("a && c = "+i+", b && d ="+j+", =>  a^3+b^3=c^3+d^3="+(powOfI+ powOfJ));
                System.out.println("a && d = "+i+", b && c ="+j+", =>  a^3+b^3=c^3+d^3="+(powOfI+ powOfJ));
                System.out.println("b && c = "+i+", a && d ="+j+", =>  a^3+b^3=c^3+d^3="+(powOfI+ powOfJ));
                System.out.println("b && d = "+i+", a && c ="+j+", =>  a^3+b^3=c^3+d^3="+(powOfI+ powOfJ));
            }
        }
    }
}