David Raleche

    
class Solution {

  public $gridNumberRows;
  public $gridNumberCols;
  public $grid;

  public function __construct() 
  {
  
  }

  public function maxConnectedColorValue(array $grid) : int
  {
     $this->grid = $grid;
    print $this->gridNumberRows = count($grid);
    $countCol = 0;
    foreach ($grid[0] as $col)
    {
      $countCol++;
    }
      
      print "\n\r";
   print  $this->gridNumberCols = $countCol;
    
      
      
    # visted cells to false
    for($r = 0; $r < $this->gridNumberRows ; $r++) {
      for($c = 0; $c < $this->gridNumberCols ; $c++)  {
        $visited[$r][$c] = false;
      }
    }

   
    $hashmapColor = [];
    for($row = 0; $row < $this->gridNumberRows ; $row++) {
      for($col = 0; $col < $this->gridNumberCols ; $col++)  {
        
        
        if($visited[$row][$col] === true) {
          continue;
        }

        
        if($grid[$row][$col] === ''){
          $visited[$row][$col] = true;
          continue;
        }

        $cellColor= $grid[$row][$col];
        $visited[$row][$col] = true;
        if(!array_key_exists($cellColor, $hashmapColor)) {
          $hashmapColor[$cellColor] = 1;
        }
        
        $this->dfs($row, $col, $visited, $hashmapColor);
        
        
        
      }
    }    
    var_dump($hashmapColor);
    return max($hashmapColor);
  }
  
    public function dfs($r, $c, &$visited, &$hashmapColor)
    {
      
      $cellColor = $this->grid[$r][$c];
        if($cellColor == '')
            return;
                
                
      # structure [R,C]
      $neighborInstruction = [[-1,0],[0,1],[1,0],[0,1]];
      
      foreach ($neighborInstruction as $neighbor) 
      {
        $neighborRow = $neighbor[0];
        $neighborCol = $neighbor[1];
        
    

          
          
          
        #if whithin boundaries
        if(((($neighborRow + $r) >= 0) && ($neighborRow + $r) < $this->gridNumberRows)
          && ($neighborCol + $r) >= 0 && ($neighborCol + $r) < $this->gridNumberCols)
        {
        
          # check same color
          if($this->grid[$neighborRow + $r][$neighborCol + $c] === $this->grid[$r][$c] 
            && $visited[$neighborRow + $r][$neighborCol + $c] === false) {
            
              if (array_key_exists($cellColor, $hashmapColor)) {
                $hashmapColor[$cellColor] += 1;
            } 

            $visited[$r][$c] = true;
            $this->dfs($neighborRow + $r, $neighborCol + $c, $visited, $hashmapColor);

          }
        }
        
      }
    }

  
}



$grid = [
['','r','b',''],
['','b','b','b'],
['','','b','']];

$solution = new Solution();
$res = $solution->maxConnectedColorValue($grid);

var_dump($res);

class Node {
  
  public $value;
  public $left;
  public $right;
  
  public function __construct(int $value) 
  {
    $this->value = $value;
   $this->left = null;
   $this->right = null;
  }
}



class Solution {

  public function bfs($node, &$hashmap, $level) 
  {
    
    if($node === null) {
      return;
    }

    
    $queue = new SplQueue();
  
    if($node->left)
    {
      $queue->enqueue($node->left);  
    }

    if($node->right)
    {
      $queue->enqueue($node->right);
    }


    
    while ($queue->count() > 0) {
    
      $childNode = $queue->pop();
      
      $hashmap[$level][] = $childNode->value;
      
        $this->bfs($childNode, $hashmap, $level + 1);
      
    }

    
  }
  
  public function AverageSumLevel($node) 
  {
    $hashmap = [];
    $level = 0;
    $hashmap[$level][] = $node->value;
    $this->bfs($node, $hashmap, $level + 1) ;
    
      
    foreach ($hashmap as $level => $values) 
    {
      $result[] = array_sum($values) / count($values);
    }
      
      
    return $result;
  }
}


$node5 = new Node(5);
$node5->left = new Node(4);
$node5->right = new Node(8);


$node5->left->left = new Node(2);
$node5->left->right = new Node(1);

$node5->right->right = new Node(9);
$node5->right->right->right = new Node(12);

