# Printing Maximum Sum Increasing Subsequence

The Maximum Sum Increasing Subsequence problem is to find the maximum sum subsequence of a given sequence such that all elements of the subsequence are sorted in increasing order.

Examples:

```Input:  [1, 101, 2, 3, 100, 4, 5]
Output: [1, 2, 3, 100]

Input:  [3, 4, 5, 10]
Output: [3, 4, 5, 10]

Input:  [10, 5, 4, 3]
Output: 

Input:  [3, 2, 6, 4, 5, 1]
Output: [3, 4, 5]
```

In previous post, we have discussed about Maximum Sum Increasing Subsequence problem. However, the post only covered code related to finding maximum sum of increasing subsequence, but not to the construction of subsequence. In this post, we will discuss how to construct Maximum Sum Increasing Subsequence itself.

Let arr[0..n-1] be the input array. We define vector L such that L[i] is itself is a vector that stores Maximum Sum Increasing Subsequence of arr[0..i] that ends with arr[i]. Therefore for an index i, L[i] can be recursively written as

```L = {arr}
L[i] = {MaxSum(L[j])} + arr[i] where j < i and arr[j] < arr[i]
= arr[i], if there is no j such that arr[j] < arr[i]
```

For example, for array [3, 2, 6, 4, 5, 1],

```L: 3
L: 2
L: 3 6
L: 3 4
L: 3 4 5
L: 1
```

Below is C++ implementation of above idea –

 `/* Dynamic Programming solution to construct ` `   ``Maximum Sum Increasing Subsequence */` `#include ` `#include ` `using` `namespace` `std; ` ` `  `// Utility function to calculate sum of all ` `// vector elements ` `int` `findSum(vector<``int``> arr) ` `{ ` `    ``int` `sum = 0; ` `    ``for` `(``int` `i: arr) ` `        ``sum += i; ` `    ``return` `sum; ` `} ` ` `  `// Function to construct Maximum Sum Increasing ` `// Subsequence ` `void` `printMaxSumIS(``int` `arr[], ``int` `n) ` `{ ` `    ``// L[i] - The Maximum Sum Increasing ` `    ``// Subsequence that ends with arr[i] ` `    ``vector > L(n); ` ` `  `    ``// L is equal to arr ` `    ``L.push_back(arr); ` ` `  `    ``// start from index 1 ` `    ``for` `(``int` `i = 1; i < n; i++) ` `    ``{ ` `        ``// for every j less than i ` `        ``for` `(``int` `j = 0; j < i; j++) ` `        ``{ ` `            ``/* L[i] = {MaxSum(L[j])} + arr[i] ` `            ``where j < i and arr[j] < arr[i] */` `            ``if` `((arr[i] > arr[j]) && ` `                ``(findSum(L[i]) < findSum(L[j]))) ` `                ``L[i] = L[j]; ` `        ``} ` ` `  `        ``// L[i] ends with arr[i] ` `        ``L[i].push_back(arr[i]); ` ` `  `        ``// L[i] now stores Maximum Sum Increasing ` `        ``// Subsequence of arr[0..i] that ends with ` `        ``// arr[i] ` `    ``} ` ` `  `    ``vector<``int``> res = L; ` ` `  `    ``// find max ` `    ``for` `(vector<``int``> x : L) ` `        ``if` `(findSum(x) > findSum(res)) ` `            ``res = x; ` ` `  `    ``// max will contain result ` `    ``for` `(``int` `i : res) ` `        ``cout << i << ``" "``; ` `    ``cout << endl; ` `} ` ` `  `// Driver function ` `int` `main() ` `{ ` `    ``int` `arr[] = { 3, 2, 6, 4, 5, 1 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``// construct and print Max Sum IS of arr ` `    ``printMaxSumIS(arr, n); ` ` `  `    ``return` `0; ` `} `

Output:

```3 4 5
```

We can optimize above DP solution by removing findSum() function. Instead, we can maintain another vector/array to store sum of maximum sum increasing subsequence that ends with arr[i]. The implementation can be seen here.

Time complexity of above Dynamic Programming solution is O(n2).
Auxiliary space used by the program is O(n2).