给定一个数组arr [] ，任务是找到删除元素的最大分数，其中可以用该元素的分数删除数组的每个元素，但是约束是如果我们删除arr [i] ，则arr [ i] + 1和arr [i] – 1被自动删除，得分为0。
例子：
Input: arr[] = {7, 2, 1, 8, 3, 3, 6, 6}
Output: 27
Explanation:
Step 0: arr[] = 7 2 1 8 3 3 6 6, Score: 0
Step 1: arr[] = 7 1 8 3 6 6, Score: 3
Step 2: arr[] = 7 1 8 6 6, Score: 6
Step 3: arr[] = 7 8 6 6, Score: 7
Step 4: arr[] = 8 6, Score: 13
Step 5: arr[] = 8 Score: 19
Step 6: arr[] = [] Score: 27
Input: arr[] = 1 2 3
Output: 4
方法：想法是使用动态编程来解决此问题。该问题的关键发现是，对于从数组中删除任何元素，元素的出现和值本身是重要的因素。
让我们以一个例子来了解观察序列是否为4 45。然后，我们有两个选择，从4到5。现在，选择4时，他的分数将为4 * 2 = 8。如果我们选择5，他的分数将是5 * 1 =5。显然，最大分数是8。
因此，对于上述序列4 4 5，freq [4] = 2，freq [5] = 1。
最后，要找到最佳分数，首先将问题分解为较小的问题将很容易。在这种情况下，我们将序列分成较小的序列，并找到最佳的解决方案。对于仅包含0的数字序列，答案将为0。类似地，如果序列仅包含数字0和1，则解决方案将为count [1] * 1。
递归关系：
dp[i] = max(dp[i – 1], dp[i – 2] + i*freq[i])
基本上，我们有2种情况，要么选择第i个元素，要么不选择第i个元素。
情况1：如果我们选择第ith个元素，则直到第ith个元素的最大分数将为dp [i-2] + i * freq [i](选择ith元素意味着删除第(i-1)个元素)
情况2：如果我们不选择第ith个元素，则直到第ith个元素的最大得分将为dp [i-1]
现在，由于我们必须最大化分数，因此我们将两者都取最大值。
下面是上述方法的实现：
C++
// C++ implementation to find the
// maximum score of the deleting a
// element from an array
#include
using namespace std;
// Function to find the maximum
// score of the deleting an element
// from an array
int findMaximumScore(vector a, int n)
{
// Creating a map to keep
// the frequency of numbers
unordered_map freq;
// Loop to iterate over the
// elements of the array
for (int i = 0; i < n; i++) {
freq[a[i]]++;
}
// Creating a DP array to keep
// count of max score at ith element
// and it will be filled
// in the bottom Up manner
vector dp(*max_element(a.begin(),
a.end())
+ 1,
0);
dp[0] = 0;
dp[1] = freq[1];
// Loop to choose the elements of the
// array to delete from the array
for (int i = 2; i < dp.size(); i++)
dp[i] = max(
dp[i - 1],
dp[i - 2] + freq[i] * i);
return dp[dp.size() - 1];
}
// Driver Code
int main()
{
int n;
n = 3;
vector a{ 1, 2, 3 };
// Function Call
cout << findMaximumScore(a, n);
return 0;
} Java
// Java implementation to find the
// maximum score of the deleting a
// element from an array
import java.util.*;
class GFG{
// Function to find the maximum
// score of the deleting an element
// from an array
static int findMaximumScore(int []a, int n)
{
// Creating a map to keep
// the frequency of numbers
@SuppressWarnings("unchecked")
HashMap freq = new HashMap();
// Loop to iterate over the
// elements of the array
for(int i = 0; i < n; i++)
{
if(freq.containsKey(a[i]))
{
freq.put(a[i],
freq.get(a[i]) + 1);
}
else
{
freq.put(a[i], 1);
}
}
// Creating a DP array to keep
// count of max score at ith element
// and it will be filled
// in the bottom Up manner
int []dp = new int[Arrays.stream(a).max().getAsInt() + 1];
dp[0] = 0;
dp[1] = freq.get(1);
// Loop to choose the elements of the
// array to delete from the array
for(int i = 2; i < dp.length; i++)
dp[i] = Math.max(dp[i - 1],
dp[i - 2] +
freq.get(i) * i);
return dp[dp.length - 1];
}
// Driver Code
public static void main(String[] args)
{
int n;
n = 3;
int []a = { 1, 2, 3 };
// Function call
System.out.print(findMaximumScore(a, n));
}
}
// This code is contributed by 29AjayKumar Python3
# Python3 implementation to find the
# maximum score of the deleting a
# element from an array
from collections import defaultdict
# Function to find the maximum
# score of the deleting an element
# from an array
def findMaximumScore(a, n):
# Creating a map to keep
# the frequency of numbers
freq = defaultdict (int)
# Loop to iterate over the
# elements of the array
for i in range (n):
freq[a[i]] += 1
# Creating a DP array to keep
# count of max score at ith element
# and it will be filled
# in the bottom Up manner
dp = [0] * (max(a) + 1)
dp[0] = 0
dp[1] = freq[1]
# Loop to choose the elements of the
# array to delete from the array
for i in range (2, len(dp)):
dp[i] = max(dp[i - 1],
dp[i - 2] +
freq[i] * i)
return dp[- 1]
# Driver Code
if __name__ == "__main__":
n = 3
a = [1, 2, 3]
# Function Call
print(findMaximumScore(a, n))
# This code is contributed by ChitranayalC#
// C# implementation to find the
// maximum score of the deleting a
// element from an array
using System;
using System.Linq;
using System.Collections.Generic;
class GFG{
// Function to find the maximum
// score of the deleting an element
// from an array
static int findMaximumScore(int []a, int n)
{
// Creating a map to keep
// the frequency of numbers
Dictionary freq = new Dictionary();
// Loop to iterate over the
// elements of the array
for(int i = 0; i < n; i++)
{
if(freq.ContainsKey(a[i]))
{
freq[a[i]] = freq[a[i]] + 1;
}
else
{
freq.Add(a[i], 1);
}
}
// Creating a DP array to keep
// count of max score at ith element
// and it will be filled
// in the bottom Up manner
int []dp = new int[a.Max() + 1];
dp[0] = 0;
dp[1] = freq[1];
// Loop to choose the elements of the
// array to delete from the array
for(int i = 2; i < dp.Length; i++)
dp[i] = Math.Max(dp[i - 1],
dp[i - 2] +
freq[i] * i);
return dp[dp.Length - 1];
}
// Driver Code
public static void Main(String[] args)
{
int n;
n = 3;
int []a = { 1, 2, 3 };
// Function call
Console.Write(findMaximumScore(a, n));
}
}
// This code is contributed by 29AjayKumar 输出：
4