$solution = new Solution();
$res = $solution->AverageSumLevel($node5);
var_dump($res);

Seating ArrangementsThere are n guests attending a dinner party, numbered from 1 to n. The ith guest has a height of arr[i-1] inches.The guests will sit down at a circular table which has n seats, numbered from 1 to n in clockwise order around the table. As the host, you will choose how to arrange the guests, one per seat. Note that there are n! possible permutations of seat assignments.Once the guests have sat down, the awkwardness between a pair of guests sitting in adjacent seats is defined as the absolute difference between their two heights. Note that, because the table is circular, seats 1 and n are considered to be adjacent to one another, and that there are therefore n pairs of adjacent guests.The overall awkwardness of the seating arrangement is then defined as the maximum awkwardness of any pair of adjacent guests. Determine the minimum possible overall awkwardness of any seating arrangement.

Signature

int minOverallAwkwardness(int[] arr)

Input

n is in the range [3, 1000]. Each height arr[i] is in the range [1, 1000].

Output

Return the minimum achievable overall awkwardness of any seating arrangement.

Example

n = 4 arr = [5, 10, 6, 8] output = 4If the guests sit down in the permutation [3, 1, 4, 2] in clockwise order around the table (having heights [6, 5, 8, 10], in that order), then the four awkwardnesses between pairs of adjacent guests will be |6-5| = 1, |5-8| = 3, |8-10| = 2, and |10-6| = 4, yielding an overall awkwardness of 4. It’s impossible to achieve a smaller overall awkwardness.Report Bug

 
function minOverallAwkwardness($arr) {
  // Write your code here

    #big O NLogN
    sort($arr);
    $n = count($arr);
    
    $temp = $arr[$n-1];
    $arr[$n-1] = $arr[$n-2] ;
    $arr[$n-2] = $temp;
    

    return smallestAwkwardnes($arr);
}

function smallestAwkwardnes($arr){
    $heap = new SplMaxHeap();
    for ($i=0; $i < count($arr) - 1; $i++) {
        $akwardness = $arr[$i] - $arr[$i + 1];
        if($akwardness < 0){
            $akwardness *= -1;
        }
        $heap->insert($akwardness);
    }
    return $heap->extract();
}

Element SwappingGiven a sequence of n integers arr, determine the lexicographically smallest sequence which may be obtained from it after performing at most k element swaps, each involving a pair of consecutive elements in the sequence.Note: A list x is lexicographically smaller than a different equal-length list y if and only if, for the earliest index at which the two lists differ, x’s element at that index is smaller than y’s element at that index.

Signature

int[] findMinArray(int[] arr, int k)

Input

n is in the range [1, 1000]. Each element of arr is in the range [1, 1,000,000]. k is in the range [1, 1000].

Output

Return an array of n integers output, the lexicographically smallest sequence achievable after at most k swaps.

Example 1

n = 3 k = 2 arr = [5, 3, 1] output = [1, 5, 3]We can swap the 2nd and 3rd elements, followed by the 1st and 2nd elements, to end up with the sequence [1, 5, 3]. This is the lexicographically smallest sequence achievable after at most 2 swaps.

Example 2

n = 5 k = 3 arr = [8, 9, 11, 2, 1] output = [2, 8, 9, 11, 1]We can swap [11, 2], followed by [9, 2], then [8, 2].

   function findMinArray(array $arr, int $k) : array
  {
    #arr = [5, 3, 1]
    # k:2, 1
    while($k >= 1) 
    {
    
      # swapping
      $temp = $arr[$k];        # temp 1      ; 1
      $arr[$k] = $arr[$k - 1]; # $arr[2] = 3; $arr[1] = 5
      $arr[$k - 1] = $temp;    # $arr[1] = 1; $arr[0] = 1

      #arr = [5, 1, 3]
      #arr = [1, 5, 3]
      $k--; #1 #0
    }
    
    return $arr;
    
