给定一个由N 个整数和一个整数K组成的数组arr[] ，任务是打印满足以下条件的子序列中可能的最大和：
- 元素arr[N – 1]和arr[0]包含在子序列中。
- 子序列中的相邻元素的距离最多为K 个索引。
例子：
Input: arr[] = {10, -5, -2, 4, 0, 3}, K = 3
Output: 17
Explanation:
One of possible way is as follows:
Include arr[0] into the subsequence. Sum = 10.
Include arr[3] in the subsequence. Therefore, sum = 10 + 4 = 14.
Include arr[5] in the subsequence. Therefore, total sum = 14 + 3 = 17.
Therefore, the maximum sum possible is 17.
Input: arr[] = {1, -5, -20, 4, -1, 3, -6, -3}, K = 2
Output: 0
朴素的方法：最简单的方法是从arr[] 中找到所有可能的子序列，相邻元素的索引之间最多有K 个差异，从索引0开始，到索引(N – 1) 结束。计算所有这些子序列的总和。最后，打印获得的所有和的最大值。
时间复杂度： O(N*2 N )
辅助空间： O(N)
高效方法：上述方法可以通过使用贪心算法和双端队列进行优化。请按照以下步骤解决问题：
- 初始化一个数组，比如dp[] ，以存储直到当前索引为止获得的最大值。
- 初始化一对双端队列，比如Q ，以存储对{dp[i], i} 。
- 将arr[0]的值分配给dp[0]并将{dp[0], 0}对推入双端队列。
- 使用变量i遍历给定数组arr[]并执行以下步骤：
- 通过arr[i]和双端队列中最大值的总和增加dp[i] ，即dp[i] = arr[i] + Q[0][0] 。
- 如果Q[-1][0]小于dp[i] ，则从末尾遍历双端队列Q并弹出最后一个元素。
- 在双端队列中附加{dp[i], i}对。
- 检查双端队列q的第一个元素的索引是否等于(i – K) ，然后从双端队列Q 中弹出第一个元素。
- 完成上述步骤后，将存储在dp[]的最后一个索引处的值，即dp[N – 1]作为结果打印出来。
下面是上述方法的实现：
C++
// CPP program for the above approach
#include
using namespace std;
// Function to find maximum sum
// of a subsequence satisfying
// the given conditions
int maxResult(int arr[], int k, int n){
// Stores the maximum sum
int dp[n] = {0};
// Starting index of
// the subsequence
dp[0] = arr[0];
// Stores the pair of maximum value
// and the index of that value
deque> q;
q.push_back({arr[0], 0});
// Traverse the array
for (int i = 1; i < n; i++)
{
// Increment the first value
// of deque by arr[i] and
// store it in dp[i]
dp[i] = arr[i] + q.front().first;
// Delete all the values which
// are less than dp[i] in deque
while (q.size() > 0 and q.back().first < dp[i])
q.pop_back();
// Append the current pair of
// value and index in deque
q.push_back({dp[i], i});
// If first value of the
// queue is at a distance > K
if (i - k == q.front().second)
q.pop_front();
}
// Return the value at the last index
return dp[n - 1];
}
// Driver Code
int main()
{
int arr[] = {10, -5, -2, 4, 0, 3};
int K = 3;
int n = sizeof(arr)/sizeof(arr[0]);
cout< Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
public class GFG {
// Pair class Store (x,y) Pair
static class Pair {
int x, y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
}
// Function to find maximum sum
// of a subsequence satisfying
// the given conditions
private static int maxResult(int[] arr, int k, int n) {
// Stores the maximum sum
int dp[] = new int[n];
// Starting index of
// the subsequence
dp[0] = arr[0];
// Stores the pair of maximum value
// and the index of that value
Deque q = new LinkedList();
q.add(new Pair(arr[0], 0));
// Traverse the array
for (int i = 1; i < n; i++)
{
// Increment the first value
// of deque by arr[i] and
// store it in dp[i]
dp[i] = arr[i] + q.peekFirst().x;
// Delete all the values which
// are less than dp[i] in deque
while (q.size() > 0 && q.peekLast().x < dp[i])
q.pollLast();
// Append the current pair of
// value and index in deque
q.add(new Pair(dp[i], i));
// If first value of the
// queue is at a distance > K
if (i - k == q.peekFirst().y)
q.pollFirst();
}
// Return the value at the last index
return dp[n - 1];
}
// Driver Code
public static void main(String[] args)
{
int arr[] = {10, -5, -2, 4, 0, 3};
int K = 3;
int n = arr.length;
System.out.println(maxResult(arr, K,n));
}
}
// This code is contributed by Dheeraj Bhagchandani. Python3
# Python program for the above approach
from collections import deque
# Function to find maximum sum
# of a subsequence satisfying
# the given conditions
def maxResult(arr, k):
# Stores the maximum sum
dp = [0]*len(arr)
# Starting index of
# the subsequence
dp[0] = arr[0]
# Stores the pair of maximum value
# and the index of that value
q = deque([(arr[0], 0)])
# Traverse the array
for i in range(1, len(arr)):
# Increment the first value
# of deque by arr[i] and
# store it in dp[i]
dp[i] = arr[i] + q[0][0]
# Delete all the values which
# are less than dp[i] in deque
while q and q[-1][0] < dp[i]:
q.pop()
# Append the current pair of
# value and index in deque
q.append((dp[i], i))
# If first value of the
# queue is at a distance > K
if i - k == q[0][1]:
q.popleft()
# Return the value at the last index
return dp[-1]
# Driver Code
arr = [10, -5, -2, 4, 0, 3]
K = 3
print(maxResult(arr, K))输出：
17时间复杂度： O(N)
辅助空间： O(N)
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