给定大小为N的数组arr [] ,任务是将数组拆分为最小数量的子集,以使每个子集中的每对元素对的差严格大于1。
注意:数组中的所有元素都是不同的。
例子:
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}
方法:这个想法是观察如果没有这样的对i , j使得| arr [i] – arr [j] | = 1 ,则可以将所有元素放在同一分区中,否则将它们分为两个分区。因此,所需的最小分区数始终为1或2 。
- 对给定的数组进行排序。
- 比较相邻元素。如果在任何时候它们的差等于1,则打印“ 2”,因为所需的子集分区数将始终为2,因为我们可以将上述对中的一个元素放入另一个子集中。
- 如果遍历所有数组都没有发现任何相邻对的差小于2,则打印“ 1”而不将数组分成子集,我们可以使所有可能的对差至少为2。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to Split the array into
// minimum number of subsets with
// difference strictly > 1
void 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 partiton
count = 2;
break;
}
}
// Print the count of partitions
cout << count << endl;
}
// Driver Code
int 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 approach
import java.util.*;
class GFG{
// Function to split the array into
// minimum number of subsets with
// difference strictly > 1
static 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 partiton
count = 2;
break;
}
}
// Print the count of partitions
System.out.print(count);
}
// Driver Code
public 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 > 1
def 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 partiton
count = 2
break
# Print the count of partitions
print(count)
# Driver Code
if __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 approach
using System;
class GFG{
// Function to split the array into
// minimum number of subsets with
// difference strictly > 1
static 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 partiton
count = 2;
break;
}
}
// Print the count of partitions
Console.Write(count);
}
// Driver Code
public 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
输出:
1
时间复杂度: O(N log N),其中N是数组的长度。
辅助空间: O(1)