Largest Independent Set Problem | DP-26

• Difficulty Level : Medium
• Last Updated : 19 Jul, 2021

Given a Binary Tree, find size of the Largest Independent Set(LIS) in it. A subset of all tree nodes is an independent set if there is no edge between any two nodes of the subset.

For example, consider the following binary tree. The largest independent set(LIS) is {10, 40, 60, 70, 80} and size of the LIS is 5. A Dynamic Programming solution solves a given problem using solutions of subproblems in bottom up manner. Can the given problem be solved using solutions to subproblems? If yes, then what are the subproblems? Can we find largest independent set size (LISS) for a node X if we know LISS for all descendants of X? If a node is considered as part of LIS, then its children cannot be part of LIS, but its grandchildren can be. Following is optimal substructure property.

1) Optimal Substructure:
Let LISS(X) indicates size of largest independent set of a tree with root X.

LISS(X) = MAX { (1 + sum of LISS for all grandchildren of X),
(sum of LISS for all children of X) }

The idea is simple, there are two possibilities for every node X, either X is a member of the set or not a member. If X is a member, then the value of LISS(X) is 1 plus LISS of all grandchildren. If X is not a member, then the value is sum of LISS of all children.

2) Overlapping Subproblems
Following is recursive implementation that simply follows the recursive structure mentioned above.

C++

 // A naive recursive implementation of// Largest Independent Set problem#include using namespace std; // A utility function to find// max of two integersint max(int x, int y){    return (x > y) ? x : y;} /* A binary tree node has data,pointer to left child and apointer to right child */class node{    public:    int data;    node *left, *right;}; // The function returns size of the// largest independent set in a given// binary treeint LISS(node *root){    if (root == NULL)    return 0;     // Calculate size excluding the current node    int size_excl = LISS(root->left) +                    LISS(root->right);     // Calculate size including the current node    int size_incl = 1;    if (root->left)        size_incl += LISS(root->left->left) +                     LISS(root->left->right);    if (root->right)        size_incl += LISS(root->right->left) +                     LISS(root->right->right);     // Return the maximum of two sizes    return max(size_incl, size_excl);} // A utility function to create a nodenode* newNode( int data ){    node* temp = new node();    temp->data = data;    temp->left = temp->right = NULL;    return temp;} // Driver Codeint main(){    // Let us construct the tree    // given in the above diagram    node *root = newNode(20);    root->left = newNode(8);    root->left->left = newNode(4);    root->left->right = newNode(12);    root->left->right->left = newNode(10);    root->left->right->right = newNode(14);    root->right = newNode(22);    root->right->right = newNode(25);     cout << "Size of the Largest"         << " Independent Set is "         << LISS(root);     return 0;} // This is code is contributed// by rathbhupendra

C

 // A naive recursive implementation of Largest Independent Set problem#include #include  // A utility function to find max of two integersint max(int x, int y) { return (x > y)? x: y; } /* A binary tree node has data, pointer to left child and a pointer to   right child */struct node{    int data;    struct node *left, *right;}; // The function returns size of the largest independent set in a given// binary treeint LISS(struct node *root){    if (root == NULL)       return 0;     // Calculate size excluding the current node    int size_excl = LISS(root->left) + LISS(root->right);     // Calculate size including the current node    int size_incl = 1;    if (root->left)       size_incl += LISS(root->left->left) + LISS(root->left->right);    if (root->right)       size_incl += LISS(root->right->left) + LISS(root->right->right);     // Return the maximum of two sizes    return max(size_incl, size_excl);}  // A utility function to create a nodestruct node* newNode( int data ){    struct node* temp = (struct node *) malloc( sizeof(struct node) );    temp->data = data;    temp->left = temp->right = NULL;    return temp;} // Driver program to test above functionsint main(){    // Let us construct the tree given in the above diagram    struct node *root         = newNode(20);    root->left                = newNode(8);    root->left->left          = newNode(4);    root->left->right         = newNode(12);    root->left->right->left   = newNode(10);    root->left->right->right  = newNode(14);    root->right               = newNode(22);    root->right->right        = newNode(25);     printf ("Size of the Largest Independent Set is %d ", LISS(root));     return 0;}

