Pairs of Positive Negative values in an array

Given an array of distinct integers, print all the pairs having positive value and negative value of a number that exists in the array. We need to print pairs in order of their occurrences. A pair whose any element appears first should be printed first.

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 : -1 1 -4 4 -8 8 -9 9

Method 1 (Simple : O(n2))
The idea is to use two nested loop. For each element arr[i], find negative of arr[i] from index i + 1 to n – 1 and store it in another array. For output, sort the stored element and print negative positive value of the stored element.



Below is the implementation of this approach:

C++

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// Simple CPP program to find pairs of positive
// and negative values present in an array.
#include <bits/stdc++.h>
using namespace std;
  
// Print pair with negative and positive value
void printPairs(int arr[], int n)
{
    vector<int> v;
  
    // For each element of array.
    for (int i = 0; i < n; i++) 
  
        // Try to find the negative value of 
        // arr[i] from i + 1 to n
        for (int j = i + 1; j < n; j++) 
  
            // If absolute values are equal print pair.
            if (abs(arr[i]) == abs(arr[j])) 
                v.push_back(abs(arr[i]));      
  
    // If size of vector is 0, therefore there is no 
    // element with positive negative value, print "0"
    if (v.size() == 0)
       return;
  
    // Sort the vector
    sort(v.begin(), v.end());
  
    // Print the pair with negative positive value.
    for (int i = 0; i < v.size(); i++)
        cout << -v[i] << " " << v[i];    
}
  
// Driven Program
int main()
{
    int arr[] = { 4, 8, 9, -4, 1, -1, -8, -9 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printPairs(arr, n);
    return 0;
}

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Java

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// Java program to find pairs of positive
// and negative values present in an array.
import java.util.*;
import java.lang.*;
  
class GFG {
  
    // Print pair with negative and positive value
    public static void printPairs(int arr[] , int n)
    {
        Vector<Integer> v = new Vector<Integer>();
        // For each element of array.
        for (int i = 0; i < n; i++) 
  
            // Try to find the negative value of 
            // arr[i] from i + 1 to n
            for (int j = i + 1; j < n; j++) 
  
                // If absolute values are equal
                // print pair.
                if (Math.abs(arr[i]) ==
                                  Math.abs(arr[j])) 
                    v.add(Math.abs(arr[i])); 
  
  
        // If size of vector is 0, therefore there
        // is no element with positive negative 
        // value, print "0"
        if (v.size() == 0)
            return;     
      
        // Sort the vector
        Collections.sort(v);
  
        // Print the pair with negative positive
        // value.
        for (int i = 0; i < v.size(); i++)
            System.out.print(-v.get(i) + " "
                                      v.get(i)); 
    }
  
    // Driven Program
    public static void main(String[] args)
    {
        int arr[] = { 4, 8, 9, -4, 1, -1, -8, -9 };
        int n = arr.length;
        printPairs(arr, n);
    }
}
  
// This code is contributed by Prasad Kshirsagar.

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Python 3

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# Simple Python 3 program to find 
# pairs of positive and negative 
# values present in an array.
  
# Print pair with negative and 
# positive value
def printPairs(arr, n):
    v = []
  
    # For each element of array.
    for i in range(n): 
  
        # Try to find the negative value 
        # of arr[i] from i + 1 to n
        for j in range( i + 1,n) :
  
            # If absolute values are 
            # equal print pair.
            if (abs(arr[i]) == abs(arr[j])) :
                v.append(abs(arr[i])) 
  
    # If size of vector is 0, therefore 
    # there is no element with positive
    # negative value, print "0"
    if (len(v) == 0):
        return;
  
    # Sort the vector
    v.sort()
  
    # Print the pair with negative 
    # positive value.
    for i in range(len( v)):
        print(-v[i], "" , v[i], end = " ") 
  
# Driver Code
if __name__ == "__main__":
    arr = [ 4, 8, 9, -4, 1, -1, -8, -9 ]
    n = len(arr)
    printPairs(arr, n)
  
