## 1006. Clumsy Factorial

Normally, the factorial of a positive integer `n` is the product of all positive integers less than or equal to `n`.  For example, `factorial(10) = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1`.

We instead make a clumsy factorial: using the integers in decreasing order, we swap out the multiply operations for a fixed rotation of operations: multiply (*), divide (/), add (+) and subtract (-) in this order.

For example, `clumsy(10) = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1`.  However, these operations are still applied using the usual order of operations of arithmetic: we do all multiplication and division steps before any addition or subtraction steps, and multiplication and division steps are processed left to right.

Additionally, the division that we use is floor division such that `10 * 9 / 8` equals `11`.  This guarantees the result is an integer.

`Implement the clumsy` function as defined above: given an integer `N`, it returns the clumsy factorial of `N`.

Example 1:

Input:

```4
```

Output:

``` 7
```

Explanation:

``` 7 = 4 * 3 / 2 + 1
```

Example 2:

Input:

```10
```

Output:

```12
```

Explanation:

```12 = 10 * 9 / 8 + 7 - 6 * 5 / 4 + 3 - 2 * 1
```

Note:

1. `1 <= N <= 10000`
2. `-2^31 <= answer <= 2^31 - 1`  (The answer is guaranteed to fit within a 32-bit integer.)

#### Thoughts:

• Split into chunks of size 4, then deal with case 3 or 2 or 1
• The first chunk has a positive sign, then negative starting from the second trunk, including the remaining (3, 2, or 1) part
• if n < 4, the sign is positive

```class Solution {

/**
* @param Integer \$N
* @return Integer
*/
function clumsy(\$n) {
\$r = 0;
\$f = 0;
\$first_positive_chunk_set = false;

if(\$n >= 4){
\$r = floor(\$n*(\$n-1)/(\$n-2)) + \$n-3;
\$n -= 4;
\$first_positive_chunk_set = true;
}

while(floor(\$n / 4) >= 1){
\$r -= floor(\$n*(\$n-1)/(\$n-2)) - \$n+3;
\$n -= 4;
}
if(\$n >= 3){
\$r -= floor(\$n*(\$n-1)/(\$n-2)) ;
\$n -= 3;
}
if(\$n >= 2){
\$r -= \$n*(\$n-1) ;
\$n -= 2;
}
if(\$n >= 1){
\$r -= \$n ;
\$n -= 1;
}
return \$first_positive_chunk_set?\$r:-\$r;
}
}```

#### Another Solution Inspired by Other

• Base on math
• Other than that(short), this solution is hard to read and understand.
```class Solution {

/**
* @param Integer \$N
* @return Integer
*/
function clumsy(\$n) {
if (\$n<5){
return  \$n+3*(\$n>2);
}
return \$n+2-3*(\$n%4==3)-(\$n%4==0);
}
}```

## 1007. Minimum Domino Rotations For Equal Row

In a row of dominoes, `A[i]` and `B[i]` represent the top and bottom halves of the `i`-th domino.  (A domino is a tile with two numbers from 1 to 6 – one on each half of the tile.)

We may rotate the `i`-th domino, so that `A[i]` and `B[i]` swap values.

Return the minimum number of rotations so that all the values in `A` are the same, or all the values in `B` are the same.

If it cannot be done, return `-1`.

Example 1:

Input:

```A = [2,1,2,4,2,2], B = [5,2,6,2,3,2]
```

Output:

```2
```

Explanation:

```The first figure represents the dominoes as given by A and B: before we do any rotations.
If we rotate the second and fourth dominoes, we can make every value in the top row equal to 2, as indicated by the second figure.
```

Example 2:

Input:

```A = [3,5,1,2,3], B = [3,6,3,3,4]
```

Output:

```-1
```

Explanation:

```In this case, it is not possible to rotate the dominoes to make one row of values equal.
```

Note:

1. `1 <= A[i], B[i] <= 6`
2. `2 <= A.length == B.length <= 20000`

#### Thoughts:

• Two possible values, A[0] and B[0]
• For each possible value, and foreach index, if one doesn’t have while the other does add 1 to counter
• There are two counters for each possible values, {B->A, A->B}
• Return the best result