  }

Suppose we have a list of N numbers, and repeat the following operation until we’re left with only a single number: Choose any two numbers and replace them with their sum. Moreover, we associate a penalty with each operation equal to the value of the new number, and call the penalty for the entire list as the sum of the penalties of each operation.For example, given the list [1, 2, 3, 4, 5], we could choose 2 and 3 for the first operation, which would transform the list into [1, 5, 4, 5] and incur a penalty of 5. The goal in this problem is to find the highest possible penalty for a given input.Signature: int getTotalTime(int[] arr)Input: An array arr containing N integers, denoting the numbers in the list.Output format: An int representing the highest possible total penalty.Constraints: 1 ≤ N ≤ 10^6 1 ≤ Ai ≤ 10^7, where *Ai denotes the ith initial element of an array. The sum of values of N over all test cases will not exceed 5 * 10^6.Example arr = [4, 2, 1, 3] output = 26First, add 4 + 3 for a penalty of 7. Now the array is [7, 2, 1] Add 7 + 2 for a penalty of 9. Now the array is [9, 1] Add 9 + 1 for a penalty of 10. The penalties sum to 26.

# int1 + int2
  #  [4, 2, 1, 3]
  # [7, 2, 1] penalty 7
  # [9, 1] penalty 7 + 9 = 16
  # 10 penalty 7 + 9 = 16 + 10 = 26
  
  #highest penalty
  
  
  #  [4, 2, 1, 3]
  # [5, 2, 3] penalty 5
  # [7,3] penalty 5 + 7 = 12
  # 10 penalty  5 + 7 = 12 + 10 = 22
  
  function getTotalTime(array $arr) : int 
  {
  $penalty = 0;
    $heap = new SplMaxHeap();
    
    # array(4, 2, 1, 3)
    foreach ($arr as $value) 
    {
      $heap->insert($value);
    }
    #heap  4,3,2,1
    
    
    while ($heap->count() > 1) # count 4; 3
    {
      $int1 = $heap->extract(); # 4 ; 7 ; 9
      $int2 = $heap->extract(); # 3 ; 2 ; 1
      $sum = ($int1 + $int2);   # 7 ; 9 ; 10
      $penalty +=  $sum;         # 7; 7+9=16;16+10 = 26 
      $heap->insert($sum);      # 7,2,1 ; 9,1; 10
    }
    
    return $penalty; #26
}

You’re given a list of n integers arr[0..(n-1)]. You must compute a list output[0..(n-1)] such that, for each index i (between 0 and n-1, inclusive), output[i] is equal to the median of the elements arr[0..i] (rounded down to the nearest integer).The median of a list of integers is defined as follows. If the integers were to be sorted, then:

• If there are an odd number of integers, then the median is equal to the middle integer in the sorted order.
• Otherwise, if there are an even number of integers, then the median is equal to the average of the two middle-most integers in the sorted order.

Signature

int[] findMedian(int[] arr)

Input

n is in the range [1, 1,000,000]. Each value arr[i] is in the range [1, 1,000,000].

Output

Return a list of n integers output[0..(n-1)], as described above.

Example 1

n = 4 arr = [5, 15, 1, 3] output = [5, 10, 5, 4]The median of [5] is 5, the median of [5, 15] is (5 + 15) / 2 = 10, the median of [5, 15, 1] is 5, and the median of [5, 15, 1, 3] is (3 + 5) / 2 = 4.

Example 2

n = 2 arr = [1, 2] output = [1, 1]The median of [1] is 1, the median of [1, 2] is (1 + 2) / 2 = 1.5 (which should be rounded down to 1).

 function findMedian(array $arr) : array
  {
      # initialization variables
      $endWindowPosition = 0;
      $output = [];
      $maxHeap = new SplMaxHeap();
  
      # iterate throug ARR
      foreach ($arr as $key => $value) # key 0,1,2,3 ; value 5,15,1,3
      {
          # detect is_odd_number
          $isEvenNumber = false;
          if ((($endWindowPosition + 1) % 2) === 0) 
          {
            $isEvenNumber = true;
              //var_dump($endWindowPosition);
          }    

          # $endWindowPosition : 0 -> $isEvenNumber: false
          # $endWindowPosition : 1 -> $isEvenNumber: true


          # building output
          if($key === 0) 
          {
            $output[] = $value; # output[5]
          }

          # EVEN
          if ($key > 0 && $isEvenNumber === true) 
          {
             # build temp arr1
              for ($i = 0; $i <= $endWindowPosition; $i++) 
              {
                $arrEvenNumber[] = $arr[$i];
              }
              # $arrEvenNumber[5,15]
              # $arrEvenNumber[5,15,1,3]
              sort($arrEvenNumber); # $arrOddNumber[1,3 5,15]
              
