Interview Note - Examples

Feb 8 2015 (Feb 8 2015) English>Interview 6 minutes read (About 879 words)

Single Number

XOR features:
a ^ b = c
a ^ c = b
b ^ c = a
a ^ 0 = a
a ^ a = 0
(a ^ b) ^ c = a ^ (b ^ c)

I: all the numbers appear twice except one
XOR all the numbers, and the result is the one
II: all the numbers appear three times except one
Use a int[32] to store each bit, if the number of this bit can be mod by 3, we set it as 0, otherwise we set it as 1.
III: all the numbers appear twice except two
XOR all the numbers so we can get the two by XOR. Since these two elements are not the same, there is at least one bit that different. We find this different position and XOR all the elements which have 1 on position different, the result is one of the two. Finally, we XOR the found one with XOR-ed two, we can get another one.

Majority Number

I: find the number appears more than half
Since it is a strict majority number, we only need to maintain a variable result and a counter. Initially, when the counter faces with a different number, minus one of counter and change the result to this new number, otherwise we only need to add 1 to the counter. And finally, the result we maintained is the result. (We need to do another pass to make sure this number is the one we are looking for)
II: find the number appears more than 1/3
Similar with I, we maintain two variables and two counters so that we can get most appeared two numbers. And then we put these two numbers back to the array to find out which one appears more than 1/3. (We can’t use counter to find out, because counters have minus operation during the traversal)

III: find the number appears more than 1/k
We maintain a HashMap<number, counter>. Since if majority, each k different numbers should have more than one element. When we get k entries in the map, we remove entries which values are 1. Finally, find the entry that have highest value(count), that key should be the majority number. In this way, we can implement this in O(n) time and O(k) extra space.

public class Solution {
    public int majorityNumber(ArrayList<Integer> nums, int k) {
        if(nums == null || nums.size() == 0) {
            return -1;
        }

        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        for(int num : nums) {
            if(map.containsKey(num)) {
                map.put(num, map.get(num) + 1);
            } else {
                // if there are k entries, check if there are any entries that
                // its value == 1, that is not satisfied with "more than 1/k"
                if(map.size() == k) {
                    Iterator<Map.Entry<Integer, Integer>> iter = map.entrySet().iterator();
                    while(iter.hasNext()) {
                        Map.Entry<Integer, Integer> entry = iter.next();
                        if(entry.getValue() - 1 == 0) {
                            // can't use: map.remove(entry.getKey());
                            // above wrong code would lead the iter can't find its next !!!
                            iter.remove();
                        } else {
                            map.put(entry.getKey(), entry.getValue() - 1);
                        }
                    }
                } else {
                    map.put(num, 1);
                }
            }
        }

        // find the one that have highest value(count), that key should be the majority number
        int value = 0;
        int result = -1;
        for(Map.Entry<Integer, Integer> entry : map.entrySet()) {
            if(entry.getValue() > value) {
                value = entry.getValue();
                result = entry.getKey();
            }
        }
        return result;
    }
}

Interview Note - Data Structure

Feb 1 2015 (Feb 1 2015) English>Interview 7 minutes read (About 1038 words)

Data Structure is a way to organize data. It provides some methods to handle data stream, e.g. insert, delete, etc.

Linear Data Structure

Queue & Stack

Min Stack
Use two stacks, one is storing the input, when calling pop() or peek(), pop from another stack, which stores the minimum values from top to bottom.

class MinStack {
    ArrayList<Integer> stack = new ArrayList<Integer>();
    ArrayList<Integer> minStack = new ArrayList<Integer>();
    public void push(int x) {
        stack.add(x);
        if(minStack.isEmpty() || minStack.get(minStack.size() - 1) >= x) {
            minStack.add(x);
        }
        return;
    }

    public void pop() {
        if(stack.isEmpty()) {
            return;
        }
        int elem = stack.remove(stack.size() - 1);
        if(!minStack.isEmpty() && minStack.get(minStack.size() - 1) == elem) {
            minStack.remove(minStack.size() - 1);
        }
        return;
    }

    public int top() {
        if(!stack.isEmpty()) {
            return stack.get(stack.size() - 1);
        }
        return 0;
    }

    public int getMin() {
        if(!minStack.isEmpty()) {
            return minStack.get(minStack.size() - 1);
        }
        return 0;
    }
}

Implement Queue by stacks
Use two stacks, one is for storing elements. When calling pop() or top(), pop the elements from the first stack and push them into the second one.

