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

• Last Updated : 16 Apr, 2021

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:

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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 ``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 first = ``new` `Vector<>();``    ``Vector 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

 ``
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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