给定一个数组arr []和一个整数K ，任务是找到最大和最小长度的子数组，以使相邻元素之间的差异最大为K。
例子：
Input: arr[] = {2, 4, 6}, K = 2
Output: 3, 3
Explanation:
Minimum and Maximum length subarray such that difference
between adjacent elements is at most is 3, which is {2, 4, 2}
Input: arr[] = {2, 3, 6, 7, 8}, K = 2
Output: 2, 3
Explanation:
Minimum length Subarray(2) => {2, 3}
Maximum length Subarray(3) => {6, 7, 8}
方法：想法是遍历数组，从每个元素开始并向其左右方向移动，直到相邻元素之间的差小于K。最后，使用下面定义的当前长度更新最大和最小长度子数组–
if (current_length maximum_length)
maximum_length = current_length
下面是上述方法的实现：
C++
// C++ program to find the minimum
// and maximum length subarray such
// that difference between adjacent
// elements is atmost K
#include
using namespace std;
// Function to find the maximum and
// minimum length subarray
void findMaxMinSubArray(int arr[],
int K, int n)
{
// Initialise minimum and maximum
// size of subarray in worst case
int min = n;
int max = 0;
// Left will scan the size of
// possible subarray in left
// of selected element
int left;
// Right will scan the size of
// possible subarray in right
// of selected element
int right;
// Temp will store size of group
// associateed with every element
int tmp;
// Loop to find size of subarray
// for every element of array
for (int i = 0; i < n; i++) {
tmp = 1;
left = i;
// Left will move in left direction
// and compare difference between
// itself and element left to it
while (left - 1 >= 0
&& abs(arr[left] - arr[left - 1])
<= K) {
left--;
tmp++;
}
// right will move in right direction
// and compare difference between
// itself and element right to it
right = i;
while (right + 1 <= n - 1
&& abs(arr[right] - arr[right + 1])
<= K) {
right++;
tmp++;
}
// if subarray of much lesser or much
// greater is found than yet
// known then update
if (min > tmp)
min = tmp;
if (max < tmp)
max = tmp;
}
// Print minimum and maximum
// possible size of subarray
cout << min << ", "
<< max << endl;
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 5, 6, 7 };
int K = 2;
int n = sizeof(arr) / sizeof(arr[0]);
// function call to find maximum
// and minimum possible sub array
findMaxMinSubArray(arr, K, n);
return 0;
} Java
// Java program to find the minimum
// and maximum length subarray such
// that difference between adjacent
// elements is atmost K
import java.io.*;
import java.util.*;
class GFG {
// Function to find the maximum and
// minimum length subarray
static void findMaxMinSubArray(int arr[],
int K, int n)
{
// Initialise minimum and maximum
// size of subarray in worst case
int min = n;
int max = 0;
// Left will scan the size of
// possible subarray in left
// of selected element
int left;
// Right will scan the size of
// possible subarray in right
// of selected element
int right;
// Temp will store size of group
// associateed with every element
int tmp;
// Loop to find size of subarray
// for every element of array
for(int i = 0; i < n; i++)
{
tmp = 1;
left = i;
// Left will move in left direction
// and compare difference between
// itself and element left to it
while (left - 1 >= 0 &&
Math.abs(arr[left] -
arr[left - 1]) <= K)
{
left--;
tmp++;
}
// Right will move in right direction
// and compare difference between
// itself and element right to it
right = i;
while (right + 1 <= n - 1 &&
Math.abs(arr[right] -
arr[right + 1]) <= K)
{
right++;
tmp++;
}
// If subarray of much lesser or much
// greater is found than yet
// known then update
if (min > tmp)
min = tmp;
if (max < tmp)
max = tmp;
}
// Print minimum and maximum
// possible size of subarray
System.out.print(min);
System.out.print(", ");
System.out.print(max);
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 2, 5, 6, 7 };
int K = 2;
int n = arr.length;
// Function call to find maximum
// and minimum possible sub array
findMaxMinSubArray(arr, K, n);
}
}
// This code is contributed by coder001Python3
# Python3 program to find the minimum
# and maximum length subarray such
# that difference between adjacent
# elements is atmost K
# Function to find the maximum and
# minimum length subarray
def findMaxMinSubArray(arr, K, n):
# Initialise minimum and maximum
# size of subarray in worst case
min = n
max = 0
# Left will scan the size of
# possible subarray in left
# of selected element
left = 0
# Right will scan the size of
# possible subarray in right
# of selected element
right = n
# Temp will store size of group
# associateed with every element
tmp = 0
# Loop to find size of subarray
# for every element of array
for i in range(0, n):
tmp = 1
left = i
# Left will move in left direction
# and compare difference between
# itself and element left to it
while (left - 1 >= 0 and
abs(arr[left] -
arr[left - 1]) <= K):
left = left - 1
tmp = tmp + 1
# Right will move in right direction
# and compare difference between
# itself and element right to it
right = i
while (right + 1 <= n - 1 and
abs(arr[right] -
arr[right + 1]) <= K):
right = right + 1
tmp = tmp + 1
# If subarray of much lesser or much
# greater is found than yet
# known then update
if (min > tmp):
min = tmp
if (max < tmp):
max = tmp
# Print minimum and maximum
# possible size of subarray
print(min, end = ', ')
print(max, end = '\n')
# Driver Code
arr = [ 1, 2, 5, 6, 7 ]
K = 2
n = len(arr)
# Function call to find maximum
# and minimum possible sub array
findMaxMinSubArray(arr, K, n)
# This code is contributed by PratikC#
// C# program to find the minimum
// and maximum length subarray such
// that difference between adjacent
// elements is atmost K
using System;
class GFG{
// Function to find the maximum and
// minimum length subarray
static void findMaxMinSubArray(int[] arr,
int K, int n)
{
// Initialise minimum and maximum
// size of subarray in worst case
int min = n;
int max = 0;
// Left will scan the size of
// possible subarray in left
// of selected element
int left;
// Right will scan the size of
// possible subarray in right
// of selected element
int right;
// Temp will store size of group
// associateed with every element
int tmp;
// Loop to find size of subarray
// for every element of array
for(int i = 0; i < n; i++)
{
tmp = 1;
left = i;
// Left will move in left direction
// and compare difference between
// itself and element left to it
while (left - 1 >= 0 &&
Math.Abs(arr[left] -
arr[left - 1]) <= K)
{
left--;
tmp++;
}
// Right will move in right direction
// and compare difference between
// itself and element right to it
right = i;
while (right + 1 <= n - 1 &&
Math.Abs(arr[right] -
arr[right + 1]) <= K)
{
right++;
tmp++;
}
// If subarray of much lesser or much
// greater is found than yet
// known then update
if (min > tmp)
min = tmp;
if (max < tmp)
max = tmp;
}
// Print minimum and maximum
// possible size of subarray
Console.Write(min);
Console.Write(", ");
Console.Write(max);
}
// Driver code
public static void Main()
{
int[] arr = { 1, 2, 5, 6, 7 };
int K = 2;
int n = arr.Length;
// Function call to find maximum
// and minimum possible sub array
findMaxMinSubArray(arr, K, n);
}
}
// This code is contributed by AbhiThakur输出：
2, 3