大小为 k 的所有子数组的最小和最大元素的总和。
给定一个由正整数和负整数组成的数组，任务是计算所有大小为 k 的子数组的最小和最大元素的总和。
例子：
Input : arr[] = {2, 5, -1, 7, -3, -1, -2}
K = 4
Output : 18
Explanation : Subarrays of size 4 are :
{2, 5, -1, 7}, min + max = -1 + 7 = 6
{5, -1, 7, -3}, min + max = -3 + 7 = 4
{-1, 7, -3, -1}, min + max = -3 + 7 = 4
{7, -3, -1, -2}, min + max = -3 + 7 = 4
Sum of all min & max = 6 + 4 + 4 + 4
= 18 如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。
这个问题主要是下面问题的扩展。
大小为 k 的所有子数组的最大值
方法一（简单）
运行两个循环以生成大小为 k 的所有子数组并找到最大值和最小值。最后，返回所有最大和最小元素的总和。
该解决方案花费的时间是 O(nk)。
方法二（使用Dequeue高效）
这个想法是使用 Dequeue 数据结构和滑动窗口的概念。我们创建了两个大小为 k ('S' , 'G') 的空双端队列，它们只存储当前窗口中没有用的元素的索引。如果一个元素不能是下一个子数组的最大值或最小值，则它是无用的。
a) In deque 'G', we maintain decreasing order of
values from front to rear
b) In deque 'S', we maintain increasing order of
values from front to rear
1) First window size K
1.1) For deque 'G', if current element is greater
than rear end element, we remove rear while
current is greater.
1.2) For deque 'S', if current element is smaller
than rear end element, we just pop it while
current is smaller.
1.3) insert current element in both deque 'G' 'S'
2) After step 1, front of 'G' contains maximum element
of first window and front of 'S' contains minimum
element of first window. Remaining elements of G
and S may store maximum/minimum for subsequent
windows.
3) After that we do traversal for rest array elements.
3.1) Front element of deque 'G' is greatest and 'S'
is smallest element of previous window
3.2) Remove all elements which are out of this
window [remove element at front of queue ]
3.3) Repeat steps 1.1 , 1.2 ,1.3
4) Return sum of minimum and maximum element of all
sub-array size k.以下是上述想法的实现
C++
// C++ program to find sum of all minimum and maximum
// elements Of Sub-array Size k.
#include
using namespace std;
// Returns sum of min and max element of all subarrays
// of size k
int SumOfKsubArray(int arr[] , int n , int k)
{
int sum = 0; // Initialize result
// The queue will store indexes of useful elements
// in every window
// In deque 'G' we maintain decreasing order of
// values from front to rear
// In deque 'S' we maintain increasing order of
// values from front to rear
deque< int > S(k), G(k);
// Process first window of size K
int i = 0;
for (i = 0; i < k; i++)
{
// Remove all previous greater elements
// that are useless.
while ( (!S.empty()) && arr[S.back()] >= arr[i])
S.pop_back(); // Remove from rear
// Remove all previous smaller that are elements
// are useless.
while ( (!G.empty()) && arr[G.back()] <= arr[i])
G.pop_back(); // Remove from rear
// Add current element at rear of both deque
G.push_back(i);
S.push_back(i);
}
// Process rest of the Array elements
for ( ; i < n; i++ )
{
// Element at the front of the deque 'G' & 'S'
// is the largest and smallest
// element of previous window respectively
sum += arr[S.front()] + arr[G.front()];
// Remove all elements which are out of this
// window
while ( !S.empty() && S.front() <= i - k)
S.pop_front();
while ( !G.empty() && G.front() <= i - k)
G.pop_front();
// remove all previous greater element that are
// useless
while ( (!S.empty()) && arr[S.back()] >= arr[i])
S.pop_back(); // Remove from rear
// remove all previous smaller that are elements
// are useless
while ( (!G.empty()) && arr[G.back()] <= arr[i])
G.pop_back(); // Remove from rear
// Add current element at rear of both deque
G.push_back(i);
S.push_back(i);
}
// Sum of minimum and maximum element of last window
sum += arr[S.front()] + arr[G.front()];
return sum;
}
// Driver program to test above functions
int main()
{
int arr[] = {2, 5, -1, 7, -3, -1, -2} ;
int n = sizeof(arr)/sizeof(arr[0]);
int k = 3;
cout << SumOfKsubArray(arr, n, k) ;
return 0;
} Java
// Java program to find sum of all minimum and maximum
// elements Of Sub-array Size k.
import java.util.Deque;
import java.util.LinkedList;
public class Geeks {
// Returns sum of min and max element of all subarrays
// of size k
public static int SumOfKsubArray(int arr[] , int k)
{
int sum = 0; // Initialize result
// The queue will store indexes of useful elements
// in every window
// In deque 'G' we maintain decreasing order of
// values from front to rear
// In deque 'S' we maintain increasing order of
// values from front to rear
Deque S=new LinkedList<>(),G=new LinkedList<>();
// Process first window of size K
int i = 0;
for (i = 0; i < k; i++)
{
// Remove all previous greater elements
// that are useless.
while ( !S.isEmpty() && arr[S.peekLast()] >= arr[i])
S.removeLast(); // Remove from rear
// Remove all previous smaller that are elements
// are useless.