The speaking of domino’s top and bottom is misleading, no domino has the same value for top and bottom. Why not just say a single digit number?

```class Solution {

/**
* @param Integer[] \$A
* @param Integer[] \$B
* @return Integer
*/
function minDominoRotations(\$A, \$B) {

\$p = [];      // Possible Values
\$p[] = \$A[0]; // Either First elem if A
\$p[] = \$B[0]; // Either First elem of B
\$r = [];      // List of mins

// Foreach Possible Values
foreach(\$p as \$s){
\$x = 0;   // Counter, swap B to match A's Value
\$y = 0;   // Counter, swap A to match B's Value
\$nf = false;    // For a possible value, not found both Swap A to B and B to A
for(\$i = 0;\$i < count(\$A); \$i ++){
if(\$s == \$A[\$i] && \$s !== \$B[\$i]){
\$y ++;
} elseif (\$s == \$B[\$i] && \$s !== \$A[\$i]){
\$x ++;
} else if(\$s !== \$A[\$i] && \$s !== \$B[\$i]) {
// If possible value not found in either array, move to next possible value.
\$nf = true;
break;
}

}

if(!\$nf){
// If found
if(\$x == 0 || \$y ==0){
// O is the best, no need to continue
return 0;
}
// Otherwise push the best counter result to array
\$r[] = min(\$x, \$y);
}

}
// When all possible values are no longer possible :(
if(empty(\$r)){
return -1;
}
// Otherwise return the best result
return min(\$r);
}
}```

## 154. Find Minimum in Rotated Sorted Array II

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e.,  `[0,1,2,4,5,6,7]` might become  `[4,5,6,7,0,1,2]`).

Find the minimum element.

The array may contain duplicates.

Example 1:

Input:

``` [1,3,5]
```

Output:

` 1`

Example 2:

Input:

``` [2,2,2,0,1]
```

Output:

` 0`

Note:

#### Thoughts:

• try binary search first, then go through the  ambiguous sections one by one.
My First Draft:
```class Solution {

/**
* @param Integer[] \$nums
* @return Integer
*/
function findMin(\$nums) {
\$last = count(\$nums)-1;
\$first = 0;
\$mid = floor((\$last + \$first) /2);
\$min = null;
while(\$nums[\$first] > \$nums[\$mid] xor \$nums[\$mid] > \$nums[\$last]){
if(\$nums[\$first] > \$nums[\$mid]){
\$last = \$mid;
\$mid = floor((\$last + \$first) /2);
} else {
\$first = \$mid +1;
\$mid = floor((\$last + \$first) /2);
}
}
if(\$first == \$last){
return \$nums[\$last];
} else {
for(\$min = \$nums[\$first];\$first <= \$last;\$first++){
if(\$nums[\$first] < \$min){
\$min = \$nums[\$first];
}
}

}
return \$min;
}

}```

sheehan‘s pretty solution:
```class Solution {
public:
int findMin(vector<int> &num) {
int lo = 0;
int hi = num.size() - 1;
int mid = 0;

while(lo < hi) {
mid = lo + (hi - lo) / 2;

if (num[mid] > num[hi]) {
lo = mid + 1;
}
else if (num[mid] < num[hi]) {
hi = mid;
}
else { // when num[mid] and num[hi] are same
hi--;
}
}
return num[lo];
}
};```

## 181. Employees Earning More Than Their Managers

The `Employee` table holds all employees including their managers. Every employee has an Id, and there is also a column for the manager Id.
```+----+-------+--------+-----------+
| Id | Name  | Salary | ManagerId |
+----+-------+--------+-----------+
| 1  | Joe   | 70000  | 3         |
| 2  | Henry | 80000  | 4         |
| 3  | Sam   | 60000  | NULL      |
| 4  | Max   | 90000  | NULL      |
+----+-------+--------+-----------+
```

Given the `Employee` table, write a SQL query that finds out employees who earn more than their managers. For the above table, Joe is the only employee who earns more than his manager.

```+----------+
| Employee |
+----------+
| Joe      |
+----------+
```

#### Thoughts:

Self  Join, nothing special

```# Write your MySQL query statement below
SELECT e1.Name as Employee FROM Employee e1  LEFT JOIN Employee e2 on e2.id =  e1.ManagerId WHERE e2.Salary < e1.Salary;```

## 104. Maximum Depth of Binary Tree

Given a binary tree, find its maximum depth.