# This code is contributed 
# by ChitraNayal

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C#

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// C# program to find pairs of positive
// and negative values present in an array.
using System;
using System.Collections.Generic;
  
class GFG 
{
  
    // Print pair with negative and positive value
    public static void printPairs(int []arr , int n)
    {
        List<int> v = new List<int>();
          
        // For each element of array.
        for (int i = 0; i < n; i++) 
  
            // Try to find the negative value of 
            // arr[i] from i + 1 to n
            for (int j = i + 1; j < n; j++) 
  
                // If absolute values are equal
                // print pair.
                if (Math.Abs(arr[i]) ==
                                Math.Abs(arr[j])) 
                    v.Add(Math.Abs(arr[i])); 
  
  
        // If size of vector is 0, therefore there
        // is no element with positive negative 
        // value, print "0"
        if (v.Count == 0)
            return;     
      
        // Sort the vector
        v.Sort();
  
        // Print the pair with negative positive
        // value.
        for (int i = 0; i < v.Count; i++)
            Console.Write(-v[i] + " "
                                    v[i]); 
    }
  
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = { 4, 8, 9, -4, 1, -1, -8, -9 };
        int n = arr.Length;
        printPairs(arr, n);
    }
}
  
// This code has been contributed by 29AjayKumar

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Output:

-1 1-4 4-8 8-9 9

Method 2 (Hahing)
The idea is to use hashing. Traverse the given array, increase the count at absolute value of hash table. If count becomes 2, store its absolute value in another vector. And finally sort the vector. If the size of the vector is 0, print “0”, else for each term in vector print first its negative value and the the positive value.

Below is the implementation of this approach:

C++

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// Efficient CPP program to find pairs of 
// positive and negative values present in
// an array.
#include <bits/stdc++.h>
using namespace std;
  
// Print pair with negative and positive value
void printPairs(int arr[], int n)
{
    vector<int> v;
    unordered_map<int, bool> cnt;
  
    // For each element of array.
    for (int i = 0; i < n; i++) {
  
        // If element has not encounter early,
        // mark it on cnt array.
        if (cnt[abs(arr[i])] == 0)
            cnt[abs(arr[i])] = 1;
  
        // If seen before, push it in vector (
        // given that elements are distinct)
        else {
            v.push_back(abs(arr[i]));
            cnt[abs(arr[i])] = 0;
        }
    }
  
    if (v.size() == 0)
        return;
  
    sort(v.begin(), v.end());
    for (int i = 0; i < v.size(); i++)
        cout << -v[i] << " " << v[i] << " ";
}
  
// Driven Program
int main()
{
    int arr[] = { 4, 8, 9, -4, 1, -1, -8, -9 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printPairs(arr, n);
    return 0;
}

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Java

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// Efficient Java program to find pairs of 
// positive and negative values present in 
// an array. 
import java.util.*;
  
class GFG 
  
// Print pair with negative 
// and positive value 
static void printPairs(int arr[], int n) 
    ArrayList<Integer> v = new ArrayList<Integer> (); 
    HashMap<Integer, 
            Integer> cnt = new HashMap<Integer, 
                                       Integer>(); 
  
    // For each element of array. 
    for (int i = 0; i < n; i++) 
    
  
        // If element has not encounter early, 
        // mark it on cnt array. 
        if (cnt.containsKey(Math.abs(arr[i]))) 
            cnt.put(Math.abs(arr[i]), 1); 
  
        // If seen before, push it in vector 
        // (given that elements are distinct) 
        else 
        
            v.add(Math.abs(arr[i])); 
            cnt.put(Math.abs(arr[i]), 0); 
        
    
  
    if (v.size() == 0
        return
  
    Collections.sort(v); 
    for (int i = 0; i < v.size(); i++) 
        System.out.print("-" + v.get(i) + 
                         " " + v.get(i) + " "); 
  
// Driver Code
public static void main(String[] args) 
    int arr[] = { 4, 8, 9, -4, 1, -1, -8, -9 }; 
    int n = arr.length; 
    printPairs(arr, n); 
}
  
// This code is contributed by Prerna Saini

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Output:

-1 1 -4 4 -8 8 -9 9

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