Find pairs of Positive and Negative values present in given array
Last Updated :
03 Oct, 2022
Given an array of distinct integers, print all the pairs having both positive and negative values of a number that exists in the array. The pairs can be printed in any order.
Examples:
Input: arr[] = {1, -3, 2, 3, 6, -1}
Output: -1 1 -3 3
Input: arr[] = {4, 8, 9, -4, 1, -1, -8, -9}
Output: -4 4 -8 8 -9 9 -1 1
Naive Approach: To solve the problem follow the below idea:
The idea is to use two nested loops. For each element arr[i], find negative of arr[i] from index i + 1 to n – 1 and store it in another array
Below is the implementation of this approach:
C++
#include <bits/stdc++.h>
using namespace std;
void printPairs( int arr[], int n)
{
vector< int > v;
for ( int i = 0; i < n; i++)
for ( int j = i + 1; j < n; j++)
if ( abs (arr[i]) == abs (arr[j]))
v.push_back( abs (arr[i]));
if (v.size() == 0)
return ;
for ( int i = 0; i < v.size(); i++)
cout << -v[i] << " " << v[i] << " " ;
}
int main()
{
int arr[] = { 4, 8, 9, -4, 1, -1, -8, -9 };
int n = sizeof (arr) / sizeof (arr[0]);
printPairs(arr, n);
return 0;
}
|
Java
import java.lang.*;
import java.util.*;
class GFG {
public static void printPairs( int arr[], int n)
{
Vector<Integer> v = new Vector<Integer>();
for ( int i = 0 ; i < n; i++)
for ( int j = i + 1 ; j < n; j++)
if (Math.abs(arr[i]) == Math.abs(arr[j]))
v.add(Math.abs(arr[i]));
if (v.size() == 0 )
return ;
for ( int i = 0 ; i < v.size(); i++)
System.out.print(-v.get(i) + " " + v.get(i)
+ " " );
}
public static void main(String[] args)
{
int arr[] = { 4 , 8 , 9 , - 4 , 1 , - 1 , - 8 , - 9 };
int n = arr.length;
printPairs(arr, n);
}
}
|
Python 3
def printPairs(arr, n):
v = []
for i in range (n):
for j in range (i + 1 , n):
if ( abs (arr[i]) = = abs (arr[j])):
v.append( abs (arr[i]))
if ( len (v) = = 0 ):
return
for i in range ( len (v)):
print ( - v[i], " ", v[i], end=" ")
if __name__ = = "__main__" :
arr = [ 4 , 8 , 9 , - 4 , 1 , - 1 , - 8 , - 9 ]
n = len (arr)
printPairs(arr, n)
|
C#
using System;
using System.Collections.Generic;
class GFG {
public static void printPairs( int [] arr, int n)
{
List< int > v = new List< int >();
for ( int i = 0; i < n; i++)
for ( int j = i + 1; j < n; j++)
if (Math.Abs(arr[i]) == Math.Abs(arr[j]))
v.Add(Math.Abs(arr[i]));
if (v.Count == 0)
return ;
for ( int i = 0; i < v.Count; i++)
Console.Write(-v[i] + " " + v[i] + " " );
}
public static void Main(String[] args)
{
int [] arr = { 4, 8, 9, -4, 1, -1, -8, -9 };
int n = arr.Length;
printPairs(arr, n);
}
}
|
Javascript
<script>
function printPairs(arr,n)
{
let v = [];
for (let i = 0; i < n; i++)
for (let j = i + 1; j < n; j++)
if (Math.abs(arr[i]) ==
Math.abs(arr[j]))
v.push(Math.abs(arr[i]));
if (v.length == 0)
return ;
for (let i = 0; i < v.length; i++)
document.write(-v[i] + " "
+ v[i] + " " );
}
let arr=[4, 8, 9, -4, 1, -1, -8, -9];
let n = arr.length;
printPairs(arr, n);
</script>
|
Output
-4 4 -8 8 -9 9 -1 1
Time Complexity: O(N2)
Auxiliary Space: O(N)
Find pairs of Positive and Negative values present in given array using hashing:
To solve the problem follow the below idea:
The idea is to use hashing to store count of absolute value of every element present in the array. If the count of any element is equal to 2, then it means that a pair has been found
Follow the given steps to solve the problem:
- Traverse the given array, and increase the count at the absolute value of the hash table.
