Given an array A of size N and an integer B, the task is to find all unique quadruplets arranged in the lexicographically increasing order such that the sum of the elements of each quadrupled is B and the gcd of the absolute values of all elements of a quadrupled is 1.
Example:
Input: A = {1, 0, -1, 0, -2, 2 }, B = 0
Output:
-2 -1 1 2
-1 0 0 1
Explanation: There are only three unique quadruplets which have sum = 0 which are {{-2, 0, 0, 2}, {-2, -1, 1, 2}, {-1, 0, 0, 1}} and out of these quadruplets only the second and the third quadrupled have gcd equal to 1.Input: A = { 1, 5, 1, 0, 6, 0 }, B = 7
Output:
0 0 1 6
0 1 1 5
Approach: The idea is to store the sum of each pair of elements in a hashmap, then iterate through all pairs of elements and then lookup the hashmap to find pairs such that the sum of the quadrupled becomes equal to B and the gcd of absolute values of all elements of the quadrupled is equal to 1. The detailed approach using hashmap has been discussed in this article. Follow the steps to solve the problem.
- Insert the sum of each pair of elements of the array into a hashmap mp.
- Initialize a set st, to store all the quadruplets.
-
Traverse the array from i = 0 to N-1
- Traverse the array from j = i+1 to N-1
- Find all pairs from the hashmap whose sum is equal to B- A[i]-A[j]. Initialize a vector of pairs v to mp[B-A[i]-A[j]].
- Traverse through the vector v using a variable k
- If v[k].first or v[k].second is equal to either i or j then continue to the next iteration.
- Store the elements of the quadrupled in a temp array. Sort the temp array. If the gcd of all elements of temp array is 1 then insert the temp array into st.
- Traverse the array from j = i+1 to N-1
- Traverse through the set st and print all the quadruplets.
// C++ code for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find all // quadruplets with sum B. void find4Sum( int A[], int N, int B)
{ // Hashmap to store sum
// of all pairs
unordered_map< int ,
vector<pair< int , int > > >
mp;
// Set to store all quadruplets
set<vector< int > > st;
// Traverse the array
for ( int i = 0; i < N - 1; i++) {
// Traverse the array
for ( int j = i + 1; j < N; j++) {
int sum = A[i] + A[j];
// Insert sum of
// current pair into
// the hashmap
mp[sum].push_back({ i, j });
}
}
// Traverse the array
for ( int i = 0; i < N - 1; i++) {
// Traverse the array
for ( int j = i + 1; j < N; j++) {
int sum = A[i] + A[j];
// Lookup the hashmap
if (mp.find(B - sum) != mp.end()) {
vector<pair< int , int > > v
= mp[B - sum];
for ( int k = 0; k < v.size(); k++) {
pair< int , int > it = v[k];
if (it.first != i && it.second != i
&& it.first != j
&& it.second != j) {
vector< int > temp;
temp.push_back(A[i]);
temp.push_back(A[j]);
temp.push_back(A[it.first]);
temp.push_back(A[it.second]);
// Stores the gcd of the
// quadrupled
int gc = abs (temp[0]);
gc = __gcd( abs (temp[1]), gc);
gc = __gcd( abs (temp[2]), gc);
gc = __gcd( abs (temp[3]), gc);
// Arrange in
// ascending order
sort(temp.begin(), temp.end());
// Insert into set if gcd is 1
if (gc == 1)
st.insert(temp);
}
}
}
}
}
// Iterate through set
for ( auto it = st.begin(); it != st.end(); it++) {
vector< int > temp = *it;
// Print the elements
for ( int i = 0; i < 4; i++) {
cout << temp[i] << " " ;
}
cout << endl;
}
} // Driver Code int main()
{ // Input
int N = 6;
int A[6]
= { 1, 0, -1, 0, -2, 2 };
int B = 0;
// Function Call
find4Sum(A, N, B);
return 0;
} |
import java.util.*;
public class Main {
// Function to find all quadruplets with sum B.