              $mediumIndex = floor(($endWindowPosition + 1) / 2); #1, 1
              $int1 = $arrEvenNumber[$mediumIndex - 1 ]; # 5, 3
              $int2 = $arrEvenNumber[$mediumIndex ]; # 15 ,5        
              
              $median = floor(($int1 + $int2) / 2); # 5+15 / 2 === 10 # 3 + 5 / 2 = 4
              
              $output[] = $median; #output[5, 10]; #output[5, 10, 5,4]
              $arrEvenNumber =  [];
          }

          # ODD
         if(($key > 0) && ($isEvenNumber === false)) {
             //var_dump($endWindowPosition);
           # build temp arr1
            for ($i = 0; $i <= $endWindowPosition; $i++) 
            {
              $arrOddNumber[] = $arr[$i]; # $arrOddNumber[5,15,1]
            }
            sort($arrOddNumber); # $arrOddNumber[1, 5,15]
            $mediumIndex = floor($endWindowPosition  / 2);
            // var_dump($endWindowPosition);
            $intOddMedian = $arrOddNumber[$mediumIndex]; # 5
            $output[] = $intOddMedian; # output[5, 10, 5]
             $arrOddNumber = [];
         }
       $endWindowPosition++; # 1, 2, 3
      }    
     
    
      return $output; 

  }

Largest Triple ProductsYou’re given a list of n integers arr[0..(n-1)]. You must compute a list output[0..(n-1)] such that, for each index i (between 0 and n-1, inclusive), output[i] is equal to the product of the three largest elements out of arr[0..i] (or equal to -1 if i < 2, as arr[0..i] then includes fewer than three elements).Note that the three largest elements used to form any product may have the same values as one another, but they must be at different indices in arr.

Signature

int[] findMaxProduct(int[] arr)

Input

n is in the range [1, 100,000]. Each value arr[i] is in the range [1, 1,000].

Output

Return a list of n integers output[0..(n-1)], as described above.

Example 1

n = 5 arr = [1, 2, 3, 4, 5] output = [-1, -1, 6, 24, 60]The 3rd element of output is 3*2*1 = 6, the 4th is 4*3*2 = 24, and the 5th is 5*4*3 = 60.

Example 2

n = 5 arr = [2, 1, 2, 1, 2] output = [-1, -1, 4, 4, 8]The 3rd element of output is 2*2*1 = 4, the 4th is 2*2*1 = 4, and the 5th is 2*2*2 = 8.

function findMaxProduct($arr) {
  // Write your code here
     $heap = new SplMaxHeap();
        $result = [-1, -1];
        $n = count($arr);
        

        for($i = 0; $i <= $n - 3; $i++)
        {
            $product = slidingWindowProduct($i, $arr, $heap);
            $result[] = $product;
        }
        
        
        return $result;
    }
    
     function slidingWindowProduct($strartIndex, $arr, &$heap)  
    {
        $product = 1;
        if($strartIndex == 0) {
            for($i = $strartIndex; $i <= $strartIndex + 2; $i++) 
            {
                $heap->insert($arr[$i]);
            }            
        } 
        else 
        {
            $heap->insert($arr[$strartIndex + 2]);
        }

        
        for($i=0; $i < 3; $i++) 
        {
            $value = $heap->extract();
            $temp[] = $value;
            $product *= $value;
        }


        while(count($temp) > 0) 
        {
            $val = array_shift($temp);
            $heap->insert($val);
        }
        
        return $product; #6
    }

class solution {
    
    public function __construct(){}
    
    public function findMaxProduct(array $arr) : array
    {
        $heap = new SplMaxHeap();
        $result = [-1, -1];
        $n = count($arr);
        

        for($i = 0; $i <= $n - 3; $i++)
        {
            $product = $this->slidingWindowProduct($i, $arr, $heap);
            $result[] = $product;
        }
        
