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)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
