Find first index of a number in the sorted array containing duplicates

This can be done easily in linear time by making a pass through the array. In coding interviews it is not likely the solution the interviewer is looking for. So we have to do better. We know that binary search provides a efficient solution for finding an element in sorted array. Lets look at the problem to see if we can modify the binary search algorithm to solve this problem. We can start with binary search to find the index. Once we find the index we have to look to the left to check if the number is present. If we look to the left linearly then the solution will be O(n). To do better than that we can use another binary search in the left and continue until we find the fist index where the number occurs.

/**
 * Find first index of the number k in the sorted
 * array x of size n
 */
int first_index(int x[], int n, int k) {
    int lo = 0;
    int hi = n - 1;

    while (lo <= hi) {
        int mid = lo + (hi - lo)/2;

        if (x[mid] == k) {
            if (mid > 0 && x[mid - 1] == k) {
                hi = mid - 1;
            } else {
                return mid;
            }

        } else if(x[mid] < k) {
            lo = mid + 1;
        } else {
            hi = mid - 1;
        }
    }

    return -1;
}

Iterative inorder traversal of a binary tree using constant space.

In order traversal of a binary tree can be performed using O(1) storage if the tree stores parent pointer. The trick is to look at possible directions we need to move in the tree.  As shown in the diagram below there are 3 directions that you can move in a binary tree. They are to the left, right or up. Given this all we need is to establish conditions that decides the movement. Consider last visited node as the node we were in before coming to the current node. Here are the cases:

1. If the last visited node is the parent node and the current node's left child exists move left.

2. If the last visited node is the left child of the current node and the current node has right child move right.

3. If the last visited node is on the left or right move up.

For the node to be the next successor for inorder traversal one of the following conditions should hold ( We will print the node value in our sample code below whenever this condition holds):

1. The last visited node was the left child.

2.  The left child of the current node doesn't exist.

It's now easy to translate these conditions into code. All you need is these conditions and some additional null checks as shown in the code below.

void inorder_constant_space(Node *root) {

    if (!root)
        return;

    Node *last = NULL;
    Node *node = root;

    while (node) {

        if (last == node->left || 
                (last == node->parent && node->left == NULL)) {
            std::cout << node->name;
        }

        if (last == node->left && node->right != NULL) {
            last = node;
            node = node->right;
        } else if(last == node->left) {
            last = node;
            node = node->parent;
        } else if (last == node->parent && node->left == NULL
                         && node->right != NULL) {
            last = node;
            node = node->right;
        } else if (last == node->parent && node->left == NULL) {
            last = node;
            node = node->parent;
        } else if (last == node->parent) {
            last = node;
            node = node->left;
        } else {
            last = node;
            node = node->parent;
        }
       
    }
}

Fill in the sibling pointer in the binary tree

You are given a binary tree which has a sibling pointer in addition to left and right pointer. The tree has left and right child set correctly but sibling pointer is not yet set. Your task is to traverse the tree and fill in the sibling pointer. A sibling is defined as a node next to the given node at the same level.

In the tree above B and C, and D and E are siblings. So sibling pointer of B should point to C and sibling pointer of D should point to E. Sibling pointer of C and E should be null because they don't have any siblings to the right.

This problem is easily solved using level order traversal as shown below.

class Node {
public:
    Node *left;
    Node *right;

    // Sibling pointer that needs to be set
    Node *sibling;

    // Value stored in the node. For simplicity

    // we assume name to be the value
    std::string name;
};

// Fill in the sibling pointer of a binary tree
void fill_node_with_next_sbiling(Node *node) {
    if (!node)
        return;

    // Initialize the queue with the root node.
    std::queue<node> queue;
    queue.push(node);

    // Set count of current level in the queue to 1.
    int cur_count = 1;
    int child_count = 0;

    Node *prev = NULL;

    while (!queue.empty()) {
        Node *n = queue.front();
        queue.pop();

        // Set the sibling pointer and set previous as current node
        if (prev) {
            prev->sibling = n;
            prev = n;
        } else {
            prev = n;
        }

        cur_count--;

        if (n->left != NULL) {
            queue.push(n->left);
            child_count++;
        }

        if (n->right != NULL) {
            queue.push(n->right);
            child_count++;
        }

        // All nodes in the current level are done.
        // Set the cur_count to next level count 
        // for processing next level
        if (cur_count == 0) {
            cur_count = child_count;
            child_count = 0;
            prev = NULL;
        }

