Find pairs of Positive and Negative values present in given array

27 Jun, 2022
Given an array of distinct integers, print all the pairs having positive value and negative value 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 
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, print negative positive value of the stored element.
Below is the implementation of this approach:  
C++
Java
Python 3
C#
Javascript
// 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 valuevoid 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;     // Print the pair with negative positive value.    for (int i = 0; i < v.size(); i++)        cout << -v[i] << " " << v[i] << " ";} // Driven Programint main(){    int arr[] = { 4, 8, 9, -4, 1, -1, -8, -9 };    int n = sizeof(arr) / sizeof(arr[0]);    printPairs(arr, n);    return 0;}
// Java program to find pairs of positive// and negative values present in an array.import java.lang.*;import java.util.*; 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;         // 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.
# 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     # Print the pair with negative    # positive value.    for i in range(len(v)):        print(-v[i], "", v[i], end=" ")  # Driver Codeif __name__ == "__main__":    arr = [4, 8, 9, -4, 1, -1, -8, -9]    n = len(arr)    printPairs(arr, n) # This code is contributed# by ChitraNayal
// 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;         // 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
<script>// Javascript program to find pairs of positive// and negative values present in an array.         // Print pair with negative and positive value    function printPairs(arr,n)    {        let  v = [];                 // For each element of array.        for (let i = 0; i < n; i++)              // Try to find the negative value of            // arr[i] from i + 1 to n            for (let j = i + 1; j < n; j++)                  // If absolute values are equal                // print pair.                if (Math.abs(arr[i]) ==                                  Math.abs(arr[j]))                    v.push(Math.abs(arr[i]));            // If size of vector is 0, therefore there        // is no element with positive negative        // value, print "0"        if (v.length == 0)            return;                      // Print the pair with negative positive        // value.        for (let i = 0; i < v.length; i++)            document.write(-v[i] + " "                          + v[i] + " ");    }         // Driven Program    let arr=[4, 8, 9, -4, 1, -1, -8, -9];    let n = arr.length;    printPairs(arr, n);         // This code is contributed by rag2127</script>
-4 4 -8 8 -9 9 -1 1 
Time Complexity: O(n*n)Auxiliary Space : O(n)
Method 2 (Hashing) 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. If the size of the vector is 0, print “0”, else for each term in vector print first its negative value and the positive value.
Below is the implementation of this approach:
C++
Java
Python3
C#
Javascript
// 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 valuevoid 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;     for (int i = 0; i < v.size(); i++)        cout << "-" << v[i] << " " << v[i] << " ";} // Driven Programint main(){    int arr[] = { 4, 8, 9, -4, 1, -1, -8, -9 };    int n = sizeof(arr) / sizeof(arr[0]);     printPairs(arr, n);     return 0;}
// 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 encounter early,            // then increment its count            if (cnt.containsKey(Math.abs(arr[i])))                cnt.put(Math.abs(arr[i]),                        cnt.get(Math.abs(arr[i])) + 1);            // If element has not seen before,            // then initialize its count to 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)                             + " ");    }    // 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
# Efficient Python3 program to find pairs of# positive and negative values present in# an array.# Print pair with negative and# positive value  def printPairs(arr, n):  s = set()ret = [] # For each element of array.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=" ")# Driver Codeif __name__ == "__main__": arr = [4, 8, 9, -4, 1, -1, -8, -9]n = len(arr) printPairs(arr, n)# This code is contributed by RohitOberoi
// Efficient 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    static void printPairs(int[] arr, int n)    {        List<int> v = new List<int>();        Dictionary<int, bool> cnt            = new Dictionary<int, bool>();         // For each element of array.        for (int i = 0; i < n; i++) {             // If element has not encounter early,            // mark it on cnt array.            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] + " ");    }    // Driver code    static void Main()    {        int[] arr = { 4, 8, 9, -4, 1, -1, -8, -9 };        int n = arr.Length;        printPairs(arr, n);    }}// This code is contributed by divyeshrabadiya07
<script>// Efficient JavaScript program to find pairs of// positive and negative values present in// an array.// Print pair with negative and positive valuefunction printPairs(arr,n){let v = new Array();let cnt = new Map();// For each element of array.for (let i = 0; i < n; i++) {// If element has not encounter early,// mark it on cnt array.if (cnt.has(Math.abs(arr[i])) == false)cnt.set(Math.abs(arr[i]) , 1);// If seen before, push it in vector (// given that elements are distinct)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] + " ");}// Driven Programlet arr = [ 4, 8, 9, -4, 1, -1, -8, -9 ];let n = arr.length;printPairs(arr, n);// This code is contributed by shinjanpatra.</script>
-4 4 -1 1 -8 8 -9 9 
Time Complexity: O(n)Auxiliary Space: O(n)
Method 3 (Using set) : 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 exits then add it on to the set.
Below is the implementation of this approach:
C++
Java
Python3
Javascript
// 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 valuevoid 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;}// Driven Programint main(){    int arr[] = { 4, 8, 9, -4, 1, -1, -8, -9 };    int n = sizeof(arr) / sizeof(arr[0]);    printPairs(arr, n);    return 0;}
// Efficient CPP program to find pairs of// positive and negative values present in// an array.import java.util.*; public class GFG {    // Print pair with negative and positive value    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;    }    // 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 Aarti_Rathi
# Print pair with negative and positive valuedef 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 # Driven Programarr = [ 4, 8, 9, -4, 1, -1, -8, -9 ]n = len(arr)printPairs(arr, n) # This code is contributed by shinjanpatra.
<script>// Efficient JavaScript program to find pairs of// positive and negative values present in// an array.// Print pair with negative and positive valuefunction 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 ;}// Driver Programlet arr = [ 4, 8, 9, -4, 1, -1, -8, -9 ];let n = arr.length;printPairs(arr, n);// This code is contributed by shinjanpatra.</script>
-4 4 -1 1 -8 8 -9 9 
Time Complexity: O(n)Auxiliary Space : O(n)
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