        
        return $result;
    }
    
    private function slidingWindowProduct($strartIndex, $arr, &$heap) : mixed 
    {
        $product = 1;
        if($strartIndex == 0) {
            for($i = $strartIndex; $i <= $strartIndex + 2; $i++) 
            {
                $heap->insert($arr[$i]);
            }            
        } 
        else 
        {
            $heap->insert($arr[$strartIndex + 2]);
        }

        
        for($i=0; $i < 3; $i++) 
        {
            $value = $heap->extract();
            $temp[] = $value;
            $product *= $value;
        }


        while(count($temp) > 0) 
        {
            $val = array_shift($temp);
            $heap->insert($val);
        }
        
        return $product; #6
    }
}


$arr_1 = array(1, 2, 3, 4, 5);
$arr_2 = array(2, 4, 7, 1, 5, 3);


$solution = new Solution();
$res =  $solution->findMaxProduct($arr_2);
var_dump($res);

You are given two strings s and t. You can select any substring of string s and rearrange the characters of the selected substring. Determine the minimum length of the substring of s such that string t is a substring of the selected substring.Signature int minLengthSubstring(String s, String t)Inputs and t are non-empty strings that contain less than 1,000,000 characters eachOutput Return the minimum length of the substring of s. If it is not possible, return -1Examples = “dcbefebce” t = “fd” output = 5Explanation: Substring “dcbef” can be rearranged to “cfdeb”, “cefdb”, and so on. String t is a substring of “cfdeb”. Thus, the minimum length required is 5.

Facebook Solution

function minLengthSubstring($s, $t) {
  // Write your code here
       $s1 = str_split($s);
        $t1 = str_split($t);
        $hashmap = [];
        $countLengthSubstring = 1;
        $startingcounting = false;
        
        foreach ($t1 as $character) {
            $hashmap[$character] += 1;
        }
        
        foreach ($s1 as $character1) {
            if(array_key_exists($character1, $hashmap )) {
                if($startingcounting == false) {
                    $startingcounting = true;                
                }
                $hashmap[$character1] -= 1;
                if($hashmap[$character1] <= 0){
                    unset($hashmap[$character1]);       
                }
            }
            
            if(count($hashmap) == 0) {
                return $countLengthSubstring;
            }
            
            if($startingcounting) {
                $countLengthSubstring++;
            }
        }
        return -1;
  
}  

Clean

class solution {
    
    public function __construct() {}
    public function minLengthSubstring(string $s11, string $t11) 
    {
        $s1 = str_split($s11);
        $t1 = str_split($t11);
        $hashmap = [];
        $countLengthSubstring = 1;
        $startingcounting = false;
        
        foreach ($t1 as $character) {
            $hashmap[$character] += 1;
        }
        
        foreach ($s1 as $character1) {
            if(array_key_exists($character1, $hashmap )) {
                if($startingcounting == false) {
                    $startingcounting = true;                
                }
                $hashmap[$character1] -= 1;
                if($hashmap[$character1] <= 0){
                    unset($hashmap[$character1]);       
                }
            }
            
            if(count($hashmap) == 0) {
                return $countLengthSubstring;
            }
            
            if($startingcounting) {
                $countLengthSubstring++;
            }
        }
        return -1;
    }
}

// $s1 = "dcbefebce";
// $t1 = "fd";

$s1 = "bfbeadbcbcbfeaaeefcddcccbbbfaaafdbebedddf";
$t1 = "cbccfafebccdccebdd";

$solution = new Solution();
print $solution->minLengthSubstring($s1, $t1);

You are given a tree that contains N nodes, each containing an integer u which corresponds to a lowercase character c in the string s using 1-based indexing.You are required to answer Q queries of type [u, c], where u is an integer and c is a lowercase letter. The query result is the number of nodes in the subtree of node u containing c.Signature int[] countOfNodes(Node root, ArrayList<Query> queries, String s)Input A pointer to the root node, an array list containing Q queries of type [u, c], and a string sConstraints N and Q are the integers between 1 and 1,000,000 u is a unique integer between 1 and N s is of the length of N, containing only lowercase letters c is a lowercase letter contained in string s Node 1 is the root of the treeOutput An integer array containing the response to each queryExample

        1(a)
        /   \
      2(b)  3(a)

s = “aba” RootNode = 1 query = [[1, ‘a’]]Note: Node 1 corresponds to first letter ‘a’, Node 2 corresponds to second letter of the string ‘b’, Node 3 corresponds to third letter of the string ‘a’.output = [2]Both Node 1 and Node 3 contain ‘a’, so the number of nodes within the subtree of Node 1 containing ‘a’ is 2.

