Given a binary tree, count the number of uni-value subtrees.

A Uni-value subtree means all nodes of the subtree have the same value.

* Method 1: Recursion from bottom to top.

```Java

/**

class Solution {

private int res = 0;

private TreeNode root;

public int countUnivalSubtrees(TreeNode root) {

if(root == null) return 0;

this.root = root;

subTreeValue(root);

return res;

}

// return a value

// if -1: means current tree is not unival

// otherwise will return the univalue

private int subTreeValue(TreeNode node){

if(node.left == null && node.right == null){ // leaf is always true.

res += 1;

return node.val;

}

int left = -1, right = -1;

if(node.left != null) left = subTreeValue(node.left);

if(node.right != null) right = subTreeValue(node.right);

if(node.left != null && node.right != null && left != -1 && right != -1){

if(left == right && left == node.val){

res += 1;

return left;

}

}else if(node.right == null && left != -1){ // right subtree is empty

if(left == node.val){

res += 1;

return left;

}

}else if(node.left == null && right != -1){

if(right == node.val){

res += 1;

return right;

}

}

return -1;

}

}

```

We want to use quad trees to store an N x N boolean grid. Each cell in the grid can only be true or false. The root node represents the whole grid. For each node, it will be subdivided into four children nodes until the values in the region it represents are all the same.

Each node has another two boolean attributes : isLeaf and val. isLeaf is true if and only if the node is a leaf node. The val attribute for a leaf node contains the value of the region it represents.

Your task is to use a quad tree to represent a given grid. The following example may help you understand the problem better:

Given the 8 x 8 grid below, we want to construct the corresponding quad tree:


The corresponding quad tree should be as following, where each node is represented as a (isLeaf, val) pair.


For the non-leaf nodes, val can be arbitrary, so it is represented as *.

Note:

1. N is less than 1000 and guaranteened to be a power of 2.

2. If you want to know more about the quad tree, you can refer to its wiki.

* Method 1: Recursion from bottom to top.

```Java

/*

class Solution {

private int[][] grid;

public Node construct(int[][] grid) {

this.grid = grid;

int len = grid.length;

return dfs(0, grid.length - 1, 0, grid.length - 1);

}

private Node dfs(int xs, int xe, int ys, int ye){

Node node = new Node();

if(xs == xe){

node.val = (grid[xs][ys] == 1);

node.isLeaf = true;

return node;

}

Node upLeft = dfs(xs, xs + (xe - xs) / 2, ys, ys + (ye - ys) / 2);

Node upRight = dfs(xs, xs + (xe - xs) / 2, ys + (ye - ys) / 2 + 1, ye);

Node bottomLeft = dfs(xs + (xe - xs) / 2 + 1, xe, ys, ys + (ye - ys) / 2);

Node bottomRight = dfs(xs + (xe - xs) / 2 + 1, xe, ys + (ye - ys) / 2 + 1, ye);

if(bottomLeft.isLeaf && bottomRight.isLeaf && upLeft.isLeaf && upRight.isLeaf

&& bottomLeft.val == bottomRight.val && bottomRight.val == upLeft.val && upLeft.val == upRight.val){

node.isLeaf = true;

node.val = bottomLeft.val;

}else{

node.bottomLeft = bottomLeft;

node.bottomRight = bottomRight;

node.topLeft = upLeft;

node.topRight = upRight;

}

return node;

}

}

```

Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.

Find all the elements of [1, n] inclusive that do not appear in this array.

Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.

* Method 1: O(n) S(n)

```Java

class Solution {

public List<Integer> findDisappearedNumbers(int[] nums) {

Set<Integer> set = new HashSet<>();

for(int num : nums) set.add(num);

List<Integer> res = new ArrayList<>();

for(int i = 1; i <= nums.length; i++)

if(!set.contains(i))

res.add(i);

return res;

}

}

```

* Method 2: O(n) + S(0)

* Since the index range is same as the length of the array, once we read a value, we change the number saved in that index - 1 to -1.

* We need to recursively to change the index.

* Finally we check the indices that are not -1.

```Java

class Solution {

public List<Integer> findDisappearedNumbers(int[] nums) {

for(int i = 0; i < nums.length; i++){

if(nums[i] == -1) continue;

int index = nums[i];

while(nums[index - 1] != -1 && nums[index - 1] != index){

int nextIndex = nums[index - 1];

nums[index - 1] = -1;

index = nextIndex;

}

if(nums[index - 1] == index) nums[index - 1] = -1;

}

List<Integer> res = new ArrayList<>();

for(int i = 0; i < nums.length; i++){

if(nums[i] > 0) res.add(i + 1);

}

return res;

}

}

```

### Reference

1. [LeetCode-448：Find All Numbers Disappeared in an Array （寻找缺失多个数字）](https://blog.csdn.net/Koala_Tree/article/details/78349644)

You are given coins of different denominations and a total amount of money. Write a function to compute the number of combinations that make up that amount. You may assume that you have infinite number of each kind of coin.

Note:

You can assume that

1. 0 <= amount <= 5000

2. 1 <= coin <= 5000

3. the number of coins is less than 500

4. the answer is guaranteed to fit into signed 32-bit integer

* Method 1: Recursion TLE

```Java

class Solution {

private int amount;

private int[] coins;

private int res;

public int change(int amount, int[] coins) {

this.amount = amount;

this.coins = coins;

dfs(0, 0);

return res;

}

private void dfs(int index, int sum){

if(sum == amount) res++;

else if(sum < amount){

for(int i = index; i < coins.length; i++){

dfs(i, sum + coins[i]);

}

}

}

}

```

* Method 2: dp

```Java

class Solution {

public int change(int amount, int[] coins) {

int[] dp = new int[amount + 1];

dp[0] = 1;

for(int coin :coins){

for(int i = 1; i <= amount; i++){

if(i - coin >= 0)

dp[i] += dp[i - coin];

}

}

return dp[amount];

}

}

```