static void find4Sum( int [] A, int N, int B)
{
// Hashmap to store sum of all pairs
Map<Integer, List<Pair<Integer, Integer> > > mp
= new HashMap<>();
// Set to store all quadruplets
Set<List<Integer> > st = new HashSet<>();
// Traverse the array
for ( int i = 0 ; i < N - 1 ; i++) {
// Traverse the array
for ( int j = i + 1 ; j < N; j++) {
int sum = A[i] + A[j];
// Insert sum of current pair into the
// hashmap
if (!mp.containsKey(sum)) {
mp.put(sum, new ArrayList<>());
}
mp.get(sum).add( new Pair<>(i, j));
}
}
// Traverse the array
for ( int i = 0 ; i < N - 1 ; i++) {
// Traverse the array
for ( int j = i + 1 ; j < N; j++) {
int sum = A[i] + A[j];
// Lookup the hashmap
if (mp.containsKey(B - sum)) {
List<Pair<Integer, Integer> > v
= mp.get(B - sum);
for (Pair<Integer, Integer> it : v) {
if (it.getKey() != i
&& it.getValue() != i
&& it.getKey() != j
&& it.getValue() != j) {
List<Integer> temp
= new ArrayList<>();
temp.add(A[i]);
temp.add(A[j]);
temp.add(A[it.getKey()]);
temp.add(A[it.getValue()]);
// Stores the gcd of the
// quadrupled
int gc = Math.abs(temp.get( 0 ));
gc = gcd(Math.abs(temp.get( 1 )),
gc);
gc = gcd(Math.abs(temp.get( 2 )),
gc);
gc = gcd(Math.abs(temp.get( 3 )),
gc);
// Arrange in ascending order
Collections.sort(temp);
// Insert into set if gcd is 1
if (gc == 1 ) {
st.add(temp);
}
}
}
}
}
}
// Iterate through set
for (List<Integer> temp : st) {
// Print the elements
for ( int i = 0 ; i < 4 ; i++) {
System.out.print(temp.get(i) + " " );
}
System.out.println();
}
}
static int gcd( int a, int b)
{
if (b == 0 )
return a;
else
return gcd(b, Math.abs(a - b));
}
static class Pair<T, U> {
T key;
U value;
public Pair(T key, U value)
{
this .key = key;
this .value = value;
}
public T getKey() { return key; }
public U getValue() { return value; }
}
// Driver Code
public static void main(String[] args)
{
// Input
int N = 6 ;
int [] A = { 1 , 0 , - 1 , 0 , - 2 , 2 };
int B = 0 ;
// Function Call
find4Sum(A, N, B);
}
} |
# Python3 code for the above approach from math import gcd
# Function to find all # quadruplets with sum B. def find4Sum( A, N, B):
# Hashmap to store sum
# of all pairs
mp = dict ()
# Set to store all quadruplets
st = set ()
# Traverse the array
for i in range (N - 1 ):
for j in range (i + 1 , N):
sum = A[i] + A[j]
# Insert sum of
# current pair into
# the hashmap
if sum not in mp:
mp[ sum ] = []
mp[ sum ].append([ i, j ])
# Traverse the array
for i in range (N - 1 ):
# Traverse the array
for j in range (i + 1 , N):
sum = A[i] + A[j]
# Lookup the hashmap
if (B - sum ) in mp:
v = mp[B - sum ]
for k in range ( len (v)):
it = v[k]
if it[ 0 ] ! = i and it[ 1 ] ! = i and it[ 0 ] ! = j and it[ 1 ] ! = j:
temp = (A[i], A[j], A[it[ 0 ]], A[it[ 1 ]])
# Stores the gcd of the
# quadrupled
gc = abs (temp[ 0 ]);
gc = gcd( abs (temp[ 1 ]), gc);
gc = gcd( abs (temp[ 2 ]), gc);
gc = gcd( abs (temp[ 3 ]), gc);
# Arrange in
# ascending order
# Insert into set if gcd is 1
if (gc = = 1 ):
st.add( tuple ( sorted (temp)))
# Iterate through set
for it in st:
temp = it;
# Print the elements
print ( * it)
# Driver Code # Input N = 6 ;
A = [ 1 , 0 , - 1 , 0 , - 2 , 2 ];
B = 0 ;
# Function Call find4Sum(A, N, B); # This code is contributed by phasing17 |
// C# code for the above approach using System;
using System.Collections.Generic;
class Program {
static int gcd( int a, int b)
{
if (a == 0) {
return b;
}
return gcd(b % a, a);
}
// Function to find all
// quadruplets with sum B.
