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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: [10]

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[0] = {arr[0]}
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[0]: 3
L[1]: 2
L[2]: 3 6
L[3]: 3 4
L[4]: 3 4 5
L[5]: 1

Below is the implementation of above idea –

C++

/* Dynamic Programming solution to construct 
   Maximum Sum Increasing Subsequence */
#include <iostream> 
#include <vector> 
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 <vector<int> > L(n); 
  
    // L[0] is equal to arr[0] 
    L[0].push_back(arr[0]); 
  
    // 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[0]; 
  
    // 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[0]); 
  
    // construct and print Max Sum IS of arr 
    printMaxSumIS(arr, n); 
  
    return 0; 
} 

Java

/* Dynamic Programming solution to construct  
Maximum Sum Increasing Subsequence */
import java.util.*; 
  
class GFG  
{ 
  
    // Utility function to calculate sum of all 
    // vector elements 
    static int findSum(Vector<Integer> arr)  
    { 
        int sum = 0; 
        for (int i : arr) 
            sum += i; 
        return sum; 
    } 
  
    // Function to construct Maximum Sum Increasing 
    // Subsequence 
    static void printMaxSumIs(int[] arr, int n)  
    { 
  
        // L[i] - The Maximum Sum Increasing 
        // Subsequence that ends with arr[i] 
        @SuppressWarnings("unchecked") 
        Vector<Integer>[] L = new Vector[n]; 
  
        for (int i = 0; i < n; i++) 
            L[i] = new Vector<>(); 
  
        // L[0] is equal to arr[0] 
        L[0].add(arr[0]); 
  
        // 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])))  
                { 
                    for (int k : L[j]) 
                        if (!L[i].contains(k)) 
                            L[i].add(k); 
                } 
            } 
  
            // L[i] ends with arr[i] 
            L[i].add(arr[i]); 
  
            // L[i] now stores Maximum Sum Increasing 
            // Subsequence of arr[0..i] that ends with 
            // arr[i] 
        } 
  
        Vector<Integer> res = new Vector<>(L[0]); 
        // res = L[0]; 
  
        // find max 
        for (Vector<Integer> x : L) 
            if (findSum(x) > findSum(res)) 
                res = x; 
  
        // max will contain result 
        for (int i : res) 
            System.out.print(i + " "); 
        System.out.println(); 
    } 
  
    // Driver Code 
    public static void main(String[] args)  
    { 
        int[] arr = { 3, 2, 6, 4, 5, 1 }; 
        int n = arr.length; 
  
        // construct and print Max Sum IS of arr 
        printMaxSumIs(arr, n); 
    } 
} 
  
// This code is contributed by 
// sanjeev2552 

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[0] is equal to arr[0] 
    L[0].append(arr[0]) 
  
    # 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[0] 
  
    # 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 

C#

/* Dynamic Programming solution to construct  
Maximum Sum Increasing Subsequence */
using System; 
using System.Collections.Generic; 
  
class GFG  
{ 
  
    // Utility function to calculate sum of all 
    // vector elements 
    static int findSum(List<int> arr)  
    { 
        int sum = 0; 
        foreach (int i in arr) 
            sum += i; 
        return sum; 
    } 
  
    // Function to construct Maximum Sum Increasing 
    // Subsequence 
    static void printMaxSumIs(int[] arr, int n)  
    { 
  
        // L[i] - The Maximum Sum Increasing 
        // Subsequence that ends with arr[i] 
        List<int>[] L = new List<int>[n]; 
  
        for (int i = 0; i < n; i++) 
            L[i] = new List<int>(); 
  
        // L[0] is equal to arr[0] 
        L[0].Add(arr[0]); 
  
        // 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])))  
                { 
                    foreach (int k in L[j]) 
                        if (!L[i].Contains(k)) 
                            L[i].Add(k); 
                } 
            } 
  
            // L[i] ends with arr[i] 
            L[i].Add(arr[i]); 
  
            // L[i] now stores Maximum Sum Increasing 
            // Subsequence of arr[0..i] that ends with 
            // arr[i] 
        } 
  
        List<int> res = new List<int>(L[0]); 
        // res = L[0]; 
  
        // find max 
        foreach (List<int> x in L) 
            if (findSum(x) > findSum(res)) 
                res = x; 
  
        // max will contain result 
        foreach (int i in res) 
            Console.Write(i + " "); 
        Console.WriteLine(); 
    } 
  
    // Driver Code 
    public static void Main(String[] args)  
    { 
        int[] arr = { 3, 2, 6, 4, 5, 1 }; 
        int n = arr.Length; 
  
        // construct and print Max Sum IS of arr 
        printMaxSumIs(arr, n); 
    } 
} 
  
// This code is contributed by PrinciRaj1992 

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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Printing Maximum Sum Increasing Subsequence

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

C++

Java

Python3

C#

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