Java

 // A naive recursive implementation of// Largest Independent Set problemclass GFG { // A utility function to find// max of two integersstatic int max(int x, int y){    return (x > y) ? x : y;} /* A binary tree node has data,pointer to left child and apointer to right child */static class Node{    int data;    Node left, right;}; // The function returns size of the// largest independent set in a given// binary treestatic int LISS(Node root){    if (root == null)    return 0;     // Calculate size excluding the current node    int size_excl = LISS(root.left) +                    LISS(root.right);     // Calculate size including the current node    int size_incl = 1;    if (root.left!=null)        size_incl += LISS(root.left.left) +                    LISS(root.left.right);    if (root.right!=null)        size_incl += LISS(root.right.left) +                    LISS(root.right.right);     // Return the maximum of two sizes    return max(size_incl, size_excl);} // A utility function to create a nodestatic Node newNode( int data ){    Node temp = new Node();    temp.data = data;    temp.left = temp.right = null;    return temp;} // Driver Codepublic static void main(String args[]) {    // Let us construct the tree    // given in the above diagram    Node root = newNode(20);    root.left = newNode(8);    root.left.left = newNode(4);    root.left.right = newNode(12);    root.left.right.left = newNode(10);    root.left.right.right = newNode(14);    root.right = newNode(22);    root.right.right = newNode(25);     System.out.println("Size of the Largest"        + " Independent Set is "        + LISS(root));    }} // This code has been contributed by 29AjayKumar

Python3

 # A naive recursive implementation of# Largest Independent Set problem # A utility function to find# max of two integersdef max(x, y):    if(x > y):        return x    else:        return y # A binary tree node has data,#pointer to left child and a#pointer to right childclass node :    def __init__(self):        self.data = 0        self.left = self.right = None # The function returns size of the# largest independent set in a given# binary treedef LISS(root):     if (root == None) :        return 0     # Calculate size excluding the current node    size_excl = LISS(root.left) + LISS(root.right)     # Calculate size including the current node    size_incl = 1    if (root.left != None):        size_incl += LISS(root.left.left) + \                    LISS(root.left.right)    if (root.right != None):        size_incl += LISS(root.right.left) + \                    LISS(root.right.right)     # Return the maximum of two sizes    return max(size_incl, size_excl) # A utility function to create a nodedef newNode( data ) :     temp = node()    temp.data = data    temp.left = temp.right = None    return temp # Driver Code # Let us construct the tree# given in the above diagramroot = newNode(20)root.left = newNode(8)root.left.left = newNode(4)root.left.right = newNode(12)root.left.right.left = newNode(10)root.left.right.right = newNode(14)root.right = newNode(22)root.right.right = newNode(25) print( "Size of the Largest"        , " Independent Set is "        , LISS(root) ) # This code is contributed by Arnab Kundu

C#

 // C# program for calculating LISS// using dynamic programmingusing System; class LisTree{    /* A binary tree node has data, pointer    to left child and a pointer to right    child */    public class node    {        public int data, liss;        public node left, right;         public node(int data)        {            this.data = data;            this.liss = 0;        }    }     // A memoization function returns size    // of the largest independent set in    // a given binary tree    static int liss(node root)    {        if (root == null)            return 0;        if (root.liss != 0)            return root.liss;        if (root.left == null && root.right == null)            return root.liss = 1;                 // Calculate size excluding the        // current node        int liss_excl = liss(root.left) + liss(root.right);                 // Calculate size including the        // current node        int liss_incl = 1;        if (root.left != null)        {            liss_incl += (liss(root.left.left) +                        liss(root.left.right));        }        if (root.right != null)        {            liss_incl += (liss(root.right.left) +                        liss(root.right.right));        }                 // Maximum of two sizes is LISS,        // store it for future uses.        return root.liss = Math.Max(liss_excl, liss_incl);    }     // Driver code    public static void Main(String[] args)    {        // Let us construct the tree given        // in the above diagram                 node root = new node(20);        root.left = new node(8);        root.left.left = new node(4);        root.left.right = new node(12);        root.left.right.left = new node(10);        root.left.right.right = new node(14);        root.right = new node(22);        root.right.right = new node(25);        Console.WriteLine("Size of the Largest Independent Set is " + liss(root));    }} // This code is contributed by Princi Singh

Javascript



Output:

Size of the Largest Independent Set is 5

Time complexity of the above naive recursive approach is exponential. It should be noted that the above function computes the same subproblems again and again. For example, LISS of node with value 50 is evaluated for node with values 10 and 20 as 50 is grandchild of 10 and child of 20.