static void find4Sum( int [] A, int N, int B)
{
// Hashmap to store sum of all pairs
Dictionary< int , List<Tuple< int , int > > > mp
= new Dictionary< int ,
List<Tuple< int , int > > >();
// Hashset to store all quadruplets
HashSet<Tuple< int , int , int , int > > st
= new HashSet<Tuple< int , int , int , int > >();
// Traverse the array
for ( int i = 0; i < N - 1; i++) {
for ( int j = i + 1; j < N; j++) {
int sum = A[i] + A[j];
// Insert sum of current pair into the
// hashmap
if (!mp.ContainsKey(sum)) {
mp[sum] = new List<Tuple< int , int > >();
}
mp[sum].Add( new Tuple< int , int >(i, j));
}
}
// Traverse the array
for ( int i = 0; i < N - 1; i++) {
// Traverse the array
for ( int j = i + 1; j < N; j++) {
int sum = A[i] + A[j];
// Lookup the hashmap
if (mp.ContainsKey(B - sum)) {
List<Tuple< int , int > > v = mp[B - sum];
foreach (Tuple< int , int > it in v)
{
if (it.Item1 != i && it.Item2 != i
&& it.Item1 != j
&& it.Item2 != j) {
Tuple< int , int , int , int > temp
= Tuple.Create(A[i], A[j],
A[it.Item1],
A[it.Item2]);
// Stores the gcd of the
// quadrupled
int gc = Math.Abs(temp.Item1);
gc = gcd(Math.Abs(temp.Item2),
gc);
gc = gcd(Math.Abs(temp.Item3),
gc);
gc = gcd(Math.Abs(temp.Item4),
gc);
// Arrange in ascending order
// Insert into hashset if gcd is
// 1
if (gc == 1) {
int [] arr = { temp.Item1,
temp.Item2,
temp.Item3,
temp.Item4 };
Array.Sort(arr);
st.Add(Tuple.Create(
arr[0], arr[1], arr[2],
arr[3]));
}
}
}
}
}
}
// Iterate through hashset
foreach (Tuple< int , int , int , int > it in st)
{
// Print the elements
Console.WriteLine( "{0} {1} {2} {3}" , it.Item1,
it.Item2, it.Item3, it.Item4);
}
}
// Driver Code
static void Main( string [] args)
{
// Input
int N = 6;
int [] A = { 1, 0, -1, 0, -2, 2 };
int B = 0;
// Function Call
find4Sum(A, N, B);
}
} // This code is contributed by Pushpesh Raj. |
// Function to find gcd of two numbers function gcd(a, b) {
if (a === 0) {
return b;
}
return gcd(b % a, a);
} // Function to find all unique quadruplets with // sum B arranged in the lexicographically increasing order function find4Sum(A, N, B) {
// Hashmap to store sum of all pairs
const mp = new Map();
// Set to store all quadruplets
const st = new Set();
// Traverse the array
for (let i = 0; i < N - 1; i++) {
for (let j = i + 1; j < N; j++) {
const sum = A[i] + A[j];
// Insert sum of current pair into the hashmap
if (!mp.has(sum)) {
mp.set(sum, []);
}
mp.get(sum).push([i, j]);
}
}
// Traverse the array
for (let i = 0; i < N - 1; i++) {
// Traverse the array
for (let j = i + 1; j < N; j++) {
const sum = A[i] + A[j];
// Lookup the hashmap
if (mp.has(B - sum)) {
const v = mp.get(B - sum);
for (let k = 0; k < v.length; k++) {
const it = v[k];
if (
it[0] !== i &&
it[1] !== i &&
it[0] !== j &&
it[1] !== j
) {
const temp = [A[i], A[j], A[it[0]], A[it[1]]];
// Stores the gcd of the quadrupled
let gc = Math.abs(temp[0]);
gc = gcd(Math.abs(temp[1]), gc);
gc = gcd(Math.abs(temp[2]), gc);
gc = gcd(Math.abs(temp[3]), gc);
// Arrange in ascending order
temp.sort((a, b) => a - b);
// Insert into set if gcd is 1 and the quadruplet
// is lexicographically increasing
if (gc === 1 && !st.has(temp.join( ' ' ))) {
st.add(temp.join( ' ' ));
console.log(...temp);
}
}
}
}
}
}
} // Driver Code // Input const N = 6; const A = [1, 0, -1, 0, -2, 2]; const B = 0; // Function Call find4Sum(A, N, B); |
-1 0 0 1 -2 -1 1 2
Time Complexity: O(N^3)
Auxiliary Space: O(N^2)