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 conditionvoid 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 Codeint 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 conditionstatic 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 Codepublic 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 conditiondef 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 Codeif __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 approachusing System;using System.Collections.Generic;class GFG{ // Function to split the given// array into two separate arrays// satisfying given conditionstatic 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 Codepublic 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 conditionfunction 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 Klet k = 5;// Given arraylet a = [ 0, 1, 3, 2, 4, 5, 6, 7, 8, 9, 10 ];// Given sizelet 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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