Since same subproblems are called again, this problem has Overlapping Subproblems property. So LISS problem has both properties (see this and this) of a dynamic programming problem. Like other typical Dynamic Programming(DP) problems, recomputations of same subproblems can be avoided by storing the solutions to subproblems and solving problems in bottom up manner.

Following are implementation of Dynamic Programming based solution. In the following solution, an additional field ‘liss’ is added to tree nodes. The initial value of ‘liss’ is set as 0 for all nodes. The recursive function LISS() calculates ‘liss’ for a node only if it is not already set.

C++

 /* Dynamic programming based programfor Largest Independent Set problem */#include using namespace std; // A utility function to find max of two integersint max(int x, int y) { return (x > y)? x: y; } /* A binary tree node has data, pointerto left child and a pointer toright child */class node{    public:    int data;    int liss;    node *left, *right;}; // A memoization function returns size// of the largest independent set in// a given binary treeint LISS(node *root){    if (root == NULL)        return 0;     if (root->liss)        return root->liss;     if (root->left == NULL && root->right == NULL)        return (root->liss = 1);     // Calculate size excluding the current node    int liss_excl = LISS(root->left) + LISS(root->right);     // Calculate size including the current node    int liss_incl = 1;    if (root->left)        liss_incl += LISS(root->left->left) + LISS(root->left->right);    if (root->right)        liss_incl += LISS(root->right->left) + LISS(root->right->right);     // Maximum of two sizes is LISS, store it for future uses.    root->liss = max(liss_incl, liss_excl);     return root->liss;} // A utility function to create a nodenode* newNode(int data){    node* temp = new node();    temp->data = data;    temp->left = temp->right = NULL;    temp->liss = 0;    return temp;} // Driver codeint main(){    // Let us construct the tree    // given in the above diagram    node *root     = newNode(20);    root->left         = newNode(8);    root->left->left     = newNode(4);    root->left->right     = newNode(12);    root->left->right->left = newNode(10);    root->left->right->right = newNode(14);    root->right         = newNode(22);    root->right->right     = newNode(25);     cout << "Size of the Largest Independent Set is " << LISS(root);     return 0;} // This code is contributed by rathbhupendra

C

 /* Dynamic programming based program for Largest Independent Set problem */#include #include  // A utility function to find max of two integersint max(int x, int y) { return (x > y)? x: y; } /* A binary tree node has data, pointer to left child and a pointer to   right child */struct node{    int data;    int liss;    struct node *left, *right;}; // A memoization function returns size of the largest independent set in//  a given binary treeint LISS(struct node *root){    if (root == NULL)        return 0;     if (root->liss)        return root->liss;     if (root->left == NULL && root->right == NULL)        return (root->liss = 1);     // Calculate size excluding the current node    int liss_excl = LISS(root->left) + LISS(root->right);     // Calculate size including the current node    int liss_incl = 1;    if (root->left)        liss_incl += LISS(root->left->left) + LISS(root->left->right);    if (root->right)        liss_incl += LISS(root->right->left) + LISS(root->right->right);     // Maximum of two sizes is LISS, store it for future uses.    root->liss = max(liss_incl, liss_excl);     return root->liss;} // A utility function to create a nodestruct node* newNode(int data){    struct node* temp = (struct node *) malloc( sizeof(struct node) );    temp->data = data;    temp->left = temp->right = NULL;    temp->liss = 0;    return temp;} // Driver program to test above functionsint main(){    // Let us construct the tree given in the above diagram    struct node *root         = newNode(20);    root->left                = newNode(8);    root->left->left          = newNode(4);    root->left->right         = newNode(12);    root->left->right->left   = newNode(10);    root->left->right->right  = newNode(14);    root->right               = newNode(22);    root->right->right        = newNode(25);     printf ("Size of the Largest Independent Set is %d ", LISS(root));     return 0;}

