给定一个由n 个整数组成的数组arr[]和一个整数k ,任务是找到子序列可能的最大和,使得子序列中没有两个元素出现在原始数组中距离≤ k 处。
例子:
Input: arr[] = {5, 3, 4, 11, 2}, k=1
Output: 16
All possible sub-sequences are {5, 4, 2}, {5, 11}, {5, 2}, {3, 11}, {3, 2}, {4, 2} and {11}
Out of which 5 + 11 = 16 gives the maximum sum.
Input: arr[] = {6, 7, 1, 3, 8, 2, 4}, k = 2
Output: 15
方法:在选择索引i处的元素时,我们有两种选择,要么将当前元素包含在子序列中,要么不包含。让dp[i]表示到目前为止到达索引i处的元素的最大总和。我们可以计算dp[i]的值如下:
dp[i] = max(dp[i – (k + 1)] + arr[i], dp[i – 1])
dp[i – (k + 1)] + arr[i] is the case when element at index i is included. In that situation, maximum value will be arr[i] + maximum value till the last included element from the array.
dp[i – 1] is the case when current element is not included and the maximum value till now will be the maximum value till the previous element.
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the maximum sum possible
int maxSum(int* arr, int k, int n)
{
if (n == 0)
return 0;
if (n == 1)
return arr[0];
if (n == 2)
return max(arr[0], arr[1]);
// dp[i] represent the maximum sum so far
// after reaching current position i
int dp[n];
// Initialize dp[0]
dp[0] = arr[0];
// Initialize the dp values till k since any
// two elements included in the sub-sequence
// must be atleast k indices apart, and thus
// first element and second element
// will be k indices apart
for (int i = 1; i <= k; i++)
dp[i] = max(arr[i], dp[i - 1]);
// Fill remaining positions
for (int i = k + 1; i < n; i++)
dp[i] = max(arr[i], dp[i - (k + 1)] + arr[i]);
// Return the maximum sum
int max = *(std::max_element(dp, dp + n));
return max;
}
// Driver code
int main()
{
int arr[] = { 6, 7, 1, 3, 8, 2, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
int k = 2;
cout << maxSum(arr, k, n);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
// Function to return the maximum sum possible
static int maxSum(int []arr, int k, int n)
{
if (n == 0)
return 0;
if (n == 1)
return arr[0];
if (n == 2)
return Math.max(arr[0], arr[1]);
// dp[i] represent the maximum sum so far
// after reaching current position i
int[] dp = new int[n];
// Initialize dp[0]
dp[0] = arr[0];
// Initialize the dp values till k since any
// two elements included in the sub-sequence
// must be atleast k indices apart, and thus
// first element and second element
// will be k indices apart
for (int i = 1; i <= k; i++)
dp[i] = Math.max(arr[i], dp[i - 1]);
// Fill remaining positions
for (int i = k + 1; i < n; i++)
dp[i] = Math.max(arr[i], dp[i - (k + 1)] + arr[i]);
// Return the maximum sum
return maximum(dp);
}
static int maximum(int[] arr)
{
int max = Integer.MIN_VALUE;
for(int i = 0; i < arr.length; i++)
{
if(arr[i] > max)
{
max = arr[i];
}
}
return max;
}
// Driver code
public static void main (String[] args)
{
int []arr = { 6, 7, 1, 3, 8, 2, 4 };
int n = arr.length;
int k = 2;
System.out.println(maxSum(arr, k, n));
}
}
// This code is contributed by mits
Python3
# Python3 implementation of the approach
# Function to return the
# maximum sum possible
def maxSum(arr, k, n) :
if (n == 0) :
return 0;
if (n == 1) :
return arr[0];
if (n == 2) :
return max(arr[0], arr[1]);
# dp[i] represent the maximum sum so far
# after reaching current position i
dp = [0] * n ;
# Initialize dp[0]
dp[0] = arr[0];
# Initialize the dp values till k since any
# two elements included in the sub-sequence
# must be atleast k indices apart, and thus
# first element and second element
# will be k indices apart
for i in range(1, k + 1) :
dp[i] = max(arr[i], dp[i - 1]);
# Fill remaining positions
for i in range(k + 1, n) :
dp[i] = max(arr[i],
dp[i - (k + 1)] + arr[i]);
# Return the maximum sum
max_element = max(dp);
return max_element;
# Driver code
if __name__ == "__main__" :
arr = [ 6, 7, 1, 3, 8, 2, 4 ];
n = len(arr);
k = 2;
print(maxSum(arr, k, n));
# This code is contributed by Ryuga
C#
// C# implementation of the approach
using System;
using System.Linq;
class GFG
{
// Function to return the maximum sum possible
static int maxSum(int []arr, int k, int n)
{
if (n == 0)
return 0;
if (n == 1)
return arr[0];
if (n == 2)
return Math.Max(arr[0], arr[1]);
// dp[i] represent the maximum sum so far
// after reaching current position i
int[] dp = new int[n];
// Initialize dp[0]
dp[0] = arr[0];
// Initialize the dp values till k since any
// two elements included in the sub-sequence
// must be atleast k indices apart, and thus
// first element and second element
// will be k indices apart
for (int i = 1; i <= k; i++)
dp[i] = Math.Max(arr[i], dp[i - 1]);
// Fill remaining positions
for (int i = k + 1; i < n; i++)
dp[i] = Math.Max(arr[i], dp[i - (k + 1)] + arr[i]);
// Return the maximum sum
int max = dp.Max();
return max;
}
// Driver code
static void Main()
{
int []arr = { 6, 7, 1, 3, 8, 2, 4 };
int n = arr.Length;
int k = 2;
Console.WriteLine(maxSum(arr, k, n));
}
}
// This code is contributed by mits
Javascript
15
时间复杂度:O(N)
辅助空间: O(N)
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