while ( !G.isEmpty() && arr[G.peekLast()] <= arr[i])
G.removeLast(); // Remove from rear
// Add current element at rear of both deque
G.addLast(i);
S.addLast(i);
}
// Process rest of the Array elements
for ( ; i < arr.length; i++ )
{
// Element at the front of the deque 'G' & 'S'
// is the largest and smallest
// element of previous window respectively
sum += arr[S.peekFirst()] + arr[G.peekFirst()];
// Remove all elements which are out of this
// window
while ( !S.isEmpty() && S.peekFirst() <= i - k)
S.removeFirst();
while ( !G.isEmpty() && G.peekFirst() <= i - k)
G.removeFirst();
// remove all previous greater element that are
// useless
while ( !S.isEmpty() && arr[S.peekLast()] >= arr[i])
S.removeLast(); // Remove from rear
// remove all previous smaller that are elements
// are useless
while ( !G.isEmpty() && arr[G.peekLast()] <= arr[i])
G.removeLast(); // Remove from rear
// Add current element at rear of both deque
G.addLast(i);
S.addLast(i);
}
// Sum of minimum and maximum element of last window
sum += arr[S.peekFirst()] + arr[G.peekFirst()];
return sum;
}
public static void main(String args[])
{
int arr[] = {2, 5, -1, 7, -3, -1, -2} ;
int k = 3;
System.out.println(SumOfKsubArray(arr, k));
}
}
//This code is contributed by Gaurav Tiwari Python
# Python3 program to find Sum of all minimum and maximum
# elements Of Sub-array Size k.
from collections import deque
# Returns Sum of min and max element of all subarrays
# of size k
def SumOfKsubArray(arr, n , k):
Sum = 0 # Initialize result
# The queue will store indexes of useful elements
# in every window
# In deque 'G' we maintain decreasing order of
# values from front to rear
# In deque 'S' we maintain increasing order of
# values from front to rear
S = deque()
G = deque()
# Process first window of size K
for i in range(k):
# Remove all previous greater elements
# that are useless.
while ( len(S) > 0 and arr[S[-1]] >= arr[i]):
S.pop() # Remove from rear
# Remove all previous smaller that are elements
# are useless.
while ( len(G) > 0 and arr[G[-1]] <= arr[i]):
G.pop() # Remove from rear
# Add current element at rear of both deque
G.append(i)
S.append(i)
# Process rest of the Array elements
for i in range(k, n):
# Element at the front of the deque 'G' & 'S'
# is the largest and smallest
# element of previous window respectively
Sum += arr[S[0]] + arr[G[0]]
# Remove all elements which are out of this
# window
while ( len(S) > 0 and S[0] <= i - k):
S.popleft()
while ( len(G) > 0 and G[0] <= i - k):
G.popleft()
# remove all previous greater element that are
# useless
while ( len(S) > 0 and arr[S[-1]] >= arr[i]):
S.pop() # Remove from rear
# remove all previous smaller that are elements
# are useless
while ( len(G) > 0 and arr[G[-1]] <= arr[i]):
G.pop() # Remove from rear
# Add current element at rear of both deque
G.append(i)
S.append(i)
# Sum of minimum and maximum element of last window
Sum += arr[S[0]] + arr[G[0]]
return Sum
# Driver program to test above functions
arr=[2, 5, -1, 7, -3, -1, -2]
n = len(arr)
k = 3
print(SumOfKsubArray(arr, n, k))
# This code is contributed by mohit kumarC#
// C# program to find sum of all minimum and maximum
// elements Of Sub-array Size k.
using System;
using System.Collections.Generic;
class Geeks
{
// Returns sum of min and max element of all subarrays
// of size k
public static int SumOfKsubArray(int []arr , int k)
{
int sum = 0; // Initialize result
// The queue will store indexes of useful elements
// in every window
// In deque 'G' we maintain decreasing order of
// values from front to rear
// In deque 'S' we maintain increasing order of
// values from front to rear
List S = new List();
List G = new List();
// Process first window of size K
int i = 0;
for (i = 0; i < k; i++)
{
// Remove all previous greater elements
// that are useless.
while ( S.Count != 0 && arr[S[S.Count - 1]] >= arr[i])
S.RemoveAt(S.Count - 1); // Remove from rear
// Remove all previous smaller that are elements
// are useless.
while ( G.Count != 0 && arr[G[G.Count - 1]] <= arr[i])
G.RemoveAt(G.Count - 1); // Remove from rear
// Add current element at rear of both deque
G.Add(i);
S.Add(i);
}
// Process rest of the Array elements
for ( ; i < arr.Length; i++ )
{
// Element at the front of the deque 'G' & 'S'
// is the largest and smallest
// element of previous window respectively
sum += arr[S[0]] + arr[G[0]];
// Remove all elements which are out of this
// window
while ( S.Count != 0 && S[0] <= i - k)
S.RemoveAt(0);
while ( G.Count != 0 && G[0] <= i - k)
G.RemoveAt(0);
// remove all previous greater element that are
// useless
while ( S.Count != 0 && arr[S[S.Count-1]] >= arr[i])
S.RemoveAt(S.Count - 1 ); // Remove from rear
// remove all previous smaller that are elements
// are useless
while ( G.Count != 0 && arr[G[G.Count - 1]] <= arr[i])
G.RemoveAt(G.Count - 1); // Remove from rear
// Add current element at rear of both deque
G.Add(i);
S.Add(i);
}
// Sum of minimum and maximum element of last window
sum += arr[S[0]] + arr[G[0]];
return sum;
}
// Driver code
public static void Main(String []args)
{
int []arr = {2, 5, -1, 7, -3, -1, -2} ;
int k = 3;
Console.WriteLine(SumOfKsubArray(arr, k));
}
}
// This code is contributed by gauravrajput1 Javascript
输出：
14