给定长度为N的数组arr []和整数K ，任务是找到最长的子数组，其中任意两个Distict值之间的差等于K。打印获得的最长子数组的长度。否则，如果未获得此类子数组，则打印-1 。
例子：
Input: arr[] = {0, 0, 1, 1, 3, 3, 3}, K = 1
Output: 4
Explanation:
The subarray {0, 0, 1, 1} is the only subarray having difference between any two distinct values equal to K( = 1). Hence, length is equal to 4.
Input: arr[] = {5, 7, 1, 1, 2, 4, 4, 4, 5, 5, 4, 5, 8, 9}, K = 1
Output: 7
Explanation:
Subarrays {1, 1, 2}, {4, 4, 4, 5, 5, 4, 5} and {8, 9} are the only subarray having difference between any two distinct values equal to K( = 1).
The longest subarray among them is {4, 4, 4, 5, 5, 4, 5}.
Hence, the length is 7.
天真的方法：
- 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且这两个值之间的差为K的子数组。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。
时间复杂度： O(N 3 )
辅助空间： O(N)
高效方法：
可以观察到，对于任何子数组而言，由两个元素之差恰好为K的元素组成，子数组必须仅由两个不同的值组成。因此，可以通过使用set来找到仅具有两个不同值且差异为K的最长子数组，从而进一步优化上述方法。请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。继续至下一个元素，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，请将该元素插入集合，然后继续增加子数组的长度。一旦找到第三个不同的元素，将当前子数组的长度与子数组的最大长度进行比较，并进行相应的更新。
- 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。
下面是上述方法的实现：
C++
// C++ implementation to find the
// longest subarray consisting of
// only two values with difference K
#include
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 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 s = new HashSet();
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 values with 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 himanshu77C#
// 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 s = new HashSet();
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：
2
时间复杂度： O(N 2 * logN)
辅助空间： O(N)