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Printing Maximum Sum Increasing Subsequence
• Difficulty Level : Medium
• Last Updated : 28 Dec, 2020

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 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 the 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<``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[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 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[] 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 res = ``new` `Vector<>(L[``0``]);` `        ``// res = L[0];`   `        ``// find max` `        ``for` `(Vector 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 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]. 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).

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 ` `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 > 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 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 L = ``new` `ArrayList();`   `    ``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 result = ``new` `ArrayList();`   `    ``// 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 arr = ``new` `ArrayList();` `    ``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`

## Python3

 `# 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`

## C#

 `/* 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> L = ``new` `List>();` `     `  `        ``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`

Output

`1 2 3 100`

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