# Maximum Sum Subsequence of length k

Given an array sequence [A1, A2 …An], the task is to find the maximum possible sum of increasing subsequence S of length k such that S1<=S2<=S3………<=Sk.

Examples:

Input :
n = 8 k = 3
A=[8 5 9 10 5 6 21 8]
Output : 40
Possible Increasing subsequence of Length 3 with maximum possible sum is 9 10 21

Input :
n = 9 k = 4
A=[2 5 3 9 15 33 6 18 20]
Output : 62
Possible Increasing subsequence of Length 4 with maximum possible sum is 9 15 18 20

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

One thing that is clearly visible that it can be easily solved with dynamic programming and this problem is a simple variation of Longest Increasing Subsequence. If you are unknown of how to calculate the longest increasing subsequence then see the implementation going to the link.

Naive Approach:
In the brute force approach, first we will try to find all the subsequences of length k and will check whether they are increasing or not. There could be nCk such sequences in the worst case when all elements are in increasing order. Now we will find the maximum possible sum for such sequences.
Time Complexity would be O((nCk)*n).

Efficient Approach:
We will be using a two-dimensional dp array in which dp[i][l] means that maximum sum subsequence of length l taking array values from 0 to i and the subsequence is ending at index ‘i’. Range of ‘l’ is from 0 to k-1. Using the approach of longer increasing subsequence on the inner loop when j<i we will check if arr[j] < arr[i] for checking if subsequence increasing.
This problem can be divided into its subproblems:

dp[i]=arr[i] for length 1 , maximum icreasing subsequence is equal to the array value
dp[i][l+1]= max(dp[i][l+1], dp[j][l]+arr[i]) for any length l between 1 to k-1

This means that if for ith position and subsequence of length l+1 , there exists some subsequence at j (j < i) of length l for which sum of dp[j][l] + arr[i] is more than its initial calculated value then update that value.
Then finally we will find the maximum value of dp[i][k] i.e for every ‘i’ if subsequence of k length is causing more sum than update the required ans.

Below is the implementation code:

 `/*C++ program to calculate the maximum sum of  ` `increasing subsequence of length k*/` `#include ` `using` `namespace` `std; ` `int` `MaxIncreasingSub(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` `    ``// In the implementation dp[n][k] represents ` `    ``// maximum sum subsequence of length k and the ` `    ``// subsequence is ending at index n. ` `    ``int` `dp[n][k + 1], ans = -1; ` ` `  `    ``// Initializing whole multidimensional  ` `    ``// dp array with value -1 ` `    ``memset``(dp, -1, ``sizeof``(dp)); ` ` `  `    ``// For each ith position increasing subsequence ` `    ``// of length 1 is equal to that array ith value ` `    ``// so initializing dp[i] with that array value ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``dp[i] = arr[i]; ` `    ``} ` ` `  `    ``// Starting from 1st index as we have calculated ` `    ``// for 0th index. Computing optimized dp values  ` `    ``// in bottom-up manner ` `    ``for` `(``int` `i = 1; i < n; i++) { ` `        ``for` `(``int` `j = 0; j < i; j++) { ` ` `  `            ``// check for increasing subsequence ` `            ``if` `(arr[j] < arr[i]) { ` `                ``for` `(``int` `l = 1; l <= k - 1; l++) { ` ` `  `                    ``// Proceed if value is pre calculated ` `                    ``if` `(dp[j][l] != -1) { ` ` `  `                        ``// Check for all the subsequences  ` `                        ``// ending at any j

 `/*Java program to calculate the maximum sum of  ` `increasing subsequence of length k*/` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` `     `  `static` `int` `MaxIncreasingSub(``int` `arr[], ``int` `n, ``int` `k)  ` `{  ` `    ``// In the implementation dp[n][k] represents  ` `    ``// maximum sum subsequence of length k and the  ` `    ``// subsequence is ending at index n.  ` `    ``int` `dp[][]=``new` `int``[n][k + ``1``], ans = -``1``;  ` ` `  `    ``// Initializing whole multidimensional  ` `    ``// dp array with value -1  ` `    ``for``(``int` `i = ``0``; i < n; i++) ` `        ``for``(``int` `j = ``0``; j < k + ``1``; j++) ` `            ``dp[i][j]=-``1``; ` ` `  `    ``// For each ith position increasing subsequence  ` `    ``// of length 1 is equal to that array ith value  ` `    ``// so initializing dp[i] with that array value  ` `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{  ` `        ``dp[i][``1``] = arr[i];  ` `    ``}  ` ` `  `    ``// Starting from 1st index as we have calculated  ` `    ``// for 0th index. Computing optimized dp values  ` `    ``// in bottom-up manner  ` `    ``for` `(``int` `i = ``1``; i < n; i++)  ` `    ``{  ` `        ``for` `(``int` `j = ``0``; j < i; j++)  ` `        ``{  ` ` `  `            ``// check for increasing subsequence  ` `            ``if` `(arr[j] < arr[i])  ` `            ``{  ` `                ``for` `(``int` `l = ``1``; l <= k - ``1``; l++)  ` `                ``{  ` ` `  `                    ``// Proceed if value is pre calculated  ` `                    ``if` `(dp[j][l] != -``1``) ` `                    ``{  ` ` `  `                        ``// Check for all the subsequences  ` `                        ``// ending at any j