public class Solution {
    private Stack<Integer> stack1;
    private Stack<Integer> stack2;

    public Solution() {
       // do initialization
       stack1 = new Stack<Integer>();
       stack2 = new Stack<Integer>();
    }
    
    public void push(int element) {
        stack1.push(element);
    }

    public int pop() {
        if(stack2.isEmpty()) {
            while(!stack1.isEmpty()) {
                stack2.push(stack1.pop());
            }
        }
        return stack2.pop();
    }

    public int top() {
        if(stack2.isEmpty()) {
            while(!stack1.isEmpty()) {
                stack2.push(stack1.pop());
            }
        }
        return stack2.peek();
    }
}

Largest Rectangle in Histogram
Brute force: totally O(n^2) windows and O(n) to find the minimun one in each window, so we need O(n^3) time.
Improve: for each number, search both side of it until find two smaller number, calculate the result, O(n^2) cost.
Best: use a stack to store the index (all increased heights), when faced with a smaller one, pop out from tha stack and calculate the area until the value bigger than the peek() one. This method cost O(n) time, with O(n) space in worst case (two passes).

Interview Note - Graph & Search

Jan 29 2015 (Jan 29 2015) English>Interview 7 minutes read (About 1048 words)

Clone Graph

/**
 * Definition for undirected graph.
 * class UndirectedGraphNode {
 *     int label;
 *     ArrayList<UndirectedGraphNode> neighbors;
 *     UndirectedGraphNode(int x) { label = x; neighbors = new ArrayList<UndirectedGraphNode>(); }
 * };
 */
public class Solution {
    public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
        if(node == null) {
            return null;
        }

        LinkedList<UndirectedGraphNode> queue = new LinkedList<UndirectedGraphNode>();
        HashMap<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<UndirectedGraphNode, UndirectedGraphNode>();
        queue.offer(node);
        map.put(node, new UndirectedGraphNode(node.label));

        while(!queue.isEmpty()) {
            UndirectedGraphNode cur = queue.poll();
            for(int i = 0; i < cur.neighbors.size(); i++) {
                // !!! if node not in map, add a new node into the map !!!
                if(!map.containsKey(cur.neighbors.get(i))) {
                    map.put(cur.neighbors.get(i), new UndirectedGraphNode(cur.neighbors.get(i).label));
                    queue.offer(cur.neighbors.get(i));
                }
                // now we are sure that the node is existed, add the neighbors to the value
                map.get(cur).neighbors.add(map.get(cur.neighbors.get(i)));
            }
        }

        return map.get(node);
    }
}

Topological Sorting
DFS: O(n) time with O(n) space for the map and the result.

/**
 * Definition for Directed graph.
 * class DirectedGraphNode {
 *     int label;
 *     ArrayList<DirectedGraphNode> neighbors;
 *     DirectedGraphNode(int x) { label = x; neighbors = new ArrayList<DirectedGraphNode>(); }
 * };
 */
public class Solution {
    /**
     * @param graph: A list of Directed graph node
     * @return: Any topological order for the given graph.
     */    
    public ArrayList<DirectedGraphNode> topSort(ArrayList<DirectedGraphNode> graph) {
        ArrayList<DirectedGraphNode> result = new ArrayList<DirectedGraphNode>();
        if(graph == null || graph.size() == 0) {
            return result;
        }
        // construct map with all nodes
        HashMap<DirectedGraphNode, Integer> map = new HashMap<DirectedGraphNode, Integer>();
        for(DirectedGraphNode node: graph) {
            // mark 0 as unsorted
            map.put(node, 0);
        }
        // find a new unsorted node to start sorting (if possible):
        while (hasUnsorted(map, graph)) {
            DirectedGraphNode node = null;
            for (DirectedGraphNode temp : graph) {
                if (map.get(temp) == 0) {
                    node = temp;
                }
            }
            // get the node and do sort(search):
            sort(map, graph, result, node);
        }
        return result;
    }
    
    // check if there is any node that not yet been sorted
    public boolean hasUnsorted(Map<DirectedGraphNode, Integer> map, ArrayList<DirectedGraphNode> graph){
        for (DirectedGraphNode node : graph) {
            if (map.get(node) == 0) {
                return true;
            }
        }
        return false;
    }
    
    // search and sort the graph
    public void sort(Map<DirectedGraphNode, Integer> map, ArrayList<DirectedGraphNode> graph, ArrayList<DirectedGraphNode> result, DirectedGraphNode node){
        if (map.get(node) != 0) {
            // if 1: System.out.println("It is not a DAG");
            // if 2: sorted
            return;
        }
        // mark 1 as visited(not yet been sorted), do with its neighbors:
        map.put(node, 1);
        for (DirectedGraphNode next : node.neighbors) {
            sort(map, graph, result, next);
        }
        // mark 2 as sorted
        // map.put(node, 2);
        result.add(0, node);
    }
}

Interview Note - Dynanmic Programming

Jan 25 2015 (Jan 25 2015) English>Interview 8 minutes read (About 1184 words)

1. Dynamic Programming

A method for solving a complex problem by breaking it down into a collection of simpler sub-problems.

