Helenyin123

    public int lengthOfLongestSubstringKDistinct(String s, int k) {
        Map<Character, Integer> map = new HashMap<>();
        if (s == null || s.length() == 0 || k == 0) return 0;
        int start = 0;
        int longest = 0;
        for (int i = 0; i < s.length(); i++) {
            if (map.containsKey(s.charAt(i)) || map.size() < k) {
                map.put(s.charAt(i), i);
            } else {
                int minPos = getMinPos(map);

Helenyin123

    public boolean hasGroupsSizeX(int[] deck) {
        if (deck == null || deck.length <= 1) return false;
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < deck.length; i++) {
            int freq = map.getOrDefault(deck[i], 0);
            map.put(deck[i], freq + 1);
        }
        int n = 0; List<Integer> lcf = new ArrayList<>();
        for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
            if (n == 0) {

Helenyin123

class Solution {
    public int superpalindromesInRange(String L, String R) {
        Long l = (long)Math.sqrt(Long.parseLong(L));
        Long r = (long)Math.sqrt(Long.parseLong(R));
        int re = 0;
        for(int odd = 0;odd <= 1;odd++){
            long core = 1;
            long num = getP(core,odd == 0);
            while(num <= r){
                if(num >= l){

Helenyin123

You will be supplied with two data files in CSV format .
The first file contains statistics about various dinosaurs. The second file contains additional data.
Given the following formula, `speed = ((STRIDE_LENGTH / LEG_LENGTH) - 1) * SQRT(LEG_LENGTH * g)`
Where g = 9.8 m/s^2 (gravitational constant)

Write a program to read in the data files from disk, it must then print the names of only the bipedal dinosaurs from fastest to slowest.
Do not print any other information.

$ cat dataset1.csv
NAME,LEG_LENGTH,DIET

Helenyin123

// store all the flower position to a TreeSet, each time a flower bloom, we check the distance of this flower to 
// two closest flower is k or not, return the first data that satisfy the condition.
// TreeSet: Time = O(NlogN) Space = O(N)
 public int kEmptySlots(int[] flowers, int k) {
    if (flowers == null || k > flowers.length - 2) return -1;
    int N = flowers.length;
    TreeSet<Integer> set = new TreeSet<>();
    for (int i = 0; i < N; i++) {
        set.add(flowers[i]); // include right and exclude left
        Integer lower = set.lower(flowers[i]);

Helenyin123

    public int numDecodings(String s) {
        if (s == null || s.length() == 0) return 0;
        int[] result = new int[]{0};
        dfs(s, 0, 1, result);
        return result[0];
    }
    private void dfs(String s, int i, int count, int[] result) {
        if (i == s.length()) {
            count = count % 1000000007;
            result[0] = (result[0] + count) % 1000000007;

Helenyin123

bhh

Helenyin123

    public int orderOfLargestPlusSign(int N, int[][] mines) {
        int[][] matrix = new int[N][N];
        for (int i = 0; i < N; i++) {
            Arrays.fill(matrix[i], 1);
        }
        for (int[] pair : mines) {
            matrix[pair[0]][pair[1]] = 0;
        }
        // calculate number of consecutive 1's in four directions: up, down, left, right
        int[][] left = new int[N][N];

Helenyin123

    public int minFactory(int[] array, int range) {
        int i = 0;
        int count = 0;
        int start = 0;
        while (i < array.length && i <= range) {
            if (array[i] == 1) {
                start = i;
            }
            i++;
        }

Helenyin123

    /**
    有一个数组，里面是26个字母，顺序是乱的，有个API叫做 swich_with_a(int pos), 这个方法的作用是调换pos位置上的字母和字母”a“的位置
   （无论”a”在哪里，这个方法会把“a“换到pos的位置）。让你实现把这26个字母排序，但是只能用到这个swith_with_a来调换位置。
    遇到一个char[i]有两个思路:1. 把char[i]换到出去，到它应该在的位置上去，但这样不能让char[i]本身是对的，走完一遍后结果还是不对
    2. 不管char[i]是什么，我只要把字母i + 'a'换进来，换到位置i上就行，这样就保证了那个位置的字母一定是对的
    */

    public char[] sortChar(char[] array) {
        // record each character's index
        int[] index = new int[26];