Printing Maximum Sum Increasing Subsequence

Last Updated : 23 Apr, 2025

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]

In previous post, we have discussed the 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 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 the 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 Code 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

Python

# 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

/* 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

JavaScript

<script>  /* Dynamic Programming solution to construct Maximum Sum Increasing Subsequence */          // Utility function to calculate sum of all     // vector elements     function findSum(arr)     {         let sum = 0;         for (let i=0;i<arr.length;i++)             sum += arr[i];         return sum;     }          // Function to construct Maximum Sum Increasing     // Subsequence     function printMaxSumIs(arr,n)     {         // L[i] - The Maximum Sum Increasing         // Subsequence that ends with arr[i]                  let L = new Array(n);           for (let i = 0; i < n; i++)             L[i] = [];           // L[0] is equal to arr[0]         L[0].push(arr[0]);           // start from index 1         for (let i = 1; i < n; i++) {               // for every j less than i             for (let 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 (let k=0;k<L[j].length;k++)                         if (!L[i].includes(L[j][k]))                             L[i].push(L[j][k]);                 }             }               // L[i] ends with arr[i]             L[i].push(arr[i]);               // L[i] now stores Maximum Sum Increasing             // Subsequence of arr[0..i] that ends with             // arr[i]         }                   let res = L[0];         // res = L[0];               // find max         for (let x=0;x<L.length;x++)             if (findSum(L[x]) > findSum(res))                 res = L[x];           // max will contain result         for (let i=0;i<res.length;i++)             document.write(res[i] + " ");         document.write("<br>");     }           // Driver Code     let arr=[3, 2, 6, 4, 5, 1];     let n = arr.length;          // construct and print Max Sum IS of arr     printMaxSumIs(arr, n);       // This code is contributed by unknown2108  </script>

Output

3 4 5

We can optimize the 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].

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

Approach 2: (Using Dynamic Programming Using O(N) space

The above approach covered how to construct a Maximum Sum Increasing Subsequence in O(N²) time and O(N²) space. In this approach, we will optimize the Space complexity and construct the Maximum Sum Increasing Subsequence in O(N²) time and O(N) space.

Let arr[0..n-1] be the input array.
We define a vector of pairs L such that L[i] first stores the Maximum Sum Increasing Subsequence of arr[0..i] that ends with arr[i] and L[i].second stores the index of the previous element used for generating the sum.
As the first element does not have any previous element hence its index would be -1 in L[0].

For example,

array = [3, 2, 6, 4, 5, 1]

L[0]: {3, -1}
L[1]: {2,  1}
L[2]: {9,  0}
L[3]: {7, 0}
L[4]: {12, 3}
L[5]: {1, 5}

As we can see above, the value of the Maximum Sum Increasing Subsequence is 12. To construct the actual Subsequence we will use the index stored in L[i].second. The steps to construct the Subsequence is shown below:

In a vector result, store the value of the element where the Maximum Sum Increasing Subsequence was found (i.e at currIndex = 4). So in the result vector, we will add arr[currIndex].
Update the currIndex to L[currIndex].second and repeat step 1 until currIndex is not -1 or it does not changes (i.e currIndex == previousIndex).
Display the elements of the result vector in reverse order.

Below is the implementation of the above idea :

