BST Sequences

A binary search tree was created by traversing through an array from left to right
and inserting each element. Given a binary search tree with distinct elements, print all possible
arrays that could have led to this tree.

ArrayList<LinkedList<Integer> > allSequences(TreeNocte node) {
        Arraylist<Linkedlist<Integer> > result = new Arraylist<Linkedlist<Integer> >();

        if (node == null) {
                result.add(new Linkedlist<Integer>());
                return result;
        }

        Linkedlist<Integer> prefix = new Linkedlist<Integer>();
        prefix.add(node.data);

        /* Recurse on left and right subtrees. */
        Arraylist<Linkedlist<Integer> > leftSeq = al1Sequences(node.left);
        ArrayList<LinkedList<Integer> > rightSeq = al1Sequences(node.right);

        /* Weave together each list from the left and right sides. */
        for (Linkedlist<Integer> left : leftSeq) {
                for (LinkedList<Integer> right : rightSeq) {
                        ArrayList<LinkedList<Integer> > weaved=
                                new Arraylist<Linkedlist<Integer> >();
                        weavelists(left, right, weaved, prefix);
                        result.addAll(weaved);
                }
        }
        return result;
}

/* Weave lists together in all possible ways. This algorithm works by removing the
 * head from one list, recursing, and then doing the same thing with the other
 * list. */
void weaveLists(LinkedList<Integer> first, LinkedList<Integer> second,
                ArrayList<LinkedList<Integer> > results, LinkedList<Integer> prefix) {
        /* One list is empty. Add remainder to [a cloned] prefix and store result. */
        if (first.size() == 0 11 second.size() == 0) {
                Linkedlist<Integer> result = (Linkedlist<Integer>)prefix.clone();
                result.addAll(first);
                result.addAll(second);
                results.add(result);
                return;
        }

        /* Recurse with head of first added to the prefix. Removing the head will damage
         * first, so we'll need to put it back where we found it afterwards. */
        int headfirst= first.removeFirst();
        prefix.addLast(headFirst);
        weavelists(first, second, results, prefix);
        prefix.removelast();
        first.addFirst(headFirst);

        /* Do the same thing with second, damaging and then restoring the list.*/
        int headSecond = second.removeFirst();
        prefix.addLast(headSecond);
        weavelists(first, second, results, prefix);
        prefix.removelast();
        second.addFirst(headSecond);
}

Paths with Sum

Related: LeetCode - Path Sum
Related: LeetCode 437 - Path Sum III
https://github.com/careercup/CtCI-6th-Edition/tree/master/Java/Ch%2004.%20Trees%20and%20Graphs/Q4_12_Paths_with_Sum

You are given a binary tree in which each node contains an integer value (which

might be positive or negative). Design an algorithm to count the number of paths that sum to a

given value. The path does not need to start or end at the root or a leaf, but it must go downwards

(traveling only from parent nodes to child nodes).

int countPathsWithSum(TreeNode root, int targetSum) {

        return countPathsWithSum(root, targetSum, 0, new HashMap<Integer, Integer>());

}

int countPathsWithSum(TreeNode node, int targetSum, int runningSum,

                      HashMap<Integer, Integer> pathCount) {

        if (node == null) return 0;   // Base case

        /* Count paths with sum ending at the current node. */

        runningSum += node.data;

        int sum= runningSum - targetSum;

        int totalPaths = pathCount.getOrDefault(sum, 0);

        /* If runningSum equals targetSum, then one additional path starts at root.

         * Add in this path.*/

        if (runningSum == targetSum) {

                totalPaths++;

        }

        /* Increment pathCount, recurse, then decrement pathCount. */

        incrementHashTable(pathCount, runningSum, 1);   // Increment pathCount

        totalPaths += countPathsWithSum(node.left, targetSum, runningSum, pathCount);

        totalPaths += countPathsWithSum(node.right, targetSum, runningSum, pathCount);

        incrementHashTable(pathCount, runningSum, -1);   // Decrement pathCount

        return totalPaths;

}

void incrementHashTable(HashMap<Integer, Integer> hashTable, int key, int delta) {

        int newCount = hashTable.getOrDefault(key, 0) + delta;

        if (newCount == 0) {//Remove when zero to reduce space usage

                hashTable.remove(key);

        } else {

                hashTable.put(key, newCount);

        }

}

Brute Force - O(nlogn)

int countPathsWithSum(TreeNode root, int targetSum) {

        if (root == null) return 0;

        /* Count paths with sum starting from the root. */

        int pathsFromRoot = countPathsWithSumFromNode(root, targetSum, 0);

