本文概述
- C ++
- Java
- Python3
- C#
例子:
输入:arr[] = {0, 0, 1, 1, 3, 3, 3}, K = 1简单的方法:
输出:4
说明:
子数组{0, 0, 1, 1}是仅有的两个子数组之间有差异的子数组等于K(= 1)的不同值。因此, 长度等于4。
输入:arr[] = {5, 7, 1, 1, 2, 2, 4, 4, 4, 5, 5, 4, 5, 8, 9}, K = 1
输出:7说明:
子数组{1、1、2}, {4、4、4、5、5、4、5}和{8、9}是仅有的两个等于K(= 1)的不同值之间具有差异的子数组。其中最长的子数组是{4, 4, 4, 5, 5, 4, 5}。因此, 长度为7。
- 一个简单的解决方案是一一考虑所有子数组, 然后查找仅包含两个不同值的子数组, 这两个值之间的差为K。不断更新获得的子数组的最大长度。
- 最后打印获得的最大长度。
辅助空间:O(N)
高效方法:
可以观察到,对于任意两个元素之差为K的元素所组成的子数组,该子数组必须只有两个不同的值。因此,可以进一步优化上述方法,使用set寻找只有两个不同值且差k的最长子数组,按照以下步骤解决问题:
- 从数组的起始索引开始第一个子数组。
- 将该元素插入集合。继续下一个元素, 并检查该元素是否与前一个元素相同或绝对差为K。
- 如果是这样, 则将该元素插入集合, 然后继续增加子数组的长度。一旦找到第三个不同的元素, 请将当前子数组的长度与子数组的最大长度进行比较, 然后进行相应更新。
- 将获得的新元素更新到集合中, 然后重复上述步骤。
- 遍历整个数组后, 打印获得的最大长度。
C ++
//C++ implementation to find the
//longest subarray consisting of
//only two values with difference K
#include <
bits/stdc++.h>
using namespace std;
//Function to return the length
//of the longest sub-array
int longestSubarray( int arr[], int n, int k)
{
int i, j, Max = 1;
//Initialize set
set<
int>
s;
for (i = 0;
i <
n - 1;
i++) {
//Store 1st element of
//sub-array into set
s.insert(arr[i]);
for (j = i + 1;
j <
n;
j++) {
//Check absolute difference
//between two elements
if ( abs (arr[i] - arr[j]) == 0
|| abs (arr[i] - arr[j]) == k) {
//If the new element is not
//present in the set
if (!s.count(arr[j])) {
//If the set contains
//two elements
if (s.size() == 2)
break ;
//Otherwise
else
s.insert(arr[j]);
}
}
else
break ;
}
if (s.size() == 2) {
//Update the maximum
//length
Max = max(Max, j - i);
//Remove the set
//elements
s.clear();
}
else
s.clear();
}
return Max;
}
//Driver Code
int main()
{
int arr[] = { 1, 0, 2, 2, 5, 5, 5 };
int N = sizeof (arr)
/sizeof (arr[0]);
int K = 1;
int length = longestSubarray(
arr, N, K);
if (length == 1)
cout <
<
-1;
else
cout <
<
length;
return 0;
}
Java
//Java implementation to find the
//longest subarray consisting of
//only two values with difference K
import java.util.*;
class GFG{
//Function to return the length
//of the longest sub-array
static int longestSubarray( int arr[], int n, int k)
{
int i, j, Max = 1 ;
//Initialize set
HashSet<
Integer>
s = new HashSet<
Integer>
();
for (i = 0 ;
i <
n - 1 ;
i++)
{//Store 1st element of
//sub-array into set
s.add(arr[i]);
for (j = i + 1 ;
j <
n;
j++)
{//Check absolute difference
//between two elements
if (Math.abs(arr[i] - arr[j]) == 0 ||
Math.abs(arr[i] - arr[j]) == k)
{//If the new element is not
//present in the set
if (!s.contains(arr[j]))
{//If the set contains
//two elements
if (s.size() == 2 )
break ;
//Otherwise
else
s.add(arr[j]);
}
}
else
break ;
}
if (s.size() == 2 )
{//Update the maximum
//length
Max = Math.max(Max, j - i);
//Remove the set
//elements
s.clear();
}
else
s.clear();
}
return Max;
}
//Driver Code
public static void main(String[] args)
{
int arr[] = { 1 , 0 , 2 , 2 , 5 , 5 , 5 };
int N = arr.length;
int K = 1 ;
int length = longestSubarray(arr, N, K);
if (length == 1 )
System.out.print(- 1 );
else
System.out.print(length);
}
}
//This code is contributed by Princi Singh
Python3
# Python3 implementation to find the
# longest subarray consisting of
# only two valueswith difference K
# Function to return the length
# of the longest sub-array
def longestSubarray (arr, n, k):
Max = 1
# Initialize set
s = set ()
for i in range (n - 1 ):
# Store 1st element of
# sub-array into set
s.add(arr[i])
for j in range (i + 1 , n):# Check absolute difference
# between two elements
if ( abs (arr[i] - arr[j]) = = 0 or
abs (arr[i] - arr[j]) = = k):
# If the new element is not
# present in the set
if ( not arr[j] in s):
# If the set contains
# two elements
if ( len (s) = = 2 ):
break
# Otherwise
else :
s.add(arr[j])else :
break
if ( len (s) = = 2 ):
# Update the maximum length
Max = max ( Max , j - i)
# Remove the set elements
s.clear()
else :
s.clear()
return Max
# Driver Code
if __name__ = = '__main__' :arr = [ 1 , 0 , 2 , 2 , 5 , 5 , 5 ]
N = len (arr)
K = 1
length = longestSubarray(arr, N, K)
if (length = = 1 ):
print ( "-1" )
else :
print (length)
# This code is contributed by himanshu77
C#
//C# implementation to find the
//longest subarray consisting of
//only two values with difference K
using System;
using System.Collections.Generic;
class GFG{
//Function to return the length
//of the longest sub-array
static int longestSubarray( int []arr, int n, int k)
{
int i, j, Max = 1;
//Initialize set
HashSet<
int>
s = new HashSet<
int>
();
for (i = 0;
i <
n - 1;
i++)
{//Store 1st element of
//sub-array into set
s.Add(arr[i]);
for (j = i + 1;
j <
n;
j++)
{//Check absolute difference
//between two elements
if (Math.Abs(arr[i] - arr[j]) == 0 ||
Math.Abs(arr[i] - arr[j]) == k)
{//If the new element is not
//present in the set
if (!s.Contains(arr[j]))
{//If the set contains
//two elements
if (s.Count == 2)
break ;
//Otherwise
else
s.Add(arr[j]);
}
}
else
break ;
}
if (s.Count == 2)
{//Update the maximum
//length
Max = Math.Max(Max, j - i);
//Remove the set
//elements
s.Clear();
}
else
s.Clear();
}
return Max;
}
//Driver Code
public static void Main(String[] args)
{
int []arr = { 1, 0, 2, 2, 5, 5, 5 };
int N = arr.Length;
int K = 1;
int length = longestSubarray(arr, N, K);
if (length == 1)
Console.Write(-1);
else
Console.Write(length);
}
}
//This code is contributed by Princi Singh
输出如下:
2
时间复杂度:O(N^2 * logN)
【最长的子数组,任意两个不同值之间的差值恰好为K】辅助空间:O(N)
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