cangoal

// There are a row of n houses, each house can be painted with one of the three colors: red, blue or green. The cost of painting each house with a certain color is different. You have to paint all the houses such that no two adjacent houses have the same color.

// The cost of painting each house with a certain color is represented by a n x 3 cost matrix. For example, costs[0][0] is the cost of painting house 0 with color red; costs[1][2] is the cost of painting house 1 with color green, and so on... Find the minimum cost to paint all houses.

// Note:
// All costs are positive integers.

public int minCost(int[][] costs) {
    // Write your code here
    if(costs.length == 0 || costs[0].length == 0) return 0;

cangoal

public class Solution {
   /**
     * @param A: An integer array
     * @return: Count the number of element before this element 'ai' is 
     *          smaller than it and return count number array
     */
    class SegmentTreeNode{
        int start, end, count;
        SegmentTreeNode left, right;
        public SegmentTreeNode(int start, int end, int count){

cangoal

// solution 1 --- brute force
public ArrayList<Integer> countOfSmallerNumber(int[] A, int[] queries) {
    // write your code here
    ArrayList<Integer> res = new ArrayList<Integer>();
    if(A == null || queries == null || queries.length == 0)
        return res;
    for(int query : queries){
        int count = 0;
        for(int num : A){
            if(num < query) count++;

cangoal

public int nthSuperUglyNumber(int n, int[] primes) {
    if(n <= 0 || primes.length == 0) return 0;
    int[] uglyNumbers = new int[n];
    int[] indexes = new int[primes.length];
    int[] factors = new int[primes.length];
    for(int i=0; i<primes.length; i++){
        factors[i] = primes[i];
    }
    
    uglyNumbers[0] = 1;

cangoal

public void heapify(int[] A) {
    // write your code here
    if(A == null || A.length == 0) return;
    
    for(int i=A.length/2 - 1; i >= 0; i--){
        int k = i;
        while(true){
            int child_1 = 2 * k + 1, child_2 = 2 * k + 2;
            if(child_1 >= A.length) break;
            else if(child_2 >= A.length){

cangoal

// solution 1
public int countOfAirplanes(List<Interval> airplanes) { 
    // write your code here
    int count = 0, max = 0;
    if(airplanes == null || airplanes.size() == 0) return max;
    Set<Integer> set = new TreeSet<Integer>();
    List<Integer> start = new ArrayList<Integer>();
    List<Integer> end = new ArrayList<Integer>();
    for(Interval itv : airplanes){
        set.add(itv.start);

cangoal

// There is a fence with n posts, each post can be painted with one of the k colors.

// You have to paint all the posts such that no more than two adjacent fence posts have the same color.

// Return the total number of ways you can paint the fence.

// Note:
// n and k are non-negative integers.

public int numWays(int n, int k) {

cangoal

// Given n nodes labeled from 0 to n - 1 and a list of undirected edges (each edge is a pair of nodes), write a function to check whether these edges make up a valid tree.

// For example:
// Given n = 5 and edges = [[0, 1], [0, 2], [0, 3], [1, 4]], return true.
// Given n = 5 and edges = [[0, 1], [1, 2], [2, 3], [1, 3], [1, 4]], return false.

// Hint:
// 1. Given n = 5 and edges = [[0, 1], [1, 2], [3, 4]], what should your return? Is this case a valid tree?
// 2. According to the definition of tree on Wikipedia: “a tree is an undirected graph in which any two vertices are connected by exactly one path. In other words, any connected graph without simple cycles is a tree.”
// Note: you can assume that no duplicate edges will appear in edges. Since all edges are undirected, [0, 1] is the same as [1, 0] and thus will not appear together inedges.

cangoal

public class NumMatrix {
    int[][] Sum;
    public NumMatrix(int[][] matrix) {
        if(matrix != null && matrix.length != 0 && matrix[0].length != 0){
            Sum = new int[matrix.length+1][matrix[0].length+1];
            for(int i=matrix.length-1; i>=0; i--){
                for(int j=matrix[0].length -1; j >= 0; j--){
                    Sum[i][j] = Sum[i][j+1] + Sum[i+1][j] + matrix[i][j] - Sum[i+1][j+1];
                }
            }

cangoal

public class Codec {

    // Encodes a tree to a single string.
    public String serialize(TreeNode root) {
        if(root == null) return null;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        StringBuilder sb = new StringBuilder();
        stack.push(root);
        while(!stack.isEmpty()){
            TreeNode cur = stack.pop();