Given an array arr[] of size N, the task is to split the array into a minimum number of subset such that each pair of elements in each subset have the difference strictly greater than 1.
Note: All elements in the array are distinct.
Examples:
Input: arr = {5, 10, 6, 50}
Output: 2
Explanation:
Possible partitions are: {5, 10, 50}, {6}Input: arr = {2, 4, 6}
Output: 1
Explanation:
Possible partitions are: {2, 4, 6}
Approach: The idea is to observe that if there is no such pair i, j such that |arr[i] – arr[j]| = 1, then it is possible to put all the elements in the same partition, otherwise divide them into two partitions. So the required minimum number of partitions is always 1 or 2.
- Sort the given array.
- Compare the adjacent elements. If at any point their difference to be equal to 1, then print “2” as the required number of subset partition will always be 2 as we can put one of the elements from the above pair into another subset.
- If we traversed all the array didn’t found any adjacent pair with a difference less than 2 then print “1” without splitting the array into subsets we can have all possible pairs difference at least 2.
Below is the implementation for the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to Split the array into// minimum number of subsets with// difference strictly > 1void split(int arr[], int n){ // Sort the array sort(arr, arr + n); int count = 1; // Traverse through the sorted array for (int i = 1; i < n; i++) { // Check the pairs of elements // with difference 1 if (arr[i] - arr[i - 1] == 1) { // If we find even a single // pair with difference equal // to 1, then 2 partitions // else only 1 partition count = 2; break; } } // Print the count of partitions cout << count << endl;}// Driver Codeint main(){ // Given array int arr[] = { 2, 4, 6 }; // Size of the array int n = sizeof(arr) / sizeof(int); // Function Call split(arr, n); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{ // Function to split the array into// minimum number of subsets with// difference strictly > 1static void split(int arr[], int n){ // Sort the array Arrays.sort(arr); int count = 1; // Traverse through the sorted array for(int i = 1; i < n; i++) { // Check the pairs of elements // with difference 1 if (arr[i] - arr[i - 1] == 1) { // If we find even a single // pair with difference equal // to 1, then 2 partitions // else only 1 partition count = 2; break; } } // Print the count of partitions System.out.print(count);}// Driver Codepublic static void main(String[] args){ // Given array int arr[] = { 2, 4, 6 }; // Size of the array int n = arr.length; // Function call split(arr, n);}}// This code is contributed by jrishabh99 |
Python3
# Python3 implementation of# the above approach# Function to Split the array into# minimum number of subsets with# difference strictly > 1def split(arr, n): # Sort the array arr.sort() count = 1 # Traverse through the sorted array for i in range(1, n): # Check the pairs of elements # with difference 1 if(arr[i] - arr[i - 1] == 1): # If we find even a single # pair with difference equal # to 1, then 2 partitions # else only 1 partition count = 2 break # Print the count of partitions print(count)# Driver Codeif __name__ == '__main__': # Given array arr = [ 2, 4, 6 ] # Size of the array n = len(arr) # Function call split(arr, n)# This code is contributed by Shivam Singh |
C#
// C# program for the above approachusing System;class GFG{ // Function to split the array into// minimum number of subsets with// difference strictly > 1static void split(int []arr, int n){ // Sort the array Array.Sort(arr); int count = 1; // Traverse through the sorted array for(int i = 1; i < n; i++) { // Check the pairs of elements // with difference 1 if (arr[i] - arr[i - 1] == 1) { // If we find even a single // pair with difference equal // to 1, then 2 partitions // else only 1 partition count = 2; break; } } // Print the count of partitions Console.Write(count);} // Driver Codepublic static void Main(string[] args){ // Given array int[] arr = new int[]{ 2, 4, 6 }; // Size of the array int n = arr.Length; // Function call split(arr, n);}} // This code is contributed by Ritik Bansal |
Javascript
<script>// Javascript program for// the above approach// Function to split the array into// minimum number of subsets with// difference strictly > 1function split(arr, n){ // Sort the array arr.sort(); let count = 1; // Traverse through the sorted array for(let i = 1; i < n; i++) { // Check the pairs of elements // with difference 1 if (arr[i] - arr[i - 1] == 1) { // If we find even a single // pair with difference equal // to 1, then 2 partitions // else only 1 partition count = 2; break; } } // Print the count of partitions document.write(count);} // Driver Code // Given array let arr = [ 2, 4, 6 ]; // Size of the array let n = arr.length; // Function call split(arr, n);</script> |
Output:
1
Time Complexity: O(N log N), where N is the length of the array.
Auxiliary Space: O(1)
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