        /* Try the nodes on the left and right. */

        int pathsOnleft = countPathsWithSum(root.left, targetSum);

        int pathsOnRight = countPathsWithSum(root.right, targetSum);

        return pathsFromRoot + pathsOnLeft + pathsOnRight;

}

int countPathsWithSumFromNode(TreeNode node, int targetSum, int currentSum) {

        if (node == null) return 0;

        currentSum += node.data;

        int totalPaths = 0;

        if (currentSum == targetSum) { // Found a path from the root

                totalPaths++;

        }

        totalPaths += countPathsWithSumFromNode(node.left, targetSum, currentSum);

        totalPaths += countPathsWithSumFromNode(node.right, targetSum, currentSum);

        return totalPaths;

}

LintCode Url Parser

http://blog.csdn.net/jmspan/article/details/51761416

Parse a html page, extract the Urls in it.

Hint: use regex to parse html.

Have you met this question in a real interview?  

Yes

Example

Given the following html page:

<html>
  <body>
    <div>
      <a href="http://www.google.com" class="text-lg">Google</a>
      <a href="http://www.facebook.com" style="display:none">Facebook</a>
    </div>
    <div>
      <a href="https://www.linkedin.com">Linkedin</a>
      <a href = "http://github.io">LintCode</a>
    </div>
  </body>
</html>

You should return the Urls in it:

[
  "http://www.google.com",
  "http://www.facebook.com",
  "https://www.linkedin.com",
  "http://github.io"
]

方法：正则表达式，重点是各种奇葩的情况。

public class HtmlParser { // Pattern pattern1 = Pattern.compile("(href\\s*=\\s*\")([^\"]*?)(\")", Pattern.CASE_INSENSITIVE); // Pattern pattern2 = Pattern.compile("(href\\s*=\\s*')([^']*?)(')", Pattern.CASE_INSENSITIVE); Pattern pattern = Pattern.compile("(href\\s*=\\s*[\"']?)([^\"'\\s>]*)([\"'>\\s])", Pattern.CASE_INSENSITIVE); /** * @param content source code * @return a list of links */ public List<String> parseUrls(String content) { // Write your code here List<String> results = new ArrayList<>(); Matcher matcher = pattern.matcher(content); match(matcher, results); return results; } private void match(Matcher matcher, List<String> results) { while (matcher.find()) { String url = matcher.group(2); if (url.length() == 0 || url.startsWith("#")) continue; results.add(url); } } }

Find frequency of each element in a limited range array in less than O(n) time - GeeksforGeeks

Find frequency of each element in a limited range array in less than O(n) time - GeeksforGeeks
Given an sorted array of positive integers, count number of occurrences for each element in the array. Assume all elements in the array are less than some constant M (much smaller than n).
Do this without traversing the complete array. i.e. expected time complexity is less than O(n).

Method 2 (Use Binary Search)
This problem can be solved in less than O(n) using a modified binary search. The idea is to recursively divide the array into two equal subarrays if its end elements are different. If both its end elements are same, that means that all elements in the subarray is also same as the array is already sorted. We then simply increment the count of the element by size of the subarray.

The time complexity of above approach is O(m log n), where m is number of distinct elements in the array of size n. Since m <= M (a constant), the time complexity of this solution is O(log n).

void findFrequencyUtil(int arr[], int low, int high,

                        vector<int>& freq)

{

    // If element at index low is equal to element

    // at index high in the array

    if (arr[low] == arr[high])

    {

        // increment the frequency of the element

        // by count of elements between high and low

        freq[arr[low]] += high - low + 1;

    }

    else

    {

        // Find mid and recurse for left and right

        // subarray

        int mid = (low + high) / 2;

        findFrequencyUtil(arr, low, mid, freq);

        findFrequencyUtil(arr, mid + 1, high, freq);

    }

}

  

// A wrapper over recursive function

// findFrequencyUtil(). It print number of

// occurrences of each element in the array.

void findFrequency(int arr[], int n)

{

    // create a empty vector to store frequencies

    // and initialize it by 0. Size of vector is

    // maximum value (which is last value in sorted

    // array) plus 1.

    vector<int> freq(arr[n - 1] + 1, 0);

      

    // Fill the vector with frequency

    findFrequencyUtil(arr, 0, n - 1, freq);

  

    // Print the frequencies

    for (int i = 0; i <= arr[n - 1]; i++)

        if (freq[i] != 0)

            cout << "Element " << i << " occurs "

                 << freq[i] << " times" << endl;

}

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