The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

Note: A leaf is a node with no children.

Example:

Given binary tree `[3,9,20,null,null,15,7]`,

```    3
/ \
9  20
/  \
15   7```

return its depth = 3.

#### Thought:

Nothing special but..

I’m using Java this time..

```/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int maxDepth(TreeNode root) {
return maxDepthHelper(root, 0);
}
public int maxDepthHelper(TreeNode node, int depth) {
if(node != null){
return Math.max(maxDepthHelper(node.left, depth+1)
, maxDepthHelper(node.right, depth+1));
}
return depth;
}
}```

## 807. Max Increase to Keep City Skyline

In a 2 dimensional array `grid`, each value `grid[i][j]` represents the height of a building located there. We are allowed to increase the height of any number of buildings, by any amount (the amounts can be different for different buildings). Height 0 is considered to be a building as well.

At the end, the “skyline” when viewed from all four directions of the grid, i.e. top, bottom, left, and right, must be the same as the skyline of the original grid. A city’s skyline is the outer contour of the rectangles formed by all the buildings when viewed from a distance. See the following example.

What is the maximum total sum that the height of the buildings can be increased?

Example:

Input:

``` grid = [[3,0,8,4],[2,4,5,7],[9,2,6,3],[0,3,1,0]]
```

Output:

``` 35
```

Explanation:

```
The grid is:
[ [3, 0, 8, 4],
[2, 4, 5, 7],
[9, 2, 6, 3],
[0, 3, 1, 0] ]

The skyline viewed from top or bottom is: [9, 4, 8, 7]
The skyline viewed from left or right is: [8, 7, 9, 3]

The grid after increasing the height of buildings without affecting skylines is:

gridNew = [ [8, 4, 8, 7],
[7, 4, 7, 7],
[9, 4, 8, 7],
[3, 3, 3, 3] ]

```

Notes:

• `1 < grid.length = grid[0].length <= 50`.
• All heights `grid[i][j]` are in the range `[0, 100]`.
• All buildings in `grid[i][j]` occupy the entire grid cell: that is, they are a `1 x 1 x grid[i][j]` rectangular prism.

#### Thoughts;

1. Vertical & Horizontal  scan, generate skyline view array
2. Max increment = min(vertical, horizontal).
3. Result = Sum (each max increment)

```class Solution {

/**
* @param Integer[][] \$grid
* @return Integer
*/
function maxIncreaseKeepingSkyline(\$grid) {
if(empty(\$grid)){
return 0;
}
\$result = 0;
\$h = array_fill(0,count(\$grid), 0);
\$v = array_fill(0,count(\$grid[0]), 0);

for(\$i = 0; \$i < count(\$grid); \$i ++){
\$h[\$i] = max(\$grid[\$i]);
for(\$j = 0; \$j < count(\$grid[0]); \$j ++){
if(\$grid[\$i][\$j] > \$v[\$j]){
\$v[\$j] = \$grid[\$i][\$j];
}
}
}

for(\$i = 0; \$i < count(\$grid); \$i ++){
for(\$j = 0; \$j < count(\$grid[0]); \$j ++){
\$max = min(\$h[\$i], \$v[\$j]);
\$result += \$max - \$grid[\$i][\$j];
}
}
return \$result;
}
}```

## 859. Buddy Strings

Given two strings `A` and `B` of lowercase letters, return `true` if and only if we can swap two letters in `A` so that the result equals `B`.

Example 1:

Input:

```A = "ab", B = "ba"
```

Output:

```true
```

Example 2:

Input:

```A = "ab", B = "ab"
```

Output:

```false
```

Example 3:

Input:

```A = "aa", B = "aa"
```

Output:

```true
```

Example 4:

Input:

```A = "aaaaaaabc", B = "aaaaaaacb"
```

Output:

```true
```

Example 5:

Input:

```A = "", B = "aa"
```

Output:

```false
```

Note:

1. `0 <= A.length <= 20000`
2. `0 <= B.length <= 20000`
3. `A` and `B` consist only of lowercase letters.

Two cases:

1. Exactly two chars are different, and two strings are same with these chars swapped
2. Two strings are exactly same, the string has duplicate char, so we can pretend to have a swap.