- If the count becomes 2, store its absolute value in another vector.
- If the size of the vector is 0, print “0”,
- else for each term in the vector print first its negative value and the positive value
Below is the implementation of this approach:
C++
#include <bits/stdc++.h>
using namespace std;
void printPairs( int arr[], int n)
{
vector< int > v;
unordered_map< int , bool > cnt;
for ( int i = 0; i < n; i++) {
if (cnt[ abs (arr[i])] == 0)
cnt[ abs (arr[i])] = 1;
else {
v.push_back( abs (arr[i]));
cnt[ abs (arr[i])] = 0;
}
}
if (v.size() == 0)
return ;
for ( int i = 0; i < v.size(); i++)
cout << "-" << v[i] << " " << v[i] << " " ;
}
int main()
{
int arr[] = { 4, 8, 9, -4, 1, -1, -8, -9 };
int n = sizeof (arr) / sizeof (arr[0]);
printPairs(arr, n);
return 0;
}
|
Java
import java.util.*;
class GFG {
static void printPairs( int arr[], int n)
{
ArrayList<Integer> v = new ArrayList<Integer>();
HashMap<Integer, Integer> cnt
= new HashMap<Integer, Integer>();
for ( int i = 0 ; i < n; i++) {
if (cnt.containsKey(Math.abs(arr[i])))
cnt.put(Math.abs(arr[i]),
cnt.get(Math.abs(arr[i])) + 1 );
else {
cnt.put(Math.abs(arr[i]), 1 );
}
if (cnt.get(Math.abs(arr[i])) == 2 ) {
v.add(Math.abs(arr[i]));
}
}
if (v.size() == 0 )
return ;
for ( int i = 0 ; i < v.size(); i++)
System.out.print( "-" + v.get(i) + " " + v.get(i)
+ " " );
}
public static void main(String[] args)
{
int arr[] = { 4 , 8 , 9 , - 4 , 1 , - 1 , - 8 , - 9 };
int n = arr.length;
printPairs(arr, n);
}
}
|
Python3
def printPairs(arr, n):
s = set ()
ret = []
for i in arr:
if abs (i) in s:
ret.append( abs (i))
else :
s.add( abs (i))
ret.sort()
for i in range ( 0 , len (ret)):
print ( - ret[i], " ", ret[i], end=" ")
if __name__ = = "__main__" :
arr = [ 4 , 8 , 9 , - 4 , 1 , - 1 , - 8 , - 9 ]
n = len (arr)
printPairs(arr, n)
|
C#
using System;
using System.Collections.Generic;
class GFG {
static void printPairs( int [] arr, int n)
{
List< int > v = new List< int >();
Dictionary< int , bool > cnt
= new Dictionary< int , bool >();
for ( int i = 0; i < n; i++) {
int absVal = Math.Abs(arr[i]);
if (!cnt.ContainsKey(absVal)) {
cnt[absVal] = true ;
}
else if (cnt[absVal] == false ) {
cnt[absVal] = true ;
}
else {
v.Add(Math.Abs(arr[i]));
cnt[absVal] = false ;
}
}
if (v.Count == 0)
return ;
v.Sort();
for ( int i = 0; i < v.Count; i++)
Console.Write(-v[i] + " " + v[i] + " " );
}
static void Main()
{
int [] arr = { 4, 8, 9, -4, 1, -1, -8, -9 };
int n = arr.Length;
printPairs(arr, n);
}
}
|
Javascript
<script>
function printPairs(arr,n)
{
let v = new Array();
let cnt = new Map();
for (let i = 0; i < n; i++) {
if (cnt.has(Math.abs(arr[i])) == false )
cnt.set(Math.abs(arr[i]) , 1);
else {
v.push(Math.abs(arr[i]));
cnt. delete (Math.abs(arr[i]));
}
}
if (v.length == 0)
return ;
v.sort((a,b)=>a-b)
for (let i = 0; i < v.length; i++)
document.write(-v[i] + " " + v[i] + " " );
}
let arr = [ 4, 8, 9, -4, 1, -1, -8, -9 ];
let n = arr.length;
printPairs(arr, n);
</script>
|
Output
-4 4 -1 1 -8 8 -9 9
Time Complexity: O(N)
Auxiliary Space: O(N)
