给定一个数组arr [] ，任务是找到所有元素的最大和，这些元素不是最长递增子序列的一部分。
例子：
Input: arr[] = {4, 6, 1, 2, 3, 8}
Output: 10
Explanation:
Elements are 4 and 6
Input: arr[] = {5, 4, 3, 2, 1}
Output: 14
Explanation:
Elements are 5, 4, 3, 2
方法：
- 这个想法是找到具有最小总和的最长的增长子序列，然后从所有元素的总和中减去它。
- 为此，我们将使用动态编程的LIS概念，将和与子序列的长度一起存储，并相应地更新最小和。
下面是上述方法的实现。
C++
// C++ program to find the Maximum sum of
// all elements which are not a part of
// longest increasing sub sequence
#include
using namespace std;
// Function to find maximum sum
int findSum(int* arr, int n)
{
int totalSum = 0;
// Find total sum of array
for (int i = 0; i < n; i++) {
totalSum += arr[i];
}
// Maintain a 2D array
int dp[2][n];
for (int i = 0; i < n; i++) {
dp[0][i] = 1;
dp[1][i] = arr[i];
}
// Update the dp array along
// with sum in the second row
for (int i = 1; i < n; i++) {
for (int j = 0; j < i; j++) {
if (arr[i] > arr[j]) {
// In case of greater length
// Update the length along
// with sum
if (dp[0][i] < dp[0][j] + 1) {
dp[0][i] = dp[0][j] + 1;
dp[1][i] = dp[1][j]
+ arr[i];
}
// In case of equal length
// find length update length
// with minimum sum
else if (dp[0][i]
== dp[0][j] + 1) {
dp[1][i]
= min(dp[1][i],
dp[1][j]
+ arr[i]);
}
}
}
}
int maxm = 0;
int subtractSum = 0;
// Find the sum that need to
// be subtracted from total sum
for (int i = 0; i < n; i++) {
if (dp[0][i] > maxm) {
maxm = dp[0][i];
subtractSum = dp[1][i];
}
else if (dp[0][i] == maxm) {
subtractSum = min(subtractSum,
dp[1][i]);
}
}
// Return the sum
return totalSum - subtractSum;
}
// Driver code
int main()
{
int arr[] = { 4, 6, 1, 2, 3, 8 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << findSum(arr, n);
return 0;
} Java
// Java program to find the Maximum sum of
// all elements which are not a part of
// longest increasing sub sequence
class GFG{
// Function to find maximum sum
static int findSum(int []arr, int n)
{
int totalSum = 0;
// Find total sum of array
for(int i = 0; i < n; i++)
{
totalSum += arr[i];
}
// Maintain a 2D array
int [][]dp = new int[2][n];
for(int i = 0; i < n; i++)
{
dp[0][i] = 1;
dp[1][i] = arr[i];
}
// Update the dp array along
// with sum in the second row
for(int i = 1; i < n; i++)
{
for(int j = 0; j < i; j++)
{
if (arr[i] > arr[j])
{
// In case of greater length
// Update the length along
// with sum
if (dp[0][i] < dp[0][j] + 1)
{
dp[0][i] = dp[0][j] + 1;
dp[1][i] = dp[1][j] + arr[i];
}
// In case of equal length
// find length update length
// with minimum sum
else if (dp[0][i] == dp[0][j] + 1)
{
dp[1][i] = Math.min(dp[1][i],
dp[1][j] + arr[i]);
}
}
}
}
int maxm = 0;
int subtractSum = 0;
// Find the sum that need to
// be subtracted from total sum
for(int i = 0; i < n; i++)
{
if (dp[0][i] > maxm)
{
maxm = dp[0][i];
subtractSum = dp[1][i];
}
else if (dp[0][i] == maxm)
{
subtractSum = Math.min(subtractSum, dp[1][i]);
}
}
// Return the sum
return totalSum - subtractSum;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 4, 6, 1, 2, 3, 8 };
int n = arr.length;
System.out.print(findSum(arr, n));
}
}
// This code is contributed by sapnasingh4991Python3
# Python3 program to find the maximum sum
# of all elements which are not a part of
# longest increasing sub sequence
# Function to find maximum sum
def findSum(arr, n):
totalSum = 0
# Find total sum of array
for i in range(n):
totalSum += arr[i]
# Maintain a 2D array
dp = [[0] * n for i in range(2)]
for i in range(n):
dp[0][i] = 1
dp[1][i] = arr[i]
# Update the dp array along
# with sum in the second row
for i in range(1, n):
for j in range(i):
if (arr[i] > arr[j]):
# In case of greaer length
# update the length along
# with sum
if (dp[0][i] < dp[0][j] + 1):
dp[0][i] = dp[0][j] + 1
dp[1][i] = dp[1][j] + arr[i]
# In case of equal length
# find length update length
# with minimum sum
elif (dp[0][i] == dp[0][j] + 1):
dp[1][i] = min(dp[1][i],
dp[1][j] +
arr[i])
maxm = 0
subtractSum = 0
# Find the sum that need to
# be subtracted from total sum
for i in range(n):
if (dp[0][i] > maxm):
maxm = dp[0][i]
subtractSum = dp[1][i]
elif (dp[0][i] == maxm):
subtractSum = min(subtractSum,
dp[1][i])
# Return the sum
return totalSum - subtractSum
# Driver code
arr = [ 4, 6, 1, 2, 3, 8 ]
n = len(arr)
print(findSum(arr, n))
# This code is contributed by himanshu77C#
// C# program to find the Maximum sum of
// all elements which are not a part of
// longest increasing sub sequence
using System;
class GFG{
// Function to find maximum sum
static int findSum(int []arr, int n)
{
int totalSum = 0;
// Find total sum of array
for(int i = 0; i < n; i++)
{
totalSum += arr[i];
}
// Maintain a 2D array
int [,]dp = new int[2, n];
for(int i = 0; i < n; i++)
{
dp[0, i] = 1;
dp[1, i] = arr[i];
}
// Update the dp array along
// with sum in the second row
for(int i = 1; i < n; i++)
{
for(int j = 0; j < i; j++)
{
if (arr[i] > arr[j])
{
// In case of greater length
// Update the length along
// with sum
if (dp[0, i] < dp[0, j] + 1)
{
dp[0, i] = dp[0, j] + 1;
dp[1, i] = dp[1, j] + arr[i];
}
// In case of equal length
// find length update length
// with minimum sum
else if (dp[0, i] == dp[0, j] + 1)
{
dp[1, i] = Math.Min(dp[1, i],
dp[1, j] + arr[i]);
}
}
}
}
int maxm = 0;
int subtractSum = 0;
// Find the sum that need to
// be subtracted from total sum
for(int i = 0; i < n; i++)
{
if (dp[0, i] > maxm)
{
maxm = dp[0, i];
subtractSum = dp[1, i];
}
else if (dp[0, i] == maxm)
{
subtractSum = Math.Min(subtractSum, dp[1, i]);
}
}
// Return the sum
return totalSum - subtractSum;
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 4, 6, 1, 2, 3, 8 };
int n = arr.Length;
Console.Write(findSum(arr, n));
}
}
// This code is contributed by sapnasingh4991输出：
10时间复杂度： O(N 2 )，其中N是数组arr []的长度
辅助空间： O(N)