1.1 When to use

One of the following three:

Maximum/Minimum Problem
Yes or No Question
Count all possible solutions

Can’t do sort or swap operation

Note: DP can’t return all results, it only returns max/min, yes/no or a certain value(like length, possible solutions, etc.).

1.2 How to think

States: what we need to store for each sub-problem (usually an array).
Function: what is the relationship between each state.
Intialization: what is the start of each state.
Answer: what is the end of each state.

Example: Number Triangle
We use an array f[i][j] to record the minimum sum from (0, 0) to (i, j). For a certain (i, j), we know it is either from (i - 1, j - 1) or (i - 1, j). So the sub-problem is to traverse all elements in the triangle and calculate all the f[i][j] = min(f[i - 1][j - 1], f[i - 1][j]) + T[i]][j]. The result will be the minimum one in f[n - 1][0, ..., n - 1].
We can also do it from bottom up. First we find the minimum of each pair in the last row, then we do above similarly but from bottom up. Finally the only one element left in the array is the result we are looking for.
A bonus point is doing this using only O(n) extra space, where n is the total number of rows in the triangle. Since each time we calculate f[i][j], the result is only related with f[i - 1][], we can only use 1D array and update the elements in it in each for loop.
Here is the code from bottom up with O(n) space cost.

public class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        if(triangle.size() == 0) {
            return 0;
        }

        int[] result = new int[triangle.size() + 1];
        
        for(int row = triangle.size() - 1; row >= 0; row--) {
            for(int col = 0; col <= row; col++) {
                result[col] = Math.min(result[col], result[col + 1]) + triangle.get(row).get(col);
            }
        }
        
        return result[0];
    }
}

Dynamic Programming uses extra space to remember the mid-result. So that it is more efficient than recursive searching, which may repeated calculate the same mid-result for many times. In the above example, if we do a brute force search, it cost O(2^n) time, where n is the height of the triangle (each row, we need to choose one from two, totally n rows). But it only costs O(n^2) in theoretical(m rows, each row has n elements, totally n^2) or O(n), which means all numbers we only need to visit once.

Interview Note - Linked List

Jan 24 2015 (Jan 24 2015) English>Interview 6 minutes read (About 857 words)

1. Introduce Dummy Node

When the head of the target list we want to return may be different from the original given list, we can use a dummy node linked to the result so that we can get what we want by calling dummy.next.

Remove Duplicates from Sorted List II:

public class Solution {
    public static ListNode deleteDuplicates(ListNode head) {
        if(head == null || head.next == null) {
            return head;
        }

        ListNode dummy = new ListNode(0);
        ListNode pre = dummy;
        ListNode cur = head;
        int dup = Integer.MIN_VALUE;

        while(cur != null && cur.next != null) {
            if(cur.val == cur.next.val) {
                dup = cur.val;
                while(cur.val == dup) {
                    cur = cur.next;
                    // if the rest of elements are all the same
                    if(cur == null) {
                        return dummy.next;
                    }
                }
            } else {
                pre.next = cur;
                pre = pre.next;
                cur = cur.next;
                // add null to pre.next: in case the rest of elements are the same and returned above
                pre.next = null;
            }
        }
        // add the last element (since cur.next != null skip this element)
        pre.next = cur;

        return dummy.next;
    }
}

Reverse Linked List II:

public class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        if(head == null) {
            return head;
        }

        // find the m position node
        int count = 1;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode pre = dummy;
        while(count < m) {
            pre = pre.next;
            count++;
        }

        // do the reverse: pre-> reverse[first->cur->next] ...
        ListNode first = pre.next;
        ListNode cur = pre.next.next;
        while(count < n) {
            ListNode next = cur.next;
            cur.next = pre.next;
            pre.next = cur;
            cur = next;
            count++;
            if(count == n) {
                first.next = next;
            }
        }

        return dummy.next;
    }
}

English