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

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

/*
 * 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

/**
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"