```class Solution {

/**
* @param String \$A
* @param String \$B
* @return Boolean
*/
function buddyStrings(\$A, \$B) {
if(empty(\$A) || empty(\$B) || count(\$A) !== count(\$B)){
return false;
}

if(\$A == \$B){
if(strlen(count_chars(\$A, 3)) != strlen(\$A)){
// Check is there're duplicate chars
return true;
}
}

\$diff_a = [];
\$diff_b = [];

for(\$i = 0; \$i < strlen(\$A); \$i ++){
if(\$A[\$i] !== \$B[\$i]){
\$diff_a[] = \$A[\$i];
\$diff_b[] = \$B[\$i];
}
}

return  count(\$diff_b) == 2
&& \$diff_a[1] == \$diff_b[0]
&& \$diff_a[0] == \$diff_b[1];
}
}```

## 987. Vertical Order Traversal of a Binary Tree

Given a binary tree, return the vertical order traversal of its nodes values.

For each node at position (X, Y), its left and right children respectively will be at positions (X-1, Y-1) and (X+1, Y-1).

Running a vertical line from X = -infinity to X = +infinity, whenever the vertical line touches some nodes, we report the values of the nodes in order from top to bottom (decreasing Y coordinates).

If two nodes have the same position, then the value of the node that is reported first is the value that is smaller.

Return an list of non-empty reports in order of X coordinate. Every report will have a list of values of nodes.

### Example 1:

Input: [3,9,20,null,null,15,7]
Output: [[9],[3,15],[20],[7]]
Explanation:
Without loss of generality, we can assume the root node is at position (0, 0):
Then, the node with value 9 occurs at position (-1, -1);
The nodes with values 3 and 15 occur at positions (0, 0) and (0, -2);
The node with value 20 occurs at position (1, -1);
The node with value 7 occurs at position (2, -2).

### Example 2:

Input: [1,2,3,4,5,6,7]
Output: [[4],[2],[1,5,6],[3],[7]]
Explanation:
The node with value 5 and the node with value 6 have the same position according to the given scheme.
However, in the report “[1,5,6]”, the node value of 5 comes first since 5 is smaller than 6.

### Note:

The tree will have between 1 and 1000 nodes.
Each node’s value will be between 0 and 1000.

## My notes.

Mind the case.

[0,8,1,null,null,3,2,null,4,5,null,null,7,6]

``````          0
8         1
3         2
4,5
6         7
``````

Rendered by Leetcode, with my lines added

## My Solution

```/**
* Definition for a binary tree node.
* class TreeNode {
*     public \$val = null;
*     public \$left = null;
*     public \$right = null;
*     function __construct(\$value) { \$this-&gt;val = \$value; }
* }
*/
class Solution {

/**
* @param TreeNode \$root
* @return Integer[][]
*/
function verticalTraversal(\$root) {
\$result = [];
\$this-&gt;traversalHelper(\$root,0,0,\$result);
// Sort by col, from left to right
ksort(\$result);
foreach(\$result as &amp;\$rs){
\$new_sub = [];
// foreach col, sort by depth from top to bottom
ksort(\$rs);
foreach(\$rs as &amp;\$r){
// Same depth, sort by value asc, according to spec
asort(\$r);
\$new_sub = array_merge(\$new_sub,\$r);
}
\$rs = \$new_sub;
}
return \$result;
}

// Recursion, go through the tree
function traversalHelper(\$root, \$pos,\$depth, &amp;\$result) {
if(\$root !== null){
if(!isset(\$result[\$pos])){
\$result[\$pos] = [];
}
if(!isset(\$result[\$pos][\$depth])){
\$result[\$pos][\$depth] = [];
}
\$result[\$pos][\$depth][] = \$root-&gt;val;

\$this-&gt;traversalHelper(\$root-&gt;left, \$pos - 1,\$depth+1, \$result);
\$this-&gt;traversalHelper(\$root-&gt;right, \$pos + 1,\$depth+1, \$result);
}
}
}```