    }
}

Convert a binary search tree into doubly linked list

Converting a binary search tree into doubly linked list is a common interview question. It's a simple question as long as we know it can be solved by extending inorder tree traversal. If you are not familiar with inorder tree traversal, I have described it here. In this problem you are given a binary search tree with left and right child and you are supposed to convert left pointer to predecessor node and right pointer to the successor node. Here is the solution to the problem:

// output list
Node *list = NULL;

void convert_to_doubly_ll(Node *root) {
    if (root == NULL)
        return;

    convert_to_doubly_ll(root->left);

    if (list) {
        list->right = root;
        root->left = list;
    } 
    
    list = root;
    
    Node *right = root->right;
    root->right = NULL;

    convert_to_doubly_ll(right);
}

If you want to do this in place, then all you have to do is pass in the previous node in each recursive call. The code is given below:

void convert_to_doubly_ll_in_place(Node *root, Node **prev) {
    if (root == NULL)
        return;

    convert_to_doubly_ll_in_place(root->left, prev);

    Node *prev_node = *prev;
    Node *right = NULL;

    if (prev_node) {
        prev_node->right = root;
        root->left = prev_node;
    } 
    
    *prev = root;
    
    right = root->right;
    root->right = NULL;

    convert_to_doubly_ll_in_place(right, prev);
}

You can call the above method like this:

Node *root = ....a tree...;
Node *prev=NULL;
convert_to_doubly_ll_in_place(root, &prev);

Tree Problems

Many of the tree problems asked in the interviews can be solved by extending traversal algorithms. There are 4 popular ways a tree can be traversed. They are preorder traversal, post order traversal, inorder and level order traversal. We will look at each of these traversals and some sample interview questions that can be answered using the given traversal.

Preorder Traversal

In preorder traversal you start from root, visit left child first and then right child. The processing for that node is done before processing it's children.

void pre_order(Node *root) {
    if (!root) {
        return;
    }
    // Process the node
    std::cout << root->name << std::endl;

    // Traverse left and right children
    pre_order(root->left);
    pre_order(root->right);
}

Sample Interview Problems:

Given a binary tree, serialize and deserialize it.

Given a root and a node in a binary tree, find the path from the root to the node.

Post Order Traversal
In post order traversal you start from root, visit left child first and then right child. The processing for the node is done after processing it's children.

void post_order(Node *root) {
    if (!root) {
        return;
    }

    // Traverse left and right children
    post_order(root->left);
    post_order(root->right);

    // Process the node.
    std::cout << root->name << std::endl;
}

Sample Interview Problems:
Given a binary tree create an in place mirror image of the tree.
Expression evaluation.

In Order Traversal
In in-order traversal you start from root, visit left subtree first and then right subtree. The processing for the node is done before you start traversing right subtree.

void in_order(Node *root) {
    if (!root) {
        return;
    }

    // Traverse left child
    post_order(root->left);

    // Process the node
    std::cout << root->name << std::end;

    // Traverse right child
    post_order(root->right);
}

Sample Interview Problems:
Convert a binary search tree into doubly linked list.

Level Order Traversal
In level-order traversal you start from root, process the root and then move to next level. In the next level you process all the nodes from left to right and then move to next level and so on.

#include <queue>
#include <iostream>
void level_order(Node *node) {
    if (!node)
        return;

    std::queue queue;

    // Count of the nodes in current level
    int cur_count = 0; 

    // Count of the nodes in the next level
    int child_count = 0; 

    queue.push(node);

    // Start with 1 as we have root in the queue
    // to begin with
    cur_count += 1;

    while (!queue.empty()) {
        Node *n = queue.front();
        queue.pop();

        // Process the node. We will just print to 
        // indicate processing.
        std::cout << n->name << " ";
        
        // Decrement of the count of nodes remaining in the
        // queue  for current level.
        cur_count--;

        if (n->left != NULL) {
            queue.push(n->left);
            child_count++;
        }

        if (n->right != NULL) {
            queue.push(n->right);
            child_count++;
        }

        // When we hit a count of 0, we start at next level
        if (cur_count == 0) {
            cur_count = child_count;
            child_count = 0;
            std::cout << std::endl;
        }
    }
}

Sample Interview Problems:
Pretty print a binary tree.
Given a binary tree, return each level as a list.
You have a binary tree.  Lets say you add an extra sibling pointer to it. Can you fill all the sibling pointers to point to it's sibling starting  from the left to the right.


I will add more sample problems as well as provide solutions to as many I can.