Divide array into two arrays which does not contain any pair with sum K

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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++

// 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;
}

Java

// 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

Python3

# 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

C#

// 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

Javascript

<script>
// Javascript program for the above approach
 
// Function to split the given
// array into two separate arrays
// satisfying given condition
function splitArray(a, n, k)
{
 
    // Stores resultant arrays
    let first = [];
    let second = [];
 
    // Traverse the array
    for (let i = 0; i < n; i++) {
 
        // If a[i] is smaller than
        // or equal to k/2
        if (a[i] <= parseInt(k / 2))
            first.push(a[i]);
        else
            second.push(a[i]);
    }
 
    // Print first array
    for (let i = 0; i < first.length;
         i++) {
        document.write(first[i] + " ");
    }
 
    // Print second array
    document.write("<br>");
    for (let i = 0; i < second.length;
         i++) {
        document.write(second[i] + " ");
    }
}
 
// Driver Code
 
// Given K
let k = 5;
 
// Given array
let a = [ 0, 1, 3, 2, 4, 5,
    6, 7, 8, 9, 10 ];
 
// Given size
let n = a.length;
splitArray(a, n, k);
     
    // This code is contributed by rishavmahato348.
</script>

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(1)

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