 `# Python program to calculate the maximum sum ` `# of increasing subsequence of length k ` ` `  `def` `MaxIncreasingSub(arr, n, k): ` `     `  `    ``# In the implementation dp[n][k] represents  ` `    ``# maximum sum subsequence of length k and the  ` `    ``# subsequence is ending at index n. ` `    ``dp ``=` `[``-``1``]``*``n ` `    ``ans ``=` `-``1` ` `  `    ``# Initializing whole multidimensional ` `    ``# dp array with value - 1 ` `    ``for` `i ``in` `range``(n): ` `        ``dp[i] ``=` `[``-``1``]``*``(k``+``1``) ` ` `  `    ``# For each ith position increasing subsequence ` `    ``# of length 1 is equal to that array ith value ` `    ``# so initializing dp[i] with that array value ` `    ``for` `i ``in` `range``(n): ` `        ``dp[i][``1``] ``=` `arr[i] ` `     `  `    ``# Starting from 1st index as we have calculated ` `    ``# for 0th index. Computing optimized dp values ` `    ``# in bottom-up manner ` `    ``for` `i ``in` `range``(``1``,n): ` `        ``for` `j ``in` `range``(i): ` `             `  `            ``# check for increasing subsequence ` `            ``if` `arr[j] < arr[i]: ` `                ``for` `l ``in` `range``(``1``,k): ` ` `  `                    ``# Proceed if value is pre calculated ` `                    ``if` `dp[j][l] !``=` `-``1``: ` `                         `  `                        ``# Check for all the subsequences ` `                        ``# ending at any j < i and try including ` `                        ``# element at index i in them for ` `                        ``# some length l. Update the maximum ` `                        ``# value for every length. ` `                        ``dp[i][l``+``1``] ``=` `max``(dp[i][l``+``1``], ` `                                        ``dp[j][l] ``+` `arr[i]) ` `     `  `    ``# The final result would be the maximum  ` `    ``# value of dp[i][k] for all different i. ` `    ``for` `i ``in` `range``(n): ` `        ``if` `ans < dp[i][k]: ` `            ``ans ``=` `dp[i][k] ` `     `  `    ``# When no subsequence of length k is ` `    ``# possible sum would be considered zero ` `    ``return` `(``0` `if` `ans ``=``=` `-``1` `else` `ans) ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` ` `  `    ``n, k ``=` `8``, ``3` `    ``arr ``=` `[``8``, ``5``, ``9``, ``10``, ``5``, ``6``, ``21``, ``8``] ` `    ``ans ``=` `MaxIncreasingSub(arr, n, k) ` `    ``print``(ans) ` ` `  `# This code is contributed by ` `# sanjeev2552 `

 `/*C# program to calculate the maximum sum of  ` `increasing subsequence of length k*/` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `static` `int` `MaxIncreasingSub(``int` `[]arr, ``int` `n, ``int` `k)  ` `{  ` `    ``// In the implementation dp[n,k] represents  ` `    ``// maximum sum subsequence of length k and the  ` `    ``// subsequence is ending at index n.  ` `    ``int` `[,]dp=``new` `int``[n, k + 1]; ` `    ``int` `ans = -1;  ` ` `  `    ``// Initializing whole multidimensional  ` `    ``// dp array with value -1  ` `    ``for``(``int` `i = 0; i < n; i++) ` `        ``for``(``int` `j = 0; j < k + 1; j++) ` `            ``dp[i, j]=-1; ` ` `  `    ``// For each ith position increasing subsequence  ` `    ``// of length 1 is equal to that array ith value  ` `    ``// so initializing dp[i,1] with that array value  ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{  ` `        ``dp[i, 1] = arr[i];  ` `    ``}  ` ` `  `    ``// Starting from 1st index as we have calculated  ` `    ``// for 0th index. Computing optimized dp values  ` `    ``// in bottom-up manner  ` `    ``for` `(``int` `i = 1; i < n; i++)  ` `    ``{  ` `        ``for` `(``int` `j = 0; j < i; j++)  ` `        ``{  ` ` `  `            ``// check for increasing subsequence  ` `            ``if` `(arr[j] < arr[i])  ` `            ``{  ` `                ``for` `(``int` `l = 1; l <= k - 1; l++)  ` `                ``{  ` ` `  `                    ``// Proceed if value is pre calculated  ` `                    ``if` `(dp[j, l] != -1) ` `                    ``{  ` ` `  `                        ``// Check for all the subsequences  ` `                        ``// ending at any j

Output:
```40
```

Time complexity: O(n^2*k)
Space complexity: O(n^2)

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