Find pairs of Positive and Negative values present in given array using set:
To solve the problem follow the below idea:
The idea is to use a set. Find the negative of the number in the set. If it exits then print both the numbers and if it does not exist then add it to the set
Below is the implementation of this approach:
C++
#include <bits/stdc++.h>
using namespace std;
void printPairs( int arr[], int n)
{
unordered_set< int > hs;
vector< int > ans;
for ( int i = 0; i < n; i++) {
if (hs.find((arr[i]) * -1) != hs.end()) {
if (arr[i] < 0) {
cout << arr[i] << " " ;
cout << (arr[i] * -1) << " " ;
}
else {
cout << (arr[i] * -1) << " " ;
cout << arr[i] << " " ;
}
}
hs.insert(arr[i]);
}
return ;
}
int main()
{
int arr[] = { 4, 8, 9, -4, 1, -1, -8, -9 };
int n = sizeof (arr) / sizeof (arr[0]);
printPairs(arr, n);
return 0;
}
|
Java
import java.util.*;
public class GFG {
public static void printPairs( int [] arr, int n)
{
HashSet<Integer> hs = new HashSet<Integer>();
ArrayList<Integer> ans = new ArrayList<Integer>();
for ( int i = 0 ; i < n; i++) {
if (hs.contains((arr[i]) * - 1 )) {
if (arr[i] < 0 ) {
System.out.print(arr[i]);
System.out.print( " " );
System.out.print((arr[i] * - 1 ));
System.out.print( " " );
}
else {
System.out.print((arr[i] * - 1 ));
System.out.print( " " );
System.out.print(arr[i]);
System.out.print( " " );
}
}
hs.add(arr[i]);
}
return ;
}
public static void main(String[] args)
{
int [] arr = { 4 , 8 , 9 , - 4 , 1 , - 1 , - 8 , - 9 };
int n = arr.length;
printPairs(arr, n);
}
}
|
Python3
def printPairs(arr, n):
hs = set ()
ans = []
for i in range (n):
if (arr[i] * - 1 ) in hs:
if (arr[i] < 0 ):
print (arr[i], end = " " )
print ((arr[i] * - 1 ), end = " " )
else :
print ((arr[i] * - 1 ), end = " " )
print (arr[i], end = " " )
hs.add(arr[i])
return
arr = [ 4 , 8 , 9 , - 4 , 1 , - 1 , - 8 , - 9 ]
n = len (arr)
printPairs(arr, n)
|
C#
using System;
using System.Collections;
using System.Collections.Generic;
public class GFG {
public static void printPairs( int [] arr, int n)
{
HashSet< int > hs = new HashSet< int >();
for ( int i = 0; i < n; i++) {
if (hs.Contains((arr[i]) * -1)) {
if (arr[i] < 0) {
Console.Write(arr[i]);
Console.Write( " " );
Console.Write((arr[i] * -1));
Console.Write( " " );
}
else {
Console.Write((arr[i] * -1));
Console.Write( " " );
Console.Write(arr[i]);
Console.Write( " " );
}
}
hs.Add(arr[i]);
}
return ;
}
static public void Main()
{
int [] arr = { 4, 8, 9, -4, 1, -1, -8, -9 };
int n = arr.Length;
printPairs(arr, n);
}
}
|
Javascript
<script>
function printPairs(arr, n)
{
let hs = new Set();
let ans = new Array();
for (let i = 0 ; i < n ; i++ ){
if ( hs.has((arr[i])*-1) == true ){
if (arr[i] < 0){
document.write(arr[i], " " );
document.write((arr[i]*-1), " " );
} else {
document.write((arr[i]*-1), " " );
document.write(arr[i], " " );
}
}
hs.add(arr[i]) ;
}
return ;
}
let arr = [ 4, 8, 9, -4, 1, -1, -8, -9 ];
let n = arr.length;
printPairs(arr, n);
</script>
|
Output
-4 4 -1 1 -8 8 -9 9
Time Complexity: O(N)
Auxiliary Space: O(N)
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