Find pairs of Positive and Negative values present in given array
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 3Input: 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++
// 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 ; // Print the pair with negative positive value. for ( int i = 0; i < v.size(); i++) cout << -v[i] << " " << v[i] << " " ; } // Driver code int main() { int arr[] = { 4, 8, 9, -4, 1, -1, -8, -9 }; int n = sizeof (arr) / sizeof (arr[0]); // Function call printPairs(arr, n); return 0; } |
Java
// 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) + " " ); } // Driver code public static void main(String[] args) { int arr[] = { 4 , 8 , 9 , - 4 , 1 , - 1 , - 8 , - 9 }; int n = arr.length; // Function call printPairs(arr, n); } } // This code is contributed by Prasad Kshirsagar. |
Python 3
# 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 Code if __name__ = = "__main__" : arr = [ 4 , 8 , 9 , - 4 , 1 , - 1 , - 8 , - 9 ] n = len (arr) # Function call printPairs(arr, n) # This code is contributed # by ChitraNayal |
C#
// 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; // Function call printPairs(arr, n); } } // This code has been contributed by 29AjayKumar |
Javascript
<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(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++
// 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 ; for ( int i = 0; i < v.size(); i++) cout << "-" << v[i] << " " << v[i] << " " ; } // Driver code int main() { int arr[] = { 4, 8, 9, -4, 1, -1, -8, -9 }; int n = sizeof (arr) / sizeof (arr[0]); // Function call printPairs(arr, n); return 0; } |
Java
// 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; // Function call printPairs(arr, n); } } // This code is contributed by Prerna Saini |
Python3
# 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 Code if __name__ = = "__main__" : arr = [ 4 , 8 , 9 , - 4 , 1 , - 1 , - 8 , - 9 ] n = len (arr) # Function call printPairs(arr, n) # This code is contributed by RohitOberoi |
C#
// 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; // Function call printPairs(arr, n); } } // This code is contributed by divyeshrabadiya07 |
Javascript
<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 = 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 Program let 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)
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++
// 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) { 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 ; } // Driver code int main() { int arr[] = { 4, 8, 9, -4, 1, -1, -8, -9 }; int n = sizeof (arr) / sizeof (arr[0]); // Function call printPairs(arr, n); return 0; } |
Java
// Java 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 ; } // Driver code public static void main(String[] args) { int [] arr = { 4 , 8 , 9 , - 4 , 1 , - 1 , - 8 , - 9 }; int n = arr.length; // Function call printPairs(arr, n); } } // This code is contributed by Aarti_Rathi |
Python3
# Python3 program to find pairs of # positive and negative values present in # an array 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 # Driver code arr = [ 4 , 8 , 9 , - 4 , 1 , - 1 , - 8 , - 9 ] n = len (arr) # Function call printPairs(arr, n) # This code is contributed by shinjanpatra. |
C#
// C# program to find pairs of positive and // negative values present in an array. using System; using System.Collections; using System.Collections.Generic; public class GFG { // Print pair with negative and positive value 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 ; } // Driver code static public void Main() { int [] arr = { 4, 8, 9, -4, 1, -1, -8, -9 }; int n = arr.Length; // Function call printPairs(arr, n); } } // This code is contributed by lokeshmvs21. |
Javascript
<script> // Efficient 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 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 Program let 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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