C++14

/* Dynamic Programming solution to construct Maximum Sum Increasing Subsequence */ #include <bits/stdc++.h> using namespace std;  // Function to construct and print the Maximum Sum // Increasing Subsequence void constructMaxSumIS(vector<int> arr, int n) {     // L[i] stores the value of Maximum Sum Increasing     // Subsequence that ends with arr[i] and the index of     // previous element used to construct the Subsequence     vector<pair<int, int> > L(n);      int index = 0;     for (int i : arr) {         L[index] = { i, index };         index++;     }      // Set L[0].second equal to -1     L[0].second = -1;      // start from index 1     for (int i = 1; i < n; i++) {         // for every j less than i         for (int j = 0; j < i; j++) {             if (arr[i] > arr[j]                 and L[i].first < arr[i] + L[j].first) {                 L[i].first = arr[i] + L[j].first;                 L[i].second = j;             }         }     }      int maxi = INT_MIN, currIndex, track = 0;      for (auto p : L) {         if (p.first > maxi) {             maxi = p.first;             currIndex = track;         }         track++;     }      // Stores the final Subsequence     vector<int> result;      // Index of previous element     // used to construct the Subsequence     int prevoiusIndex;      while (currIndex >= 0) {         result.push_back(arr[currIndex]);         prevoiusIndex = L[currIndex].second;          if (currIndex == prevoiusIndex)             break;          currIndex = prevoiusIndex;     }      for (int i = result.size() - 1; i >= 0; i--)         cout << result[i] << " "; }  // Driver Code int main() {     vector<int> arr = { 1, 101, 2, 3, 100, 4, 5 };     int n = arr.size();      // Function call     constructMaxSumIS(arr, n);      return 0; }

Java

// Dynamic Programming solution to construct // Maximum Sum Increasing Subsequence  import java.util.*; import java.awt.Point;  class GFG{      // Function to construct and print the Maximum Sum // Increasing Subsequence static void constructMaxSumIS(List<Integer> arr, int n) {          // L.get(i) stores the value of Maximum Sum Increasing     // Subsequence that ends with arr.get(i) and the index of     // previous element used to construct the Subsequence     List<Point> L = new ArrayList<Point>();      int index = 0;     for(int i : arr)     {         L.add(new Point(i, index));         index++;     }      // Set L[0].second equal to -1     L.set(0, new Point(L.get(0).x, -1));      // Start from index 1     for(int i = 1; i < n; i++)     {                  // For every j less than i         for(int j = 0; j < i; j++)         {             if (arr.get(i) > arr.get(j) &&                 L.get(i).x < arr.get(i) +                 L.get(j).x)              {                 L.set(i, new Point(arr.get(i) +                                      L.get(j).x, j));             }         }     }      int maxi = -100000000, currIndex = 0, track = 0;      for(Point p : L)      {         if (p.x > maxi)         {             maxi = p.x;             currIndex = track;         }         track++;     }      // Stores the final Subsequence     List<Integer> result = new ArrayList<Integer>();      // Index of previous element     // used to construct the Subsequence     int prevoiusIndex;      while (currIndex >= 0)      {         result.add(arr.get(currIndex));         prevoiusIndex = L.get(currIndex).y;          if (currIndex == prevoiusIndex)             break;          currIndex = prevoiusIndex;     }      for(int i = result.size() - 1; i >= 0; i--)         System.out.print(result.get(i) + " "); }   // Driver Code public static void main(String []s) {     List<Integer> arr = new ArrayList<Integer>();     arr.add(1);     arr.add(101);     arr.add(2);     arr.add(3);     arr.add(100);     arr.add(4);     arr.add(5);          int n = arr.size();      // Function call     constructMaxSumIS(arr, n);  } }  // This code is contributed by rutvik_56

Python

# Dynamic Programming solution to construct # Maximum Sum Increasing Subsequence  import sys  # Function to construct and print the Maximum Sum # Increasing Subsequence def constructMaxSumIS(arr, n) :      # L[i] stores the value of Maximum Sum Increasing     # Subsequence that ends with arr[i] and the index of     # previous element used to construct the Subsequence     L = []      index = 0     for i in arr :         L.append([i, index])         index += 1      # Set L[0].second equal to -1     L[0][1] = -1      # start from index 1     for i in range(1, n) :                # for every j less than i         for j in range(i) :             if (arr[i] > arr[j] and L[i][0] < arr[i] + L[j][0]) :                 L[i][0] = arr[i] + L[j][0]                 L[i][1] = j      maxi, currIndex, track = -sys.maxsize, 0, 0      for p in L :         if (p[0] > maxi) :             maxi = p[0]             currIndex = track              track += 1      # Stores the final Subsequence     result = []      while (currIndex >= 0) :         result.append(arr[currIndex])         prevoiusIndex = L[currIndex][1]          if (currIndex == prevoiusIndex) :             break          currIndex = prevoiusIndex      for i in range(len(result) - 1, -1, -1) :         print(result[i] , end = " ")   arr = [ 1, 101, 2, 3, 100, 4, 5 ] n = len(arr)  # Function call constructMaxSumIS(arr, n)  # This code is contributed by divyeshrabadiya07

