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]

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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 the implementation of above idea –

C++

 /* 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 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 res = L;        // find max     for (vector 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; }

Python3

 # Dynamic Programming solution to construct # Maximum Sum Increasing Subsequence */    # Utility function to calculate sum of all # vector elements def findSum(arr):        summ = 0     for i in arr:         summ += i     return summ    # Function to construct Maximum Sum Increasing # Subsequence def printMaxSumIS(arr, n):            # L[i] - The Maximum Sum Increasing     # Subsequence that ends with arr[i]     L = [[] for i in range(n)]        # L is equal to arr     L.append(arr)        # start from index 1     for i in range(1, n):                    # for every j less than i         for j in range(i):                            # L[i] = {MaxSum(L[j])} + arr[i]             # where j < i and arr[j] < arr[i]             if ((arr[i] > arr[j]) and                 (findSum(L[i]) < findSum(L[j]))):                 for e in L[j]:                     if e not in L[i]:                         L[i].append(e)            # L[i] ends with arr[i]         L[i].append(arr[i])            # L[i] now stores Maximum Sum Increasing         # Subsequence of arr[0..i] that ends with         # arr[i]        res = L        # find max     for x in L:         if (findSum(x) > findSum(res)):             res = x        # max will contain result     for i in res:         print(i, end = " ")    # Driver Code arr = [3, 2, 6, 4, 5, 1] n = len(arr)    # construct and prMax Sum IS of arr printMaxSumIS(arr, n)    # This code is contributed by Mohit Kumar

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).

This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : mohit kumar 29

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