S = "0110", N = 3

true

S = "0110", N = 4

false

Thoughts:

class Solution {

    /**
     * @param String $S
     * @param Integer $N
     * @return Boolean
     */
    function queryString($S, $N) {
        while($N > 1){
            $binary = decbin($N);
            if(strpos($S, $binary) === false){
                return false;
            }
            $N --;
        }
        return true;
    }
}

[8,1,5,2,6]

11
Explanation: i = 0, j = 2, A[i] + A[j] + i - j = 8 + 5 + 0 - 2 = 11

Thoughts:

Scan from left to right, record:

• best score so far()
• the best left element which could generate the largest score with the next element.
  • As we move on to the next, gap ++, impact/score — with the prev. best left element

class Solution {

    /**
     * @param Integer[] $A
     * @return Integer
     */
    function maxScoreSightseeingPair($A) {
        $bestResult = 0;    // Best Score (A[i] + A[j] + i - j)
        $bestAI = 0;        // The best choice of A[i] so far
        foreach($A as $a){
            // Moved to next, gap++, then the "sightseeing value" --
            $bestAI --; 
            // Best result = 
            $bestResult = max($bestResult, $bestAI + $a );
            $bestAI = max($a, $bestAI);
        }
        return $bestResult;
    }
}

well sure I can combine $bestAI –; into other expressions. Just to make it clear in this way.

Given a positive integer K, you need find the smallest positive integer N such that N is divisible by K, and N only contains the digit 1.

Return the length of N.  If there is no such N, return -1.

Example 1:

Input:

 1

Output:

 1

Explanation:

 The smallest answer is N = 1, which has length 1.

Example 2:

Input:

 2

Output:

 -1

Explanation:

 There is no such positive integer N divisible by 2.

Example 3:

Input:

 3

Output:

 3

Explanation:

 The smallest answer is N = 111, which has length 3.

Note:

• 1 <= K <= 10^5

Thoughts:

Two tricky parts.

1. $n = ($n * 10 +1) % $K;Suppose r = (aK + b) when r > K, next r supposed to be 10r+1 = 10aK + 10b + 1. Since 10aK is divisible by K, we just need to consider 10b + 1. b is r % K, so we can update r = r % K. [By https://leetcode.com/wangqiuc/]
2. The exit condition for loop, $n <= $K, actually, there always exist an answer, we can use while true instead.

Proof  [By https://leetcode.com/starsthu2016/]

It is obvious that if K is the multiple of 2 or multiple of 5, N is definitely not multiple of K because the ones digit of N is 1. Thus, return -1 for this case.

If K is neither multiple of 2 nor multiple of 5, we have this theorem.

Theorem: there must be one number from the K-long candidates list [1, 11, 111, …, 111..111(with K ones)], which is the multiple of K.

Why? Let’s think in the opposite way. Is it possible that none of the K candidates is the multiple of K?

If true, then the module of these K candidate numbers (mod K) must be in [1, .., K-1] (K-1 possible module values). Thus, there must be 2 candidate numbers with the same module. Let’s denote these two numbers as N_i with i ones, and N_j with j ones and i<j.

Thus N_j-N_i = 1111…1000…0 with (j-i) ones and i zeros. N_j-N_i = 111..11 (j-i ones) * 100000..000 (i zeros). We know that N_i and N_j have the same module of K. Then N_j-N_i is the multiple of K. However, 100000..000 (i zeros) does not contain any factors of K (K is neither multiple of 2 nor multiple of 5). Thus, 111..11 (j-i ones) is the multiple of K. This is contradictory to our assumption that all K candidates including 111..11 (j-i ones) are not multiple of K.

Finally, we know that there is at least one number, from the first K candidates, which is the multiple of K.

class Solution {

    /**
     * @param Integer $K
     * @return Integer
     */
    function smallestRepunitDivByK($K) {
        if($K % 2 == 0 || $K % 5 == 0){
            return -1;
        }
        $n = 1;
        $r = 1;
        while($n <= $K ){
            if($n % $K == 0){
                return  $r;
            }
            $n = ($n * 10 +1) % $K;
            $r ++;
        }
        return -1;
    }
}

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);
    }
}

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

2

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.

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

-1

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

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);
    }
}

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

 35

 
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] ]

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;
    }
}

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->val = $value; }
 * }
 */
class Solution {

    /**
     * @param TreeNode $root
     * @return Integer[][]
     */
    function verticalTraversal($root) {
        $result = [];
        $this->traversalHelper($root,0,0,$result);
        // Sort by col, from left to right
        ksort($result);
        foreach($result as &$rs){
            $new_sub = [];
            // foreach col, sort by depth from top to bottom
            ksort($rs);
            foreach($rs as &$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, &$result) {
        if($root !== null){
            if(!isset($result[$pos])){
                $result[$pos] = [];
            } 
            if(!isset($result[$pos][$depth])){
                $result[$pos][$depth] = [];
            }
            $result[$pos][$depth][] = $root->val;
            
            $this->traversalHelper($root->left, $pos - 1,$depth+1, $result);
            $this->traversalHelper($root->right, $pos + 1,$depth+1, $result);
        }
    }
}