bittib

/**
Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that:

Only one letter can be changed at a time
Each intermediate word must exist in the dictionary
For example,

Given:
start = "hit"
end = "cog"

bittib

/*
 * Time complexity : O(n), Space Complexity : O(n), 
 * We could just use a n size char array instead of using nRows' StringBuilder array. See the convert0 for the detail
 */
public static String convert(String s, int nRows){
    if (s == null || s.length() <= 1 || nRows <= 1)
        return s;
    StringBuilder[] sbs = new StringBuilder[nRows];

    for (int i=0; i<nRows; i++){

bittib

public static int solve(int N){
    int[] primes = new int[N];
    int cnt = 0;
    boolean[] isp = new boolean[N+1];
    Arrays.fill(isp, true);
    isp[0] = false;
    isp[1] = false;

    for (int i=2; i<=N; i++){
        if (isp[i]){

bittib

static int daysSinceYearStart(int y, int m, int d, boolean flag){
    boolean leapYear = isLeapYear(y);
	int total = leapYear ? 366 : 365;
	int days = d;
	days += DayOfMonth[m-1];
	if (leapYear && m > 2)
		days++;
	return flag ? days : total - days;
}
static int dateDiff(int y1, int m1, int d1, int y2, int m2, int d2){

bittib

/*
Implement wildcard pattern matching with support for '?' and '*'.

'?' Matches any single character.
'*' Matches any sequence of characters (including the empty sequence).

The matching should cover the entire input string (not partial).

The function prototype should be:
bool isMatch(const char *s, const char *p)

bittib

/*
Given a set of distinct integers, S, return all possible subsets.

Note:

Elements in a subset must be in non-descending order.
The solution set must not contain duplicate subsets.
For example,
If S = [1,2,3], a solution is:

bittib

/*
Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that:
•Only one letter can be changed at a time
•Each intermediate word must exist in the dictionary

For example,

Given:
start = "hit"
end = "cog"

bittib

public static void morrisInOrderTraverse(TreeNode root){
    	TreeNode ptr = root;
	while (ptr != null){
		if (ptr.left == null){
			visit(ptr);
			ptr = ptr.right;
		}else{
			TreeNode node = ptr.left;
			while (node.right != null && node.right != ptr)
				node = node.right;

bittib

/*
 * Divide two integers without using multiplication, division and mod operator.
 */
public int divide(int dividend, int divisor) {
    if (divisor == 0) throw new IllegalArgumentException("divisor cannot be 0");
    if (dividend == 0) return 0;
    boolean neg = (dividend > 0 != divisor > 0);
    long dend = dividend;
    long dsor = divisor;
    dend = Math.abs(dend);

bittib

/*
Given a binary tree, flatten it to a linked list in-place.

For example,
Given

         1
        / \
       2   5
      / \   \