class Node{
    public $val;
    public $children ;
    function __construct($val){
        $this->val = $val;  
        $this->children = array();
    }
}

// Add any helper functions you may need here


function countOfNodes($root, $queries, $s) {
      #initialization variables
        $countOfNodes = array();
        foreach  ($queries as $query) {
            $nodeValueToBeSearched = $query[0]; #1, 2, 3
            $countOfNodes[$nodeValueToBeSearched] = 1;
        }
        $parents = array();

        
        #mapping string
        $index = 1;   
        $hashmapString = array();
        $array_string = str_split($s);
        foreach($array_string as $character) {
            $hashmapString[$index] = $character;
                $index += 1;
       } 
        
        
        #depth first search
        dfs($root, $queries, $countOfNodes, $hashmapString, $parents);
        
        
        #create result
        foreach ($countOfNodes as $value) {
            $result[] = $value;
        }
        
        return $result;
    }
    
     function dfs($node, $queries, &$countOfNodes , $hashmapString, $parents) 
    {
 
        
        #populate countOfNodes
        foreach  ($queries as $query) {
            $nodeValueToBeSearched = $query[0]; #1, 2, 3
            $letterToBeSearched = $query[1];    #a, b, a
            


            if ($hashmapString[$node->val] === $letterToBeSearched) {
                if (in_array($nodeValueToBeSearched, $parents)){
                    $countOfNodes[$nodeValueToBeSearched] += 1; 


                    #$countOfNodes[1] = 1+1+1
                    #$countOfNodes[2] = 1
                }
            }            
        }
        
        #exit condition
        if(empty($node->children)){
            return;
        }
        
        
        #populating visit array
        foreach($node->children as $childNode) {
            $visit[] = $childNode;
        }
        $parents[] = $node->val;
        
        while (count($visit) > 0) {
            $nodeChild = array_shift($visit);
            
            dfs($nodeChild, 
                       $queries, 
                       $countOfNodes, 
                       $hashmapString, 
                        $parents
                      );
        }
        
        
    }

There is a binary tree with N nodes. You are viewing the tree from its left side and can see only the leftmost nodes at each level. Return the number of visible nodes.Note: You can see only the leftmost nodes, but that doesn’t mean they have to be left nodes. The leftmost node at a level could be a right node.Signature int visibleNodes(Node root) {Input The root node of a tree, where the number of nodes is between 1 and 1000, and the value of each node is between 0 and 1,000,000,000Output An int representing the number of visible nodes.Example

            8  <------ root
           / \
         3    10
        / \     \
       1   6     14
          / \    /
         4   7  13            

output = 4

<?php

// Add any extra import statements you may need here


class TreeNode{
  public $val;
  public $left;
  public $right;
  public function __construct($val=0) {
    $this->val = $val;
    $this->left = NULL;
    $this->right = NULL;
  }
}

// Add any helper functions you may need here


function visibleNodes($root) {
        $result = [];
        
        $numberVisibleNode = 1;
 
        dfs($root, $result, $numberVisibleNode);
       
        return  max($result);;
    }
    
     function dfs(TreeNode $currentNode, array &$result,$numberVisibleNode )  {
        if(($currentNode->left === null)  && ($currentNode->right === null)) {
                   
           // var_dump($currentNode->val.' '.$numberVisibleNode);
            $result[$currentNode->val] = $numberVisibleNode;
            return ;
        }
        
        if($currentNode->left != null) {
          //  var_dump('leftside '.$currentNode->val.' '.$numberVisibleNode);
            $numberVisibleNodeLeft = $numberVisibleNode;
            $numberVisibleNodeLeft++; #2
            dfs($currentNode->left, $result, $numberVisibleNodeLeft);
        }
        
        if(($currentNode->right !== null)) {
           // var_dump('righttside '.$currentNode->val.' '.$numberVisibleNode);
            $numberVisibleNodeRight = $numberVisibleNode;
            $numberVisibleNodeRight++;#5,
            dfs($currentNode->right, $result, $numberVisibleNodeRight);
        }
    }