/* Dynamic Programming solution to construct Maximum Sum Increasing Subsequence */ using System; using System.Collections.Generic;  class GFG  {          // Function to construct and print the Maximum Sum     // Increasing Subsequence     static void constructMaxSumIS(List<int> arr, int n)     {                // L[i] stores the value of Maximum Sum Increasing         // Subsequence that ends with arr[i] and the index of         // previous element used to construct the Subsequence         List<Tuple<int, int>> L = new List<Tuple<int, int>>();               int index = 0;         foreach(int i in arr) {             L.Add(new Tuple<int, int>(i, index));             index++;         }               // Set L[0].second equal to -1         L[0] = new Tuple<int, int>(L[0].Item1, -1);               // start from index 1         for (int i = 1; i < n; i++)         {                        // for every j less than i             for (int j = 0; j < i; j++)             {                 if (arr[i] > arr[j] &&                     L[i].Item1 < arr[i] +                     L[j].Item1)                  {                     L[i] = new Tuple<int,                    int>(arr[i] + L[j].Item1, j);                 }             }         }               int maxi = Int32.MinValue,        currIndex = 0, track = 0;               foreach(Tuple<int, int> p in L)          {             if (p.Item1 > maxi)             {                 maxi = p.Item1;                 currIndex = track;             }             track++;         }               // Stores the final Subsequence         List<int> result = new List<int>();               // Index of previous element         // used to construct the Subsequence         int prevoiusIndex;               while (currIndex >= 0)          {             result.Add(arr[currIndex]);             prevoiusIndex = L[currIndex].Item2;                   if (currIndex == prevoiusIndex)                 break;                   currIndex = prevoiusIndex;         }               for (int i = result.Count - 1; i >= 0; i--)             Console.Write(result[i] + " ");     }      static void Main()   {     List<int> arr = new List<int>(new                                    int[] { 1, 101, 2, 3, 100, 4, 5 });     int n = arr.Count;       // Function call     constructMaxSumIS(arr, n);    } }  // This code is contributed by divyesh072019

JavaScript

<script>  // Dynamic Programming solution to construct // Maximum Sum Increasing Subsequence   // Function to construct and print the Maximum Sum // Increasing Subsequence function constructMaxSumIS(arr, n){      // L[i] stores the value of Maximum Sum Increasing     // Subsequence that ends with arr[i] and the index of     // previous element used to construct the Subsequence     let L = []      let index = 0     for(let i of arr){         L.push([i, index])         index += 1     }      // Set L[0].second equal to -1     L[0][1] = -1      // start from index 1     for(let i=1;i<n;i++){                // for every j less than i         for(let j=0;j<i;j++){             if (arr[i] > arr[j] && L[i][0] < arr[i] + L[j][0]){                 L[i][0] = arr[i] + L[j][0]                 L[i][1] = j             }         }     }      let maxi = Number.MIN_VALUE, currIndex = 0, track = 0      for(let p of L){         if (p[0] > maxi){             maxi = p[0]             currIndex = track         }              track += 1     }      // Stores the final Subsequence     let result = []      while (currIndex >= 0){         result.push(arr[currIndex])         let prevoiusIndex = L[currIndex][1]          if (currIndex == prevoiusIndex)             break          currIndex = prevoiusIndex     }      for(let i=result.length - 1;i>=0;i--)          document.write(result[i] ," ") }   let arr = [ 1, 101, 2, 3, 100, 4, 5 ] let n = arr.length  // Function call constructMaxSumIS(arr, n)  // This code is contributed by shinjanpatra  </script>

Output

1 2 3 100

Time Complexity: O(N²)
Space Complexity: O(N)

Printing Longest Increasing Subsequence (LIS)

Aditya Goel

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Printing Maximum Sum Increasing Subsequence

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