Java

 // Java program for calculating LISS// using dynamic programming public class LisTree{    /* A binary tree node has data, pointer       to left child and a pointer to right       child */    static class node    {        int data, liss;        node left, right;         public node(int data)        {            this.data = data;            this.liss = 0;        }    }     // A memoization function returns size    // of the largest independent set in    // a given binary tree    static int liss(node root)    {        if (root == null)            return 0;        if (root.liss != 0)            return root.liss;        if (root.left == null && root.right == null)            return root.liss = 1;                 // Calculate size excluding the        // current node        int liss_excl = liss(root.left) + liss(root.right);                 // Calculate size including the        // current node        int liss_incl = 1;        if (root.left != null)        {            liss_incl += (liss(root.left.left) + liss(root.left.right));        }        if (root.right != null)        {            liss_incl += (liss(root.right.left) + liss(root.right.right));        }                 // Maximum of two sizes is LISS,        // store it for future uses.        return root.liss = Math.max(liss_excl, liss_incl);    }     public static void main(String[] args)    {        // Let us construct the tree given        // in the above diagram                 node root = new node(20);        root.left = new node(8);        root.left.left = new node(4);        root.left.right = new node(12);        root.left.right.left = new node(10);        root.left.right.right = new node(14);        root.right = new node(22);        root.right.right = new node(25);        System.out.println("Size of the Largest Independent Set is " + liss(root));    }} // This code is contributed by Rishabh Mahrsee

Python3

 # Python3 program for calculating LISS# using dynamic programming # A binary tree node has data,# pointer to left child and a# pointer to right childclass node:    def __init__(self, data):                 self.data = data        self.left = self.right = None        self.liss = 0 # A memoization function returns size# of the largest independent set in# a given binary treedef liss(root):         if root == None:        return 0         if root.liss != 0:        return root.liss         if (root.left == None and        root.right == None):        root.liss = 1        return root.liss     # Calculate size excluding the    # current node    liss_excl = (liss(root.left) +                 liss(root.right))     # Calculate size including the    # current node    liss_incl = 1    if root.left != None:        liss_incl += (liss(root.left.left) +                      liss(root.left.right))             if root.right != None:        liss_incl += (liss(root.right.left) +                      liss(root.right.right))             # Maximum of two sizes is LISS,    # store it for future uses.    root.liss = max(liss_excl, liss_incl)         return root.liss     # Driver Code # Let us construct the tree given# in the above diagramroot = node(20)root.left = node(8)root.left.left = node(4)root.left.right = node(12)root.left.right.left = node(10)root.left.right.right = node(14)root.right = node(22)root.right.right = node(25) print("Size of the Largest Independent "\      "Set is ", liss(root)) # This code is contributed by nishthagoel712

C#

 // C# program for calculating LISS// using dynamic programmingusing System;     public class LisTree{    /* A binary tree node has data, pointer    to left child and a pointer to right    child */    public class node    {        public int data, liss;        public node left, right;         public node(int data)        {            this.data = data;            this.liss = 0;        }    }     // A memoization function returns size    // of the largest independent set in    // a given binary tree    static int liss(node root)    {        if (root == null)            return 0;        if (root.liss != 0)            return root.liss;        if (root.left == null && root.right == null)            return root.liss = 1;                 // Calculate size excluding the        // current node        int liss_excl = liss(root.left) + liss(root.right);                 // Calculate size including the        // current node        int liss_incl = 1;        if (root.left != null)        {            liss_incl += (liss(root.left.left) + liss(root.left.right));        }        if (root.right != null)        {            liss_incl += (liss(root.right.left) + liss(root.right.right));        }                 // Maximum of two sizes is LISS,        // store it for future uses.        return root.liss = Math.Max(liss_excl, liss_incl);    }     // Driver code    public static void Main(String[] args)    {        // Let us construct the tree given        // in the above diagram                 node root = new node(20);        root.left = new node(8);        root.left.left = new node(4);        root.left.right = new node(12);        root.left.right.left = new node(10);        root.left.right.right = new node(14);        root.right = new node(22);        root.right.right = new node(25);        Console.WriteLine("Size of the Largest Independent Set is " + liss(root));    }} /* This code is contributed by PrinciRaj1992 */

Javascript



Output:

Size of the Largest Independent Set is 5

Time Complexity: O(n) where n is the number of nodes in given Binary tree.
Following extensions to above solution can be tried as an exercise.
1) Extend the above solution for n-ary tree.
2) The above solution modifies the given tree structure by adding an additional field ‘liss’ to tree nodes. Extend the solution so that it doesn’t modify the tree structure.
3) The above solution only returns size of LIS, it doesn’t print elements of LIS. Extend the solution to print all nodes that are part of LIS.