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Divide array into two arrays which does not contain any pair with sum K
  • Last Updated : 13 Nov, 2020

Given an array arr[] consisting of N non-negative distinct integers and an integer K, the task is to distribute the array in two arrays such that both the arrays does not contain a pair with sum K.

Examples:

Input: arr[] = {1, 0, 2, 3, 4, 7, 8, 9}, K = 4
Output: 
3, 2, 4, 7, 8, 9 
0, 1 
Explanation: Pairs (1, 3) and (0, 4) from the given array cannot be placed in the same array. Therefore, 0, 1 can be placed in an array and 3, 4 can be placed in the other array. The remaining array elements can be placed in any of the two arrays.

Input: arr[] = {0, 1, 2, 4, 5, 6, 7, 8, 9}, K = 7
Output: 
0, 1, 2, 4 
5, 6, 7, 9, 8 
 

Approach: The idea is to traverse the array and place the array elements greater than K / 2 in one array and the remaining elements in the other array. Follow the steps below to solve the problem:



  • Initialize two separate vectors first and second to store the two distributed arrays.
  • Since all the array elements are distinct, elements greater than K/2 can be stored in one array and the remaining elements in the other.
  • Traverse the given array and for each element, check if arr[i] is greater than K/2 or not. If found to be true, insert that element into vector second. Otherwise, insert it into vector first.
  • After complete traversal of the array, print elements in both the vectors.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to split the given
// array into two separate arrays
// satisfying given condition
void splitArray(int a[], int n,
                int k)
{
    // Stores resultant arrays
    vector<int> first, second;
 
    // Traverse the array
    for (int i = 0; i < n; i++) {
 
        // If a[i] is smaller than
        // or equal to k/2
        if (a[i] <= k / 2)
            first.push_back(a[i]);
        else
            second.push_back(a[i]);
    }
 
    // Print first array
    for (int i = 0; i < first.size();
         i++) {
        cout << first[i] << " ";
    }
 
    // Print second array
    cout << "\n";
    for (int i = 0; i < second.size();
         i++) {
        cout << second[i] << " ";
    }
}
 
// Driver Code
int main()
{
 
    // Given K
    int k = 5;
 
    // Given array
    int a[] = { 0, 1, 3, 2, 4, 5,
                6, 7, 8, 9, 10 };
 
    // Given size
    int n = sizeof(a)
            / sizeof(int);
 
    splitArray(a, n, k);
 
    return 0;
}

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Java

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// Java program for the above approach
import java.util.*;
 
class GFG{
  
// Function to split the given
// array into two separate arrays
// satisfying given condition
static void splitArray(int a[], int n,
                       int k)
{
     
    // Stores resultant arrays
    Vector<Integer> first = new Vector<>();
    Vector<Integer> second = new Vector<>();
 
    // Traverse the array
    for(int i = 0; i < n; i++)
    {
         
        // If a[i] is smaller than
        // or equal to k/2
        if (a[i] <= k / 2)
            first.add(a[i]);
        else
            second.add(a[i]);
    }
  
    // Print first array
    for(int i = 0; i < first.size(); i++)
    {
        System.out.print(first.get(i) + " ");
    }
  
    // Print second array
    System.out.println();
    for(int i = 0; i < second.size(); i++)
    {
        System.out.print(second.get(i) + " ");
    }
}
  
// Driver Code
public static void main(String[] args)
{
     
    // Given K
    int k = 5;
  
    // Given array
    int a[] = { 0, 1, 3, 2, 4, 5,
                6, 7, 8, 9, 10 };
  
    int n = a.length;
     
    splitArray(a, n, k);
}
}
 
// This code is contributed by code_hunt

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Python3

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# Python3 program for the above approach
 
# Function to split the given
# array into two separate arrays
# satisfying given condition
def splitArray(a, n, k):
     
    # Stores resultant arrays
    first = []
    second = []
 
    # Traverse the array
    for i in range(n):
         
        # If a[i] is smaller than
        # or equal to k/2
        if (a[i] <= k // 2):
            first.append(a[i])
        else:
            second.append(a[i])
 
    # Print first array
    for i in range(len(first)):
        print(first[i], end = " ")
 
    # Print second array
    print("\n", end = "")
    for i in range(len(second)):
        print(second[i], end = " ")
 
# Driver Code
if __name__ == '__main__':
     
    # Given K
    k = 5
     
    # Given array
    a =  [ 0, 1, 3, 2, 4, 5,
           6, 7, 8, 9, 10 ]
            
    n =  len(a)
     
    splitArray(a, n, k)
     
# This code is contributed by bgangwar59

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C#

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// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
   
// Function to split the given
// array into two separate arrays
// satisfying given condition
static void splitArray(int[] a, int n,
                       int k)
{
     
    // Stores resultant arrays
    List<int> first = new List<int>();
    List<int> second = new List<int>();
 
    // Traverse the array
    for(int i = 0; i < n; i++)
    {
         
        // If a[i] is smaller than
        // or equal to k/2
        if (a[i] <= k / 2)
            first.Add(a[i]);
        else
            second.Add(a[i]);
    }
   
    // Print first array
    for(int i = 0; i < first.Count; i++)
    {
        Console.Write(first[i] + " ");
    }
   
    // Print second array
    Console.WriteLine();
    for(int i = 0; i < second.Count; i++)
    {
        Console.Write(second[i] + " ");
    }
}
   
// Driver Code
public static void Main()
{
     
    // Given K
    int k = 5;
   
    // Given array
    int[] a = { 0, 1, 3, 2, 4, 5,
                6, 7, 8, 9, 10 };
   
    int n = a.Length;
      
    splitArray(a, n, k);
}
}
   
// This code is contributed by susmitakundugoaldanga

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Output: 

0 1 2 
3 4 5 6 7 8 9 10










 

Time Complexity: O(N), where N is the size of the given array.
Auxiliary Space: O(N)

 

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