Print all the pairs that contains the positive and negative values of an element
Given an array of distinct integers, print all the pairs having a positive value and negative value of a number that exists in the array.
Note: Order of the pairs doesn’t matter.
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
A naive approach is to run two loops i.e. Consider each element of the array using the outer loop and search for its corresponding positive/negative value in the array using an inner loop. Similarly, find all the pairs. Time Complexity of this approach will be O( n2).
A better approach is to use sorting i.e. first sort the array and then for each negative element, do a binary search to find its counterpart (+ve number). If found, print that pair. If the current element is positive then break that loop as after that there will be all the positive numbers.
C++
// CPP program to find pairs of positive// and negative values present in an array.#include <bits/stdc++.h>using namespace std; void printPairs(int arr[], int n){ bool pair_exists = false; // Sort the array sort(arr, arr + n); // Traverse the array for (int i = 0; i < n; i++) { // For every arr[i] < 0 element, // do a binary search for arr[i] > 0. if (arr[i] < 0) { // If found, print the pair. if (binary_search(arr, arr + n, -arr[i])) { cout << arr[i] << ", " << -arr[i] << endl; pair_exists = true; } } else break; } if (pair_exists == false) cout << "No such pair exists";} // Driver codeint main(){ int arr[] = { 4, 8, 9, -4, 1, -1, -8, -9 }; int n = sizeof(arr) / sizeof(arr[0]); 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{static void printPairs(int arr[], int n){ boolean pair_exists = false; // Sort the array Arrays.sort(arr); // Traverse the array for (int i = 0; i < n; i++) { // For every arr[i] < 0 element, // do a binary search for arr[i] > 0. if (arr[i] < 0) { // If found, print the pair. if (java.util.Arrays.binarySearch(arr, -arr[i])!=-1) { System.out.println(arr[i] + ", " + -arr[i] ); pair_exists = true; } } else break; } if (pair_exists == false) System.out.println("No such pair exists");} // Driver codepublic 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 Arnab Kundu |
Python 3
# Python3 program to find pairs of positive # and negative values present in an array. # function for binary searchdef binary_search(a, x, lo=0, hi=None): if hi is None: hi = len(a) while lo < hi: mid = (lo+hi)//2 midval = a[mid] if midval < x: lo = mid+1 elif midval > x: hi = mid else: return mid return -1 def printPairs(arr, n): pair_exists = False # Sort the array arr.sort() # Traverse the array for i in range(n): # For every arr[i] < 0 element, # do a binary search for arr[i] > 0. if (arr[i] < 0): # If found, print the pair. if (binary_search(arr,-arr[i])): print(arr[i] , ", " , -arr[i]) pair_exists = True else: break if (pair_exists == False): print("No such pair exists") # 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 ash264 |
C#
// C# program to find pairs // of positive and negative // values present in an array. using System;public class GFG{ static void printPairs(int []arr, int n) { bool pair_exists = false; // Sort the array Array.Sort(arr); // Traverse the array for (int i = 0; i < n; i++) { // For every arr[i] < 0 element, // do a binary search for arr[i] > 0. if (arr[i] < 0) { // If found, print the pair. if (Array.BinarySearch(arr, -arr[i])!=-1) { Console.WriteLine(arr[i] + ", " + -arr[i] ); pair_exists = true; } } else break; } if (pair_exists == false) Console.WriteLine("No such pair exists"); } // Driver code public 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 29AjayKumar |
Output:
-9, 9 -8, 8 -4, 4 -1, 1
Time Complexity: O(nlogn)
An efficient Approach is to use hashing. Below are the required steps:
- Start traversing the array.
- Store all the psoitve values in an unordered_set.
- Check for each negative element, if their corresponding positive element exists in the set or not.
- If yes, print the pair
- Also, maintain a flag to check if no such pair exists.
C++
// CPP program to find pairs of positive// and negative values present in an array#include <bits/stdc++.h>using namespace std; // Function to print pairs of positive// and negative values present in the arrayvoid printPairs(int arr[], int n){ unordered_set<int> pairs; bool pair_exists = false; // Store all the positive elements // in the unordered_set for (int i = 0; i < n; i++) if (arr[i] > 0) pairs.insert(arr[i]); // Start traversing the array for (int i = 0; i < n; i++) { // Check if the positive value of current // element exists in the set or not if (arr[i] < 0) if (pairs.find(-arr[i]) != pairs.end()) { // Print that pair cout << arr[i] << ", " << -arr[i] << endl; pair_exists = true; } } if (pair_exists == false) cout << "No such pair exists";} // Driver codeint main(){ int arr[] = { 4, 8, 9, -4, 1, -1, -8, -9 }; int n = sizeof(arr) / sizeof(arr[0]); 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{ // Function to print pairs of positive // and negative values present in the array static void printPairs(int arr[], int n) { Set<Integer> pairs = new HashSet<Integer>(); boolean pair_exists = false; // Store all the positive elements // in the unordered_set for (int i = 0; i < n; i++) if (arr[i] > 0) pairs.add(arr[i]); // Start traversing the array for (int i = 0; i < n; i++) { // Check if the positive value of current // element exists in the set or not if (arr[i] < 0) if (pairs.contains(-arr[i])) { // Print that pair System.out.println(arr[i] + ", " + -arr[i]); pair_exists = true; } } if (pair_exists == false) System.out.println("No such pair exists"); } // 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 Arnab Kundu |
Python3
# Python3 program to find pairs of positive # and negative values present in an array # Function to print pairs of positive # and negative values present in the array def printPairs(arr, n): pairs = set() pair_exists = False # Store all the positive elements # in the unordered_set for i in range(0, n): if arr[i] > 0: pairs.add(arr[i]) # Start traversing the array for i in range(0, n): # Check if the positive value of current # element exists in the set or not if arr[i] < 0: if (-arr[i]) in pairs: # Print that pair print("{}, {}".format(arr[i], -arr[i])) pair_exists = True if pair_exists == False: print("No such pair exists") # 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 Rituraj Jain |
C#
// C# program to find pairs of positive // and negative values present in an array using System;using System.Collections.Generic; class GFG{ // Function to print pairs of positive // and negative values present in the array static void printPairs(int []arr, int n) { HashSet<int> pairs = new HashSet<int>(); bool pair_exists = false; // Store all the positive elements // in the unordered_set for (int i = 0; i < n; i++) if (arr[i] > 0) pairs.Add(arr[i]); // Start traversing the array for (int i = 0; i < n; i++) { // Check if the positive value of current // element exists in the set or not if (arr[i] < 0) if (pairs.Contains(-arr[i])) { // Print that pair Console.WriteLine(arr[i] + ", " + -arr[i]); pair_exists = true; } } if (pair_exists == false) Console.WriteLine("No such pair exists"); } // 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 PrinciRaj1992 |
Output:
-4, 4 -1, 1 -8, 8 -9